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1. Groups

1.1 Laws of Composition

Definition 1.1. A law of composition (or binary operation) on a set \(S\) is a rule that assigns to each ordered pair \((a,b)\) of elements of \(S\) a unique element of \(S\), denoted by \(a \cdot b\) or simply \(ab\).

Examples.

1.2 Groups and Subgroups

Groups

Definition 1.2. A group is a set \(G\) with a law of composition that is associative, has an identity element \(e\), and in which every element has an inverse.

A group is abelian (or commutative) if its law of composition is commutative.

Important examples.

Subgroups

Definition 1.3. A subgroup \(H\) of a group \(G\) is a subset of \(G\) that contains the identity, is closed under the group operation, and is closed under inverses.

Theorem 1.1 (Subgroup Criterion). A nonempty subset \(H \subseteq G\) is a subgroup if and only if \(a b^{-1} \in H \qquad \text{for all } a,b \in H.\)

Proof idea. If the condition holds and \(a \in H\), then \(aa^{-1}=e \in H\). Then, for any \(a \in H\), we have \(ea^{-1}=a^{-1} \in H\).

Examples and facts.

Subgroups of \(\mathbb{Z}\)

The integers \(\mathbb{Z}\) form an abelian group under addition, with identity \(0\).

Theorem 1.2. Every subgroup of \(\mathbb{Z}\) is of the form \(d\mathbb{Z}\) for some nonnegative integer \(d\). If the subgroup is nonzero, then \(d\) is its smallest positive element; if the subgroup is \(\{0\}\), then \(d=0\).

2. Cyclic Groups and Homomorphisms

2.1 Cyclic Groups

For \(g \in G\), define \(g^0 := e_G\). The subgroup generated by \(g\) is \(\langle g \rangle := \{g^k : k \in \mathbb{Z}\}.\)

Definition. A group \(G\) is cyclic if there exists \(g \in G\) such that \(G = \langle g \rangle.\)

The order of an element \(g\) is the smallest positive integer \(n\) such that \(g^n=e_G\), if such an \(n\) exists. Otherwise, \(g\) has infinite order. If \(G=\langle g \rangle\), then \(|G|\) is equal to the order of \(g\).

2.2 Homomorphisms

Definition 2.2. A homomorphism \(\phi:G \to G'\) between groups is a map that preserves the operation: \(\phi(a \cdot b) = \phi(a)\cdot \phi(b).\)

Proposition 2.3. Homomorphisms preserve identities and inverses: \(\phi(e_G)=e_{G'}, \qquad \phi(g^{-1})=\phi(g)^{-1}.\)

Definitions.

Examples.

  1. The determinant map \(\det:\operatorname{GL}_n(\mathbb{R}) \to \mathbb{R}^{\times}\)

    is a homomorphism because \(\det(AB)=\det(A)\det(B)\). Its kernel is \(\operatorname{SL}_n(\mathbb{R})\).

  2. The sign map \(\operatorname{sgn}:S_n \to \{1,-1\}\)

    has kernel \(A_n\), the alternating group of even permutations.

  3. The exponential map \(\exp:(\mathbb{R},+) \to \mathbb{R}_{>0}^{\times}, \qquad x \mapsto e^x\)

    is a homomorphism.

  4. The absolute value map \(|\cdot|:\mathbb{C}^{\times} \to \mathbb{R}_{>0}^{\times}\)

    is a homomorphism because \(|ab|=|a||b|\).

  5. For a fixed element \(a \in G\), the map \(\phi:(\mathbb{Z},+) \to G, \qquad n \mapsto a^n\)

    is a homomorphism. Its image is \(\langle a \rangle\), and its kernel is \(n\mathbb{Z}\), where \(n\) is the order of \(a\).

Important properties.

2.3 Isomorphisms

Definition 2.3. An isomorphism is a bijective homomorphism. Groups \(G\) and \(G'\) are isomorphic, written \(G \cong G'\), if there exists an isomorphism \(G \to G'\).

Lemma 2.5. The inverse of an isomorphism is also an isomorphism.

Facts and examples.

3. Equivalence Relations and Cosets

3.1 Equivalence Relations and Partitions

Definition 3.1. An equivalence relation \(\sim\) on a set \(S\) satisfies:

  1. Reflexivity: \(a \sim a\).
  2. Symmetry: \(a \sim b \Rightarrow b \sim a\).
  3. Transitivity: \(a \sim b\) and \(b \sim c \Rightarrow a \sim c\).

The equivalence class of \(a\) is \([a] = \{b \in S : b \sim a\}.\)

Any element of \([a]\) is called a representative of the class.

Equivalence classes partition \(S\): they are pairwise disjoint and together cover \(S\). Conversely, every partition of \(S\) defines an equivalence relation.

Examples.

3.2 Cosets

Let \(H \le G\) be a subgroup.

Cosets partition the group \(G\). Two left cosets are equal precisely when \(aH=bH \iff a^{-1}b \in H.\)

All cosets of \(H\) have the same size, namely \(|H|\). The index of \(H\) in \(G\), denoted \([G:H]\), is the number of distinct left cosets of \(H\) in \(G\).

Theorem 3.1 (Lagrange’s Theorem). If \(G\) is finite and \(H \le G\), then \(|G| = |H|[G:H].\)

Corollary 3.2. The order of an element divides the order of the group.

Proposition 3.3. For a subgroup \(H \le G\), the following are equivalent:

  1. \(H \triangleleft G\).
  2. \(gHg^{-1}=H\) for all \(g \in G\).
  3. \(gH=Hg\) for all \(g \in G\).
  4. The left cosets of \(H\) are the same as the right cosets of \(H\).

Here \(H \triangleleft G\) means that \(H\) is a normal subgroup of \(G\). Equivalently, \(ghg^{-1} \in H \qquad \text{for all } h \in H,\ g \in G.\)

3.3 Modular Arithmetic

For \(n \in \mathbb{Z}_{>0}\), define congruence modulo \(n\) by \(a \equiv b \pmod n \iff n \mid (a-b).\)

This is an equivalence relation. The subgroup \(n\mathbb{Z}\) of \(\mathbb{Z}\) gives the same equivalence relation: two integers are equivalent modulo \(n\) exactly when they lie in the same coset of \(n\mathbb{Z}\).

The number of units in \(\mathbb{Z}/n\mathbb{Z}\) is given by Euler’s \(\varphi\)-function.

4. Correspondence, Products, and Quotients

4.1 The Correspondence Theorem

Let \(\phi:G \to G'\) be a homomorphism. Then the subgroups of \(G\) containing \(\ker \phi\) correspond to the subgroups of \(\operatorname{Im}\phi\).

Theorem 4.1 (Correspondence Theorem). If \(N=\ker \phi\), then the map \(H \mapsto \phi(H)\) is a bijection between:

Moreover, this correspondence preserves normality.

Proof.

Examples.

4.2 Product Groups

Given groups \(G\) and \(G'\), define \(G \times G' := \{(g,g') : g \in G,\ g' \in G'\}.\)

This becomes a group under the componentwise operation \((g,g')\cdot(h,h') = (gh,g'h').\)

The identity is \((e,e')\), and inverses are given by \((g,g')^{-1} = (g^{-1},{g'}^{-1}).\)

There are natural homomorphisms:

Example 4.3. \(\mathbb{R}^2=\mathbb{R}\times\mathbb{R}\) under addition.

If \(G=G'\), then \(\{(g,g):g \in G\}\) is the diagonal subgroup.

The images \(G \times \{e'\}\) and \(\{e\} \times G'\) are normal subgroups of \(G \times G'\). They satisfy \((G \times \{e'\}) \cap (\{e\} \times G') = \{(e,e')\},\) and \((G \times \{e'\})(\{e\} \times G') = G \times G'.\)

Proposition 4.3 (Internal Direct Product). Let \(H\) and \(K\) be normal subgroups of \(G\) such that \(H \cap K=\{e\}, \qquad HK=G.\)

Then \(G \cong H \times K.\)

Proof. Define \(f:H \times K \to G, \qquad f(h,k)=hk.\)

Then \(f\) is a homomorphism. It is injective because if \(hk=e\), then \(k=h^{-1} \in H \cap K\), so \(h=k=e\). It is surjective because \(HK=G\).

4.3 Quotient Groups

Theorem 4.4. If \(N \triangleleft G\), then the set of cosets \(G/N\) forms a group under \((aN)(bN)=abN.\)

Proof sketch. We need to check that the operation is well defined. If \(a'=an_1, \qquad b'=bn_2 \qquad (n_1,n_2 \in N),\)

then \(a'b' = an_1bn_2 = ab(b^{-1}n_1b)n_2.\)

Since \(N\) is normal, \(b^{-1}n_1b \in N\). Thus \((b^{-1}n_1b)n_2 \in N\), so \(a'b'N=abN\).

Associativity comes from \(G\), the identity is \(eN\), and inverses are given by \((aN)^{-1}=a^{-1}N.\)

Example 4.4. The quotient group \(\mathbb{Z}/n\mathbb{Z}\). In an abelian group, every subgroup is normal.

Theorem 4.5 (First Isomorphism Theorem). For a homomorphism \(\phi:G \to G'\), \(G/\ker \phi \cong \operatorname{Im}\phi.\)

The kernel is normal because, for \(n \in \ker\phi\), \(\phi(gng^{-1})=\phi(g)\phi(n)\phi(g)^{-1}=e.\)

Proof. Let \(N=\ker \phi\). Define \(G/N \to \operatorname{Im}\phi, \qquad gN \mapsto \phi(g).\)

For any quotient group \(G/N\), there is a canonical projection \(\pi:G \twoheadrightarrow G/N, \qquad g \mapsto gN,\) which is a homomorphism with kernel \(N\).

Example. The alternating group \(A_n\) is a normal subgroup of \(S_n\), and \(S_n/A_n \cong \mathbb{Z}/2\mathbb{Z}.\)

5. Rings and Polynomial Rings

5.1 Rings, Subrings, and Units

Definition 5.1. A ring \(R\) is a set with two laws of composition, addition \(+\) and multiplication \(\times\), such that:

The ring \(R\) is commutative if multiplication is commutative.

If every nonzero \(a \in R\) has a multiplicative inverse, then \(R\) is a field.

Example 5.1. The integers \(\mathbb{Z}\), rationals \(\mathbb{Q}\), reals \(\mathbb{R}\), and complex numbers \(\mathbb{C}\) are rings under the usual operations. The set \(\mathbb{Z}[i]=\{a+bi : a,b \in \mathbb{Z}\}\)

of Gaussian integers is a subring of \(\mathbb{C}\).

Definition 5.2. A subring \(S \subseteq R\) is a subset containing \(1\) that is closed under \(+\), \(-\), and \(\times\).

A unit in \(R\) is an element \(u\) with a multiplicative inverse \(v\), so that \(uv=1=vu.\)

The set of units is denoted \(R^\times\).

The characteristic of \(R\) is the smallest positive integer \(n\) such that \(n \cdot 1 = \underbrace{1+\cdots+1}_{n \text{ times}} = 0,\)

or \(0\) if no such \(n\) exists. Then \(n \cdot a = (n \cdot 1)a = 0.\)

Example 5.2. \(\operatorname{char}\mathbb{Z}=0, \qquad \operatorname{char}(\mathbb{Z}/n\mathbb{Z})=n.\)

5.2 Polynomial Rings

Definition 5.3. Let \(R\) be a ring. A polynomial in one variable \(x\) with coefficients in \(R\) is a formal expression \(f(x)=a_nx^n+\cdots+a_1x+a_0, \qquad a_i \in R.\)

The set \(R[x]\) of all such polynomials forms a ring under termwise addition and the usual product.

The degree \(\deg f\) is the largest \(n\) with \(a_n \ne 0\), or \(-\infty\) if \(f=0\).

Proposition 5.2. If \(R\) has no zero divisors, then \(R[x]\) has no zero divisors.

Proof. If \(f,g \ne 0\) have leading coefficients \(a_m,b_n\), then the leading term of \(fg\) is \(a_mb_nx^{m+n} \ne 0.\)

Theorem 5.3 (Division Algorithm). Let \(f \in R[x]\) be monic, and let \(g \in R[x]\). There exist unique \(q,r \in R[x]\) such that \(g=fq+r, \qquad \deg r < \deg f,\) or \(r=0\).

The proof is by induction on \(\deg g\).

Corollary 5.4. The Division Algorithm works if the leading coefficient of \(f\) is a unit. In particular, for \(a \in R\), \(f(a)\) is the remainder when \(f(x)\) is divided by \(x-a\).

6. Ideals and Quotient Rings

6.1 Homomorphisms and Ideals

Definition 6.1. A ring homomorphism \(\phi:R \to R'\) preserves addition, multiplication, and the multiplicative identity: \(\phi(a+b)=\phi(a)+\phi(b), \qquad \phi(ab)=\phi(a)\phi(b), \qquad \phi(1_R)=1_{R'}.\)

Its kernel is \(\ker \phi = \{r \in R : \phi(r)=0\}.\)

If \(\phi\) is bijective, it is an isomorphism, and we write \(R \cong R'\).

Definition 6.2. An ideal \(I \subseteq R\) is an additive subgroup such that \(rI \subseteq I, \qquad Ir \subseteq I \qquad \text{for all } r \in R.\)

Kernels are ideals. Conversely, every ideal is a kernel via a quotient map.

Example 6.1. There is a unique homomorphism \(\mathbb{Z} \to R, \qquad 1 \mapsto 1_R.\)

It defines the characteristic of \(R\). If \(\operatorname{char}R=n>0\), the image is \(\mathbb{Z}/n\mathbb{Z}\) and the kernel is \(n\mathbb{Z}\). If \(\operatorname{char}R=0\), the image is \(\mathbb{Z}\) and the kernel is \(0\).

Theorem 6.1 (Substitution Principle). Let \(\phi:R \to R'\) be a homomorphism of commutative rings, and let \(\alpha \in R'\). There is a unique homomorphism \(\Phi:R[x] \to R'\)

extending \(\phi\) on constants and sending \(x \mapsto \alpha\).

Proof. Define \(\Phi\!\left(\sum a_i x^i\right)=\sum \phi(a_i)\alpha^i.\)

Then \(\Phi\) preserves addition and multiplication by direct expansion. Uniqueness is clear from the requirement that \(\Phi\) extend \(\phi\) and send \(x\) to \(\alpha\).

The substitution principle also extends to multivariable polynomial rings.

Example 6.2. \(R[x_1,x_2] \cong R[x_1][x_2].\)

Use the substitution principle with the identity on \(R\), sending \(x_1 \mapsto x_1\) and \(x_2 \mapsto x_2\).

The ideal generated by \(a_1,\ldots,a_k\) is \((a_1,\ldots,a_k)=\sum Ra_iR.\)

Definition 6.3. The principal ideal generated by \(a\) is \((a)=\left\{\sum r_i a r_i' : r_i,r_i' \in R\right\}.\)

Proposition 6.2. A field has only the trivial ideals \(\{0\}\) and \(F\).

Proof. If an ideal contains a nonzero element \(r \in F\), then \(r\) is a unit. For every \(s \in F\), \((sr^{-1})r=s,\)

so the ideal is all of \(F\).

Corollary 6.3. Field homomorphisms are injective.

Proposition 6.4. Ideals in \(\mathbb{Z}\) are principal: \((n)=n\mathbb{Z}.\)

Proposition 6.5. Ideals in \(F[x]\) are principal.

Example 6.3. Let \(\Phi:\mathbb{Q}[x] \to \mathbb{C}, \qquad x \mapsto \sqrt[3]{2}.\)

The kernel is principal, generated by the monic polynomial of lowest degree in \(\mathbb{Q}[x]\) that has \(\sqrt[3]{2}\) as a root: \(x^3-2.\)

6.2 Quotient Rings and the Correspondence Theorem

Theorem 6.6. Let \(I\) be an ideal of \(R\). The set of cosets \(\overline{R}=R/I\) is a ring under \((a+I)+(b+I)=(a+b)+I, \qquad (a+I)(b+I)=ab+I.\)

The canonical map \(\pi:R \to R/I, \qquad a \mapsto a+I,\) is a surjective homomorphism with kernel \(I\).

Theorem 6.8 (First Isomorphism Theorem). If \(f:R \to R'\) is a ring homomorphism, then \(R/\ker f \cong \operatorname{Im}f.\)

Theorem 6.9 (Correspondence Theorem). Let \(\phi:R \to \mathcal{R}\) be a surjective homomorphism with kernel \(K\). There is a bijection between ideals of \(\mathcal{R}\) and ideals of \(R\) containing \(K\).

Moreover, if \(I\) corresponds to \(\mathcal{I}\), then \(R/I \cong \mathcal{R}/\mathcal{I}.\)

Example 6.4 (Artin 11.4.5). \(\mathbb{Z}[i]/(2+i) \cong \mathbb{F}_5.\)

7. Ring Extensions and Product Rings

7.1 Adjoining Elements

A ring extension is a ring containing another ring as a subring. Adjoining an element \(\alpha\) to a ring \(R\) means constructing an extension \(R[\alpha]\) containing \(R\) and \(\alpha\), often with \(\alpha\) satisfying a polynomial relation \(f(\alpha)=0\).

Proposition (Artin 11.5.5). Let \(f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0\)

be monic in \(R[x]\). Let \(R[\alpha] = R[x]/(f)\)

be the ring obtained by adjoining an element satisfying \(f(\alpha)=0\).

Then:

  1. The set \(\{1,\alpha,\ldots,\alpha^{n-1}\}\) is a basis of \(R[\alpha]\) over \(R\): every element of \(R[\alpha]\) can be written uniquely as a linear combination of this basis, with coefficients in \(R\).

  2. Addition of two linear combinations is vector addition.

  3. Multiplication is computed by reducing modulo \(f\). If \(\beta_1=g_1(\alpha)\) and \(\beta_2=g_2(\alpha)\), divide \(g_1g_2\) by \(f\): \(g_1g_2=fq+r, \qquad \deg r<n.\)

    Then \(\beta_1\beta_2=r(\alpha).\)

We know that \(R\) is contained in the polynomial ring \(R[x]\) as the subring of constant polynomials. We also have the canonical map \(\pi:R[x] \to R'=R[x]/(f).\)

Restricting to constant polynomials gives a homomorphism \(\psi:R \to R'.\)

Its kernel is the set of constant polynomials in the ideal: \(\ker \psi = R \cap (f).\)

The kernel is zero when \(f\) is monic or when \(R\) is a domain.

7.2 Reducible Polynomials and Product Rings

Proposition (Artin 11.6.1). Let \(R\) and \(R'\) be rings.

  1. The product set \(R \times R'\) is a ring called the product ring, with componentwise addition and multiplication: \((x,x')+(y,y')=(x+y,x'+y'),\) \((x,x')(y,y')=(xy,x'y').\)

  2. The additive and multiplicative identities in \(R \times R'\) are \((0,0)\) and \((1,1)\).

  3. The projections \(\pi:R \times R' \to R, \qquad \pi(x,x')=x,\) \(\pi':R \times R' \to R', \qquad \pi'(x,x')=x',\) are ring homomorphisms. Their kernels are \(\{0\}\times R'\) and \(R \times \{0\}\).

  4. The kernel \(\{0\}\times R'\) is a ring with multiplicative identity \(e'=(0,1)\). It is not a subring of \(R \times R'\) unless \(R\) is the zero ring.

Definition. An idempotent element \(e\) of a ring \(S\) is an element satisfying \(e^2=e.\)

Proposition (Artin 11.6.2). Let \(e\) be an idempotent element of a ring \(S\).

  1. The element \(e'=1-e\) is also idempotent, \(e+e'=1\), and \(ee'=0\).
  2. With the operations inherited from \(S\), the principal ideal \(eS\) is a ring with identity element \(e\), and multiplication by \(e\) defines a ring homomorphism \(S \to eS\).
  3. The ideal \(eS\) is not a subring of \(S\) unless \(e=1\) and \(e'=0\).
  4. The ring \(S\) is isomorphic to the product ring \(eS \times e'S.\)

Proposition (Chinese Remainder Theorem). Let \(I,J \triangleleft R\) with \(I+J=R\). Then \(R/(I \cap J) \cong R/I \times R/J\) via \(r+(I \cap J) \mapsto (r+I,\ r+J).\)

8. Fraction Fields and Maximal Ideals

8.1 Fraction Fields

Definition. An integral domain, or simply a domain, is a commutative ring with identity and no zero divisors: \(ab=0\) implies \(a=0\) or \(b=0\).

Definition. Let \(R\) be an integral domain. The fraction field (or field of fractions) \(K\) of \(R\) consists of equivalence classes of pairs \((a,b)\) with \(a,b \in R\) and \(b \ne 0\), where \((a,b) \sim (c,d) \iff ad=bc.\)

Denote the class of \((a,b)\) by \(a/b\).

Operations are defined by \(\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}, \qquad \frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}.\)

The relation is reflexive, symmetric, and transitive.

Theorem. \(K\) is a field containing \(R\) via the embedding \(r \mapsto \frac{r}{1}.\)

Every nonzero element of \(R\) becomes invertible in \(K\).

Proposition (Universal Property). Let \(\phi:R \to L\) be a ring homomorphism to a field \(L\), and suppose \(\phi\) sends nonzero elements of \(R\) to nonzero elements of \(L\). Then there is a unique homomorphism \(\Phi:K \to L\) extending \(\phi\): \(\Phi(a/b)=\phi(a)\phi(b)^{-1}.\)

Proposition. Any field containing \(R\) as a subring contains a subfield isomorphic to \(K\).

8.2 Maximal Ideals

Definition. An ideal \(M \triangleleft R\) in a commutative ring with \(1\) is maximal if \(M \ne R\) and there is no ideal \(I\) such that \(M \subsetneq I \subsetneq R.\)

Theorem. An ideal \(M\) is maximal if and only if \(R/M\) is a field.

Examples.

Proposition. If \(R\) is a PID, every nonzero prime ideal \((p)\) is maximal.

Note. \(F[x]\) is a PID. It is also a UFD.

8.3 Maximal Ideals in \(\mathbb{C}[x_1,\ldots,x_n]\)

Lemma.

  1. Let \(R\) be a ring that contains \(\mathbb{C}\) as a subring. The operations on \(R\) make \(R\) into a complex vector space.

  2. For any maximal ideal \(M\) of \(\mathbb{C}[x_1,\ldots,x_n]\), the field \(\mathcal{F}=\mathbb{C}[x_1,\ldots,x_n]/M\)

    is spanned by a countable set of elements as a vector space over \(\mathbb{C}\).

  3. Let \(V\) be a vector space over a field. If \(V\) is spanned by a countable set of vectors, then every independent subset of \(V\) is finite or countably infinite.

  4. When the fraction field \(\mathbb{C}(x)\) is made into a vector space over \(\mathbb{C}\), the uncountable set of rational functions \(\{(x-\alpha)^{-1} : \alpha \in \mathbb{C}\}\)

    is linearly independent.

Theorem (Hilbert’s Nullstellensatz). The maximal ideals of the polynomial ring \(\mathbb{C}[x_1,\ldots,x_n]\) are in bijective correspondence with points of \(\mathbb{C}^n\).

A point \(a=(a_1,\ldots,a_n) \in \mathbb{C}^n\) corresponds to the kernel \(M_a\) of the substitution map \(S_a:\mathbb{C}[x_1,\ldots,x_n] \to \mathbb{C}, \qquad x_s \mapsto a_s.\)

The kernel \(M_a\) is generated by the \(n\) linear polynomials \(x_s-a_s.\)

9. Module Basics

In linear algebra, we study vector spaces over a field \(F\). The core idea of Chapter 14 is to ask what happens when the field of scalars \(F\) is replaced by a general ring \(R\).

Assume \(R\) is a ring with identity \(1 \ne 0\). We do not require \(R\) to be commutative unless explicitly stated.

Definition 9.1. Let \(R\) be a ring. A left \(R\)-module (or simply an \(R\)-module) is an abelian group \((V,+)\) equipped with scalar multiplication \(R \times V \to V, \qquad (r,v) \mapsto r \cdot v,\) satisfying the following axioms for all \(r,s \in R\) and \(v,w \in V\):

  1. \(r(v+w)=rv+rw\).
  2. \((r+s)v=rv+sv\).
  3. \((rs)v=r(sv)\).
  4. \(1\cdot v=v\).

Remark 9.2. These axioms are exactly the same as those for a vector space, except that \(R\) is a ring. Since we cannot always divide by scalars, the geometry and structure change fundamentally. For example, span and linear independence behave differently.

Example 9.3. Every vector space over a field \(F\) is an \(F\)-module.

Example 9.4 (Abelian Groups as \(\mathbb{Z}\)-modules). Let \(V\) be any abelian group. We can make \(V\) into a \(\mathbb{Z}\)-module. For \(n \in \mathbb{Z}\) and \(v \in V\), define \(n \cdot v=v+\cdots+v\) if \(n>0\), define \(n \cdot v=0\) if \(n=0\), and define \(n \cdot v=-(v+\cdots+v)\) if \(n<0\).

Example 9.5 (\(F[x]\)-modules). What does it mean to have an \(F[x]\)-module \(V\)?

First, \(V\) is an abelian group. Since \(F \subseteq F[x]\) as constant polynomials, \(V\) is an \(F\)-module, hence a vector space over \(F\). Scalar multiplication by \(x \in F[x]\) gives a map \(T:V \to V, \qquad T(v)=x \cdot v.\)

By the module axioms, \(T(v+w)=T(v)+T(w), \qquad T(cv)=cT(v) \quad (c \in F).\)

Thus \(T\) is a linear operator on \(V\).

Conversely, given any \(F\)-vector space \(V\) and linear operator \(T:V \to V\), define an \(F[x]\)-module structure by \((a_nx^n+\cdots+a_1x+a_0)v = a_nT^n(v)+\cdots+a_1T(v)+a_0v.\)

Conclusion. An \(F[x]\)-module is precisely a pair \((V,T)\) consisting of an \(F\)-vector space \(V\) and a linear operator \(T\) on \(V\).

Definition 9.6. A submodule \(W\) of an \(R\)-module \(V\) is a nonempty subset \(W \subseteq V\) that is closed under addition and scalar multiplication.

Example 9.7. Consider the ring \(R\) as a module over itself. A submodule of \(R\) is a subset closed under addition and left scalar multiplication by elements of \(R\). This is precisely a left ideal of \(R\).

Definition 9.8. Let \(V\) and \(W\) be \(R\)-modules. A map \(\phi:V \to W\) is an \(R\)-module homomorphism (or \(R\)-linear map) if it preserves addition and scalar multiplication: \(\phi(v_1+v_2)=\phi(v_1)+\phi(v_2), \qquad \phi(rv)=r\phi(v).\)

As with vector spaces and groups:

If \(W \subseteq V\) is a submodule, the quotient group \(V/W\) becomes an \(R\)-module by \(r(v+W)=rv+W.\)

Similarly, for any two modules \(V\) and \(W\), the product \(V \times W\) is an \(R\)-module.

The isomorphism theorems carry over perfectly. The first isomorphism theorem says that if \(\phi:V \to W\) is a homomorphism, then \(V/\ker\phi \cong \operatorname{Im}\phi.\)

10. Free Modules

A central question is whether modules have bases like vector spaces do. The short answer is no. Modules that do have bases are special and easier to understand.

Definition 10.1. Let \(V\) be an \(R\)-module. A subset \(B=\{b_1,\ldots,b_n\} \subseteq V\)

is called a basis for \(V\) if every element \(v \in V\) can be written uniquely as \(v=r_1b_1+r_2b_2+\cdots+r_nb_n, \qquad r_i \in R.\)

If \(V\) has a basis, then it is called a free module over \(R\).

Example 10.2 (Free Module). The set \(R^n\) of column vectors of size \(n\) with entries in \(R\) is an \(R\)-module. The standard vectors \(e_1=(1,0,\ldots,0)^t,\ldots,e_n=(0,\ldots,0,1)^t\)

form a basis for \(R^n\). Thus \(R^n\) is a free module.

Theorem 10.1. Let \(F\) be a free \(R\)-module with basis \(B=\{b_1,\ldots,b_n\}\). Then \(F \cong R^n.\)

Proof. Define \(R^n \to F, \qquad (r_1,\ldots,r_n) \mapsto \sum_i r_i b_i.\)

This is a module homomorphism. It is surjective by definition of a basis and injective by uniqueness of expression.

More generally, maps \(R^n \to R^m\) can be represented by \(m \times n\) matrices with entries in \(R\), just as in linear algebra. Conversely, if \(A\) is an \(m \times n\) matrix, then \(X \mapsto AX\) is an \(R\)-module homomorphism \(R^n \to R^m\).

Corollary 10.2. Every finitely generated \(R\)-module \(M\) is isomorphic to a quotient of a free module.

Proof. Suppose \(M\) is generated by \(m_1,\ldots,m_r\). Let \(R^r\) have standard basis \(e_1,\ldots,e_r\). There is a unique homomorphism \(\phi:R^r \to M, \qquad \phi(e_i)=m_i.\)

Since \(m_1,\ldots,m_r\) generate \(M\), the map \(\phi\) is surjective. Let \(K=\ker \phi\). By the first isomorphism theorem, \(R^r/K \cong M.\)

The kernel \(K\) is called the module of relations among the generators \(m_i\).

Example 10.3 (Torsion Module). Let \(R=\mathbb{Z}\) and consider the \(\mathbb{Z}\)-module \(M=\mathbb{Z}/2\mathbb{Z}.\)

Then \(M\) is not a free \(\mathbb{Z}\)-module. If \(M\) had a basis, it would contain the unique nonzero element \(1\) of \(M\). But \(2 \cdot 1 = 0,\)

which contradicts uniqueness of expression.

Theorem 10.3 (Smith Normal Form). Let \(A\) be an integer matrix. There exist products \(Q\) and \(P\) of elementary integer matrices such that \(A'=Q^{-1}AP\)

is diagonal, with positive diagonal entries \(d_i\) satisfying \(d_1 \mid d_2 \mid \cdots \mid d_k.\)

Corollary 10.4. Let \(W\) be a free abelian group of rank \(m\), and let \(U\) be a subgroup of \(W\). Then \(U\) is a free abelian group of rank at most \(m\).

Combining this with Corollary 10.2, any finitely generated \(R\)-module \(M\) can be presented as the cokernel of a map \(R^m \to R^n.\)

For a homomorphism \(\phi:M \to N\), \(M/\ker \phi \cong \operatorname{Im}\phi, \qquad \operatorname{coker}\phi := N/\operatorname{Im}\phi.\)

11. Structure Theorems

11.1 Structure of Finitely Generated Abelian Groups

Recall that an abelian group is exactly the same thing as a \(\mathbb{Z}\)-module. A group \(V\) is finitely generated if there exist \(v_1,\ldots,v_n \in V\) such that every \(v \in V\) can be written as an integer linear combination \(v=c_1v_1+c_2v_2+\cdots+c_nv_n.\)

Theorem 11.1 (Structure Theorem for Finitely Generated Abelian Groups). Let \(V\) be a finitely generated abelian group. Then \(V\) is isomorphic to a direct sum of cyclic groups: \(V \cong \mathbb{Z}^{\oplus r} \oplus \mathbb{Z}/d_1\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/d_k\mathbb{Z},\)

where \(r \ge 0\) and \(d_1,\ldots,d_k\) are integers greater than \(1\) satisfying \(d_1 \mid d_2 \mid \cdots \mid d_k.\)

Furthermore, the rank \(r\) and the sequence \(d_1,\ldots,d_k\) are uniquely determined by \(V\).

Proof idea. Any such \(V\) can be presented as the cokernel of \(\phi:\mathbb{Z}^m \to \mathbb{Z}^n.\)

Let \(A\) be the matrix of \(\phi\). Smith Normal Form transforms \(A\) into a diagonal matrix \(D\) via elementary row and column operations. With respect to adapted bases, the quotient module decomposes as \(V \cong \mathbb{Z}^n/\operatorname{Im}\phi \cong \frac{\mathbb{Z}e_1' \oplus \cdots \oplus \mathbb{Z}e_n'} {\mathbb{Z}d_1e_1' \oplus \cdots \oplus \mathbb{Z}d_ke_k'}\)

and therefore \(V \cong \mathbb{Z}/d_1\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/d_k\mathbb{Z} \oplus \mathbb{Z}^{n-k}.\)

Thus \(r=n-k\). Uniqueness follows from the uniqueness of Smith Normal Form.

The theorem splits the group into two parts:

  1. The free part, \(\mathbb{Z}^{\oplus r}\), consists of the torsion-free part. The number \(r\) is called the free rank of \(V\).

  2. The torsion part, \(\mathbb{Z}/d_1\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/d_k\mathbb{Z},\)

    consists of all elements of finite order. The integers \(d_i\) are called the invariant factors of \(V\). The largest factor \(d_k\) is called the exponent of the torsion part.

11.2 The Analogue for \(F[x]\)

The integers \(\mathbb{Z}\) form a PID. Another important PID is the polynomial ring \(F[x]\).

The Smith Normal Form and the structure theorem for \(\mathbb{Z}\)-modules hold verbatim if we replace \(\mathbb{Z}\) by any PID, such as \(F[x]\).

Let \(V\) be a finite-dimensional vector space over \(F\), and let \(T:V \to V\) be a linear operator. Recall that \(V\) becomes an \(F[x]\)-module by \((a_nx^n+\cdots+a_1x+a_0)\cdot v = (a_nT^n+\cdots+a_1T+a_0I)v.\)

Theorem 11.2 (Structure Theorem for Finitely Generated \(F[x]\)-modules). Let \(V\) be a finitely generated \(F[x]\)-module. Then \(V \cong F[x]^{\oplus r} \oplus F[x]/(p_1) \oplus \cdots \oplus F[x]/(p_k),\)

where \(r \ge 0\) and \(p_1,\ldots,p_k\) are nonconstant polynomials satisfying \(p_1 \mid p_2 \mid \cdots \mid p_k.\)

If \(V\) is finite-dimensional as a vector space over \(F\), then \(r=0\), because \(F[x]\) is infinite-dimensional over \(F\).

Each cyclic summand \(W_i \cong F[x]/(p_i(x))\)

captures one invariant factor of the linear operator \(T\).

12. Fields and Algebraic Extensions

12.1 Fields and Extensions

Fields have no nontrivial proper ideals, so any unital ring homomorphism between fields \(\phi:F \to K\) must be injective. Thus, the primary relationship between fields is inclusion.

Definition 12.1. Let \(F\) and \(K\) be fields. If \(F\) is a subring of \(K\), then \(F\) is a subfield of \(K\), and \(K\) is an extension field of \(F\). We write \(K/F \qquad \text{or simply} \qquad F \subseteq K.\)

Every field contains a unique smallest field, called its prime subfield.

Consequently, any field problem can ultimately be traced back to extensions of \(\mathbb{Q}\) or \(\mathbb{F}_p\).

12.2 Adjunction of Elements

Suppose \(K/F\) is a field extension and \(\alpha \in K\). We are often interested in the smallest subfield of \(K\) containing both \(F\) and \(\alpha\).

Definition 12.2. Let \(K/F\) be a field extension, and let \(\alpha_1,\ldots,\alpha_n \in K\).

  1. \(F[\alpha_1,\ldots,\alpha_n]\) denotes the smallest subring of \(K\) containing \(F\) and all \(\alpha_i\). It consists of all polynomial expressions in the \(\alpha_i\) with coefficients in \(F\).
  2. \(F(\alpha_1,\ldots,\alpha_n)\) denotes the smallest subfield of \(K\) containing \(F\) and all \(\alpha_i\). It consists of all rational expressions in the \(\alpha_i\) with coefficients in \(F\).

An extension generated by a single element, \(F(\alpha)/F\), is called a simple extension.

Example. Consider \(\mathbb{R}/\mathbb{Q}\) and \(\alpha=\sqrt{2}\). The ring \(\mathbb{Q}[\sqrt{2}]\) consists of all numbers of the form \(a+b\sqrt{2}, \qquad a,b \in \mathbb{Q}.\)

In fact, \(\mathbb{Q}[\sqrt{2}]=\mathbb{Q}(\sqrt{2}),\)

because \(\frac{1}{a+b\sqrt{2}} = \frac{a}{a^2-2b^2} - \frac{b}{a^2-2b^2}\sqrt{2}.\)

12.3 Algebraic and Transcendental Elements

Let \(K/F\) be a field extension and \(\alpha \in K\). We analyze the relationship between \(\alpha\) and \(F\) by considering polynomials over \(F\) that have \(\alpha\) as a root.

Define the evaluation homomorphism \(\phi_\alpha:F[x] \to K, \qquad f(x) \mapsto f(\alpha).\)

The image of \(\phi_\alpha\) is exactly \(F[\alpha]\). By the first isomorphism theorem, \(F[x]/\ker\phi_\alpha \cong F[\alpha].\)

Since \(F[x]\) is a PID, the ideal \(\ker\phi_\alpha\) is either zero or generated by a single monic polynomial.

Definition 12.3.

  1. \(\alpha\) is transcendental over \(F\) if \(\ker\phi_\alpha=0\). Equivalently, no nonzero polynomial in \(F[x]\) has \(\alpha\) as a root.
  2. \(\alpha\) is algebraic over \(F\) if \(\ker\phi_\alpha \ne 0\). Equivalently, some nonzero polynomial in \(F[x]\) has \(\alpha\) as a root.

If \(\alpha\) is transcendental over \(F\), then \(F[\alpha] \cong F[x], \qquad F(\alpha) \cong F(x).\)

But \(F[x]\) is strictly contained in \(F(x)\), so \(F[\alpha] \subsetneq F(\alpha).\)

12.4 Minimal Polynomials

If \(\alpha\) is algebraic, then \(\ker\phi_\alpha=(m(x))\) for a unique monic polynomial \(m(x) \in F[x]\).

Definition. This polynomial \(m(x)\) is called the minimal polynomial of \(\alpha\) over \(F\). It is often denoted \(m_{\alpha,F}(x).\)

The minimal polynomial has several immediate properties:

  1. \(m(x)\) is the monic polynomial of lowest degree in \(F[x]\) having \(\alpha\) as a root.
  2. If \(g(x) \in F[x]\) has \(\alpha\) as a root, then \(m(x) \mid g(x)\) in \(F[x]\).
  3. \(m(x)\) is irreducible over \(F\).

For irreducibility, suppose \(m(x)=f(x)g(x)\). Then \(m(\alpha)=f(\alpha)g(\alpha)=0.\)

Since \(K\) is a field, either \(f(\alpha)=0\) or \(g(\alpha)=0\). Because \(m(x)\) generates the kernel, it must divide either \(f(x)\) or \(g(x)\), which is impossible unless one factor is a unit.

Theorem 12.1. Let \(K/F\) be a field extension, and let \(\alpha \in K\) be algebraic over \(F\). Then \(F[\alpha]=F(\alpha).\)

In particular, \(F[\alpha]\) is a field.

Proof. Let \(m(x)\) be the minimal polynomial of \(\alpha\) over \(F\). We have \(F[x]/(m(x)) \cong F[\alpha].\)

Since \(F[\alpha]\) is a subring of the field \(K\), it has no zero divisors. Therefore, \((m(x))\) is a prime ideal. In a PID, every nonzero prime ideal is maximal. Since \(\alpha\) is algebraic, \(m(x)\) is nonzero, so \((m(x))\) is maximal. Thus \(F[x]/(m(x)) \cong F[\alpha]\) is a field. By definition, \(F[\alpha]=F(\alpha)\).

By applying this repeatedly to algebraic elements, \(F[\alpha_1,\ldots,\alpha_n]=F(\alpha_1,\ldots,\alpha_n).\)

Example 12.2. Let \(\alpha=\sqrt{2}+\sqrt{3}.\)

Then \(\alpha^2=5+2\sqrt{6},\)

so \(\alpha^4-10\alpha^2+1=0.\)

Thus a candidate minimal polynomial over \(\mathbb{Q}\) is \(m_{\alpha,\mathbb{Q}}(x)=x^4-10x^2+1.\)

Checking irreducibility shows that it is indeed minimal, and \(\mathbb{Q}[\sqrt{2}+\sqrt{3}] = \mathbb{Q}(\sqrt{2}+\sqrt{3}).\) Corollary 12.3. If \(\alpha\) is algebraic over \(F\) and \(f(x)\) is the minimal polynomial of \(\alpha\) over \(F\) with degree \(n\), then \(\{1,\alpha,\ldots,\alpha^{n-1}\}\)

is a basis for \(F(\alpha)\) as a vector space over \(F\).

Corollary 12.4. Let \(K/F\) and \(L/F\) be field extensions. Suppose \(\alpha \in K\) and \(\beta \in L\) are algebraic over \(F\). There is an isomorphism of fields \(\sigma:F(\alpha) \to F(\beta)\) that is the identity on \(F\) and satisfies \(\sigma(\alpha)=\beta\) if and only if \(\alpha\) and \(\beta\) have the same minimal polynomial over \(F\).

Proposition (Eisenstein Criterion). Let \(f(x)=a_nx^n+\cdots+a_1x+a_0 \in \mathbb{Z}[x].\)

If there exists a prime \(p\) such that:

  1. \(p \nmid a_n\),
  2. \(p \mid a_i\) for all \(0 \le i < n\),
  3. \(p^2 \nmid a_0\),

then \(f(x)\) is irreducible over \(\mathbb{Q}\).

Example (Cyclotomic Polynomials). The \(n\)-th cyclotomic polynomial \(\Phi_n(x)\) is the polynomial whose roots are the primitive \(n\)-th roots of unity. It is irreducible over \(\mathbb{Q}\), and \(x^n-1=\prod_{d \mid n}\Phi_d(x).\)

For a prime \(p\), \(\Phi_p(x)=\frac{x^p-1}{x-1}=x^{p-1}+\cdots+x+1.\)

13. Degree of Field Extensions

13.1 Degree and Algebraicity

A field extension \(K/F\) can be viewed as a vector space over \(F\).

Definition. The degree of the field extension \(K/F\), denoted \([K:F]\), is the dimension of \(K\) as a vector space over \(F\).

Corollary 13.2. Let \(\alpha\) be algebraic over \(F\), with minimal polynomial \(m_{\alpha,F}(x)\). Then \([F(\alpha):F]=\deg m_{\alpha,F}.\)

Moreover, if \(n=\deg m_{\alpha,F}\), then \(\{1,\alpha,\ldots,\alpha^{n-1}\}\)

is a basis for \(F(\alpha)\) over \(F\).

Theorem 13.3. If \(K/F\) is a finite extension, then it is algebraic.

Proof. Let \([K:F]=n<\infty\), and pick \(\beta \in K\). Consider \(1,\beta,\beta^2,\ldots,\beta^n.\)

These are \(n+1\) elements in an \(n\)-dimensional vector space over \(F\), so they are linearly dependent. Thus there exist \(c_0,\ldots,c_n \in F\), not all zero, such that \(\sum_{i=0}^n c_i\beta^i=0.\)

Therefore, \(\beta\) is a root of the nonzero polynomial \(f(x)=\sum_{i=0}^n c_ix^i.\)

So \(\beta\) is algebraic over \(F\).

13.2 The Tower Law

Theorem 13.4 (Tower Law). Let \(F \subseteq K \subseteq L\) be a tower of fields. Then \([L:F]=[L:K][K:F].\)

If either \([L:K]\) or \([K:F]\) is infinite, then \([L:F]\) is infinite.

Corollary 13.5. Let \(K/F\) be an extension. The set of all elements of \(K\) that are algebraic over \(F\) forms a subfield of \(K\).

Proof idea. If \(\alpha,\beta \in K\) are algebraic over \(F\), then \(F(\alpha)/F\) is finite. Since \(\beta\) is algebraic over \(F\), it is also algebraic over \(F(\alpha)\), so \(F(\alpha,\beta)/F\) is finite by the Tower Law. Hence \(\alpha+\beta\), \(\alpha-\beta\), \(\alpha\beta\), and \(\alpha/\beta\) (when \(\beta \ne 0\)) are algebraic over \(F\).

Example 13.2. What is the degree of \(\mathbb{Q}(\sqrt{2},\sqrt{3})\) over \(\mathbb{Q}\)?

By the Tower Law, \([\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})] [\mathbb{Q}(\sqrt{2}):\mathbb{Q}].\)

The element \(\sqrt{3}\) is a root of \(x^2-3 \in \mathbb{Q}(\sqrt{2})[x]\), and \(x^2-3\) remains irreducible over \(\mathbb{Q}(\sqrt{2})\). Therefore the degree is \(4\), with basis \(\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}.\)

14. Splitting Fields and Finite Fields

14.1 Kronecker’s Theorem

Theorem 14.1 (Kronecker’s Theorem). Let \(F\) be a field, and let \(f(x) \in F[x]\) be a nonconstant polynomial. There exists an extension field \(K\) of \(F\) in which \(f(x)\) has a root.

Proof. Since \(F[x]\) is a PID, \(f(x)\) has an irreducible factor \(p(x) \in F[x]\). Since \(p(x)\) is irreducible, the ideal \(I=(p(x))\) is maximal, so \(K=F[x]/(p(x))\)

is a field.

We claim that \(\alpha=x+I\) is a root of \(p(x)\) in \(K\). Indeed, \(p(\alpha)=p(x+I)=p(x)+I=0+I=0.\)

Thus we have constructed an extension \(K\) where \(f(x)\) has a root.

14.2 Splitting Fields

Definition 14.1. Let \(F\) be a field and \(f(x) \in F[x]\). An extension \(K/F\) is a splitting field for \(f(x)\) over \(F\) if:

  1. \(f(x)\) factors completely into linear factors in \(K[x]\): \(f(x)=c(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n), \qquad \alpha_i \in K.\)

  2. The field is generated by these roots: \(K=F(\alpha_1,\alpha_2,\ldots,\alpha_n).\)

By applying Kronecker’s theorem repeatedly, we can construct a splitting field \(K/F\).

Theorem 14.2. For any field \(F\) and nonconstant \(f(x) \in F[x]\), a splitting field for \(f(x)\) exists and is unique up to isomorphism.

Proof idea for uniqueness. We prove uniqueness by induction on \(\deg f\). Suppose \(K\) and \(K'\) are both splitting fields for \(f(x)\) over \(F\). Let \(\phi:F \to F\) be the identity map.

Let \(p(x)\) be an irreducible factor of \(f(x)\) in \(F[x]\). Since \(f(x)\) splits in \(K\) and \(K'\), the polynomial \(p(x)\) has roots \(\alpha \in K\) and \(\alpha' \in K'\). Then \(F(\alpha) \cong F[x]/(p(x)) \cong F(\alpha'),\)

giving an isomorphism \(\widetilde{\phi}:F(\alpha) \to F(\alpha')\) that is the identity on \(F\).

Now view \(f(x)\) over the larger base field \(F(\alpha)\). Since we can factor out \((x-\alpha)\), the degree of \(f(x)/(x-\alpha)\) is smaller than \(\deg f\). By induction, \(\widetilde{\phi}\) extends to an isomorphism \(\Phi:K \to K'.\)

14.3 Recognizing Multiple Roots

Lemma 14.3. For \(f \in F[x]\), an element \(\alpha\) in an extension \(K/F\) is a multiple root of \(f\) if and only if it is a common root of \(f\) and \(f'\).

Corollary 14.4. Let \(f\) be an irreducible polynomial in \(F[x]\). Then \(f\) has no multiple roots in any field extension of \(F\) unless \(f'\) is the zero polynomial. In particular, if \(\operatorname{char}F=0\), then \(f\) has no multiple roots in any field extension of \(F\).

14.4 Finite Fields

For a finite field \(F\), we have \(\operatorname{char}F=p \ne 0\). The field contains \(\mathbb{F}_p\) as its prime subfield. It is also a finite-dimensional vector space over \(\mathbb{F}_p\), say of dimension \(r\). Thus the order of any finite field is a prime power: \(|F|=p^r.\)

Theorem 14.5 (Existence and Uniqueness). For every prime power \(q=p^r\), there exists exactly one finite field of order \(q\) up to isomorphism, denoted by \(\mathbb{F}_q\).

Proof. Let \(F\) be a field with \(q=p^r\) elements. Consider the multiplicative group \(F^\times = F \setminus \{0\}.\)

It has order \(q-1\). By Lagrange’s theorem, every \(\alpha \in F^\times\) satisfies \(\alpha^{q-1}=1.\)

Multiplying by \(\alpha\), every element of \(F\) satisfies \(\alpha^q-\alpha=0.\)

Thus every element of \(F\) is a root of \(x^q-x \in \mathbb{F}_p[x].\)

Since this polynomial has degree \(q\), the elements of \(F\) are exactly the roots of \(x^q-x\). Therefore \(F\) is the splitting field of \(x^q-x\) over \(\mathbb{F}_p\). Since splitting fields are unique up to isomorphism, there is at most one field of order \(q\).

For existence, take a splitting field of \(x^q-x\) over \(\mathbb{F}_p\). Its derivative is \(qx^{q-1}-1=-1,\)

so \(x^q-x\) has \(q\) distinct roots. These roots are closed under addition, subtraction, multiplication, and inversion, so they form a field with \(q\) elements. Hence \(\mathbb{F}_q\) exists.

15. Finite Fields and Primitive Elements

15.1 The Multiplicative Group of a Finite Field

Theorem 15.1. The multiplicative group \(F^\times\) of a finite field \(F\) is cyclic.

Proof. Let \(G=\mathbb{F}_q^\times\backslash \{0\}.\)

Then \(G\) is a finite abelian group of order \(q-1\). By the structure theorem for finite abelian groups, \(G \cong \mathbb{Z}/d_1\mathbb{Z} \times \mathbb{Z}/d_2\mathbb{Z} \times \cdots \times \mathbb{Z}/d_k\mathbb{Z},\)

where \(d_1 \mid d_2 \mid \cdots \mid d_k.\)

The maximal order of any element in \(G\) is the exponent \(d_k\). Thus \(\alpha^{d_k}=1 \qquad \text{for all } \alpha \in G.\)

Consequently, every element of \(G\) is a root of the polynomial \(x^{d_k}-1.\)

A polynomial of degree \(d_k\) over a field has at most \(d_k\) roots. Since all \(q-1\) elements of \(G\) are roots, we must have \(q-1 \le d_k.\)

On the other hand, \(d_k\) is the order of an element in a group of order \(q-1\), so by Lagrange’s theorem, \(d_k \mid q-1.\)

Therefore \(d_k=q-1\). Hence there exists an element of order \(q-1\) in \(G\), so \(G\) is cyclic.

A generator of \(\mathbb{F}_q^\times\) is called a primitive root of the finite field.

Example 15.1. Recall that \(\mathbb{F}_4=\mathbb{F}_2[x]/(x^2+x+1)=\mathbb{F}_2(\alpha),\)

where \(\alpha^2+\alpha+1=0.\)

Thus \(\alpha^2=\alpha+1, \qquad \alpha^3=\alpha(\alpha+1)=1.\)

So \(\alpha\) is a primitive root for \(\mathbb{F}_4\), and \(\mathbb{F}_4=\{0,1,\alpha,\alpha+1\}.\)

Example 15.2. For \(q=8\), we need an irreducible cubic over \(\mathbb{F}_2\). The polynomial \(x^3+x+1\)

has no roots in \(\mathbb{F}_2\), so it is irreducible over \(\mathbb{F}_2\). Let \(\beta\) be a root: \(\beta^3+\beta+1=0.\)

Then \(\mathbb{F}_8=\{a+b\beta+c\beta^2 : a,b,c \in \mathbb{F}_2\}.\)

The group \(\mathbb{F}_8^\times\) has order \(7\), which is prime. Therefore every nonidentity element of \(\mathbb{F}_8^\times\) is a primitive root.

15.2 Subfields and the Frobenius Automorphism

Definition 15.1. Let \(F\) be a field of characteristic \(p>0\). The map \(\phi:F \to F, \qquad \phi(x)=x^p\)

is called the Frobenius map.

It satisfies \(\phi(x+y)=\phi(x)+\phi(y), \qquad \phi(xy)=\phi(x)\phi(y).\)

If \(F\) is finite, then \(\phi\) is injective and therefore surjective, so it is an automorphism. Its fixed field is precisely \(\mathbb{F}_p\).

Theorem 15.2 (Subfield Structure). Let \(\mathbb{F}_q\) be a finite field of order \(q=p^r\). Any subfield of \(\mathbb{F}_q\) has order \(p^k\) for some integer \(k\) with \(k \mid r\).

Conversely, for every \(k \mid r\), the field \(\mathbb{F}_q\) contains exactly one subfield of order \(p^k\).

Proof. If \(K \subseteq \mathbb{F}_q\) is a subfield, then \(|K|=p^k\) for some \(k\). Since \(\mathbb{F}_q\) is a vector space over \(K\), say of dimension \(d\), \(|\mathbb{F}_q|=|K|^d=(p^k)^d=p^{kd}.\)

Thus \(r=kd\), so \(k \mid r\).

Conversely, assume \(k \mid r\). Then \(p^k-1 \mid p^r-1.\)

It follows that \(x^{p^k-1}-1 \mid x^{p^r-1}-1.\)

Multiplying by \(x\), we get \(x^{p^k}-x \mid x^{p^r}-x.\)

Since \(x^{p^r}-x\) splits into distinct linear factors in \(\mathbb{F}_q\), its divisor \(x^{p^k}-x\) also splits completely into distinct linear factors in \(\mathbb{F}_q\).

The roots of \(x^{p^k}-x\) form a subfield of \(\mathbb{F}_q\), and this subfield has \(p^k\) elements. Uniqueness follows because any subfield of order \(p^k\) consists exactly of the roots of \(x^{p^k}-x\).

Corollary 15.3 (Irreducible Factors of \(x^q-x\)). Let \(F=\mathbb{F}_p\) and \(q=p^r\). The irreducible factors of \(x^q-x \in F[x]\) are precisely the monic irreducible polynomials in \(F[x]\) whose degrees divide \(r\).

Proof. Let \(f(x) \in F[x]\) be a monic irreducible polynomial of degree \(k\), and let \(\alpha\) be a root of \(f(x)\) in some splitting field. Since \(f(x)\) is irreducible, it is the minimal polynomial of \(\alpha\) over \(F\), and \([F(\alpha):F]=k.\)

Suppose \(f(x)\) divides \(x^q-x\). Then \(\alpha\) is a root of \(x^q-x\), so \(\alpha^q=\alpha.\)

Thus \(\alpha \in \mathbb{F}_q\), and \(F(\alpha)\) is a subfield of \(\mathbb{F}_q\). By the Tower Law, \(k=[F(\alpha):F] \mid [\mathbb{F}_q:F]=r.\)

Conversely, suppose \(k \mid r\). The field extension \(F(\alpha)\) has order \(p^k\), so \(F(\alpha) \cong \mathbb{F}_{p^k}.\)

By the subfield structure theorem, \(\mathbb{F}_q\) contains a subfield isomorphic to \(\mathbb{F}_{p^k}\). Identifying \(F(\alpha)\) with this subfield, we see that \(\alpha \in \mathbb{F}_q\). Hence \(\alpha^q-\alpha=0.\)

Since \(f(x)\) is the minimal polynomial of \(\alpha\) over \(F\), it must divide \(x^q-x\).

15.3 Primitive Elements

A simple extension \(K=F(\gamma)\) is generated by a single element.

Lemma (GCD in \(F[x]\)). Two polynomials \(f,g \in F[x]\), not both zero, have a unique greatest common divisor \(d\). Moreover, there exist \(r,s \in F[x]\) such that \(rf+sg=d\).

Theorem 15.4 (Primitive Element Theorem). Let \(K/F\) be a finite extension. If \(\operatorname{char}F=0\), or if \(F\) is a finite field, then there exists \(\gamma \in K\) such that \(K=F(\gamma)\). Such a \(\gamma\) is called a primitive element.

Proof in characteristic \(0\). It is enough to prove the theorem for \(K=F(\alpha,\beta)\); the general case follows by induction. Let \(f(x)\) be the minimal polynomial of \(\alpha\) over \(F\), with roots \(\alpha=\alpha_1,\alpha_2,\ldots,\alpha_n\) in some splitting field. Let \(g(x)\) be the minimal polynomial of \(\beta\) over \(F\), with roots \(\beta=\beta_1,\ldots,\beta_m\). Since \(\operatorname{char}F=0\), these roots are distinct.

We look for a primitive element of the form \(\gamma=\alpha+c\beta\), where \(c \in F\). We choose \(c\) so that \(\gamma\) does not accidentally equal \(\alpha_i+c\beta_j\) unless \((i,j)=(1,1)\). If \(\alpha+c\beta=\alpha_i+c\beta_j\), then \(c=(\alpha-\alpha_i)/(\beta_j-\beta)\). Since \(F\) is infinite and there are only finitely many forbidden values of \(c\), we can choose \(c \in F\) avoiding all of them.

Let \(\gamma=\alpha+c\beta\). Clearly \(F(\gamma)\subseteq F(\alpha,\beta)\). It remains to show that \(\beta \in F(\gamma)\). Consider the polynomials \(g(x)\) and \(h(x)=f(\gamma-cx) \in F(\gamma)[x]\). The element \(\beta\) is a common root of \(g(x)\) and \(h(x)\), so \(x-\beta\) is a common factor.

In fact, \(x-\beta\) is the greatest common divisor. If \(\beta_j\) is another root of \(g(x)\) and \(h(\beta_j)=0\), then \(f(\gamma-c\beta_j)=0\), so \(\gamma-c\beta_j=\alpha_i\) for some \(i\). This is exactly the type of equality avoided by the choice of \(c\). Hence the only common root is \(\beta\).

By the GCD lemma, \(x-\beta=rg+sh\) for some \(r,s \in F(\gamma)[x]\). Therefore \(\beta \in F(\gamma)\). Since \(\alpha=\gamma-c\beta\), we also have \(\alpha \in F(\gamma)\), so \(F(\gamma)\supseteq F(\alpha,\beta)\). Thus \(F(\gamma)=F(\alpha,\beta)\).

If \(F\) is finite, then \(K\) is finite. The group \(K^\times\) is cyclic; if \(\gamma\) generates \(K^\times\), then every nonzero element of \(K\) is a power of \(\gamma\), so \(K=F(\gamma)\).

Remark. In fact, it is enough for \(K/F\) to be a finite separable extension. Separable means that the minimal polynomial of every \(\alpha \in K\) has distinct roots. The proof also shows that primitive elements are generic.

16. Group Actions

16.1 Group Operations

Definition 16.1. Let \(G\) be a group and \(S\) a set. An operation, or action, of \(G\) on \(S\) is a map \(G \times S \to S\), denoted \((g,s)\mapsto g\cdot s\), satisfying:

  1. Identity: \(1\cdot s=s\) for all \(s \in S\), where \(1\) is the identity of \(G\).
  2. Associativity: \((gh)\cdot s=g\cdot(h\cdot s)\) for all \(g,h \in G\) and \(s \in S\).

When such an operation is given, we say that \(G\) acts on \(S\). The action partitions \(S\) into disjoint equivalence classes: define \(s\sim s'\) if there exists \(g \in G\) such that \(g\cdot s=s'\). Each equivalence class is called an orbit.

Definition 16.2. The orbit of \(s \in S\) is the equivalence class \(O_s=\{g\cdot s : g \in G\}\). If there is only one orbit, so that \(O_s=S\) for some \(s\) and hence for all \(s \in S\), then \(G\) is said to act transitively on \(S\).

Definition 16.3. The stabilizer of \(s \in S\), denoted \(G_s\), is the set of elements of \(G\) that fix \(s\): \(G_s=\{g \in G : g\cdot s=s\}\).

Lemma 16.1. For any \(s \in S\), the stabilizer \(G_s\) is a subgroup of \(G\).

Proof. The identity satisfies \(1\cdot s=s\), so \(1 \in G_s\). If \(g,h \in G_s\), then \((gh)\cdot s=g\cdot(h\cdot s)=g\cdot s=s\), so \(gh \in G_s\). Finally, if \(g \in G_s\), then \(g\cdot s=s\), so \(s=g^{-1}\cdot(g\cdot s)=(g^{-1}g)\cdot s\), hence \(g^{-1}\cdot s=s\). Thus \(g^{-1} \in G_s\).

16.2 The Action on Cosets

Let \(H\) be a subgroup of \(G\). The set of left cosets of \(H\) in \(G\) is denoted by \(G/H\). The group \(G\) acts naturally on \(G/H\) by left multiplication.

Proposition 16.2. The map \(G \times (G/H) \to G/H\) given by \(g(aH)=(ga)H\) is a transitive group action. The stabilizer of the identity coset \(1H=H\) is exactly \(H\).

Proof. The action axioms are immediate. For transitivity, given \(aH\) and \(bH\), choose \(g=ba^{-1}\); then \(g(aH)=bH\). The identity satisfies \(1\cdot(aH)=aH\). Finally, \(g\) stabilizes \(H\) if and only if \(g\cdot H=H\), equivalently \(gH=H\), which holds if and only if \(g \in H\).

Theorem 16.3. Suppose \(G\) acts transitively on a set \(S\). Let \(s \in S\), and let \(G_s\) be its stabilizer. There is a bijection \(\phi:G/G_s \to S\) given by \(\phi(gG_s)=g\cdot s\).

Proof. First, \(\phi\) is well defined. If \(aG_s=bG_s\), then \(b^{-1}a \in G_s\). Thus \((b^{-1}a)\cdot s=s\), so \(a\cdot s=b\cdot s\). Next, \(\phi\) is injective: if \(\phi(aG_s)=\phi(bG_s)\), then \(a\cdot s=b\cdot s\), so \(b^{-1}a\cdot s=s\). Hence \(b^{-1}a \in G_s\), and therefore \(aG_s=bG_s\). Finally, \(\phi\) is surjective because the action is transitive: for every \(s' \in S\), there exists \(g \in G\) such that \(g\cdot s=s'\), and then \(\phi(gG_s)=s'\).

Corollary. For any action of \(G\) on \(S\) and any \(s \in S\), the group \(G\) acts transitively on the orbit \(O_s\), and there is a bijection \(G/G_s \to O_s\).

16.3 Counting Formula

Theorem 16.4 (Orbit-Stabilizer Formula). Let \(G\) be a finite group acting on a set \(S\). For any \(s \in S\), the size of the orbit \(O_s\) is equal to the index of the stabilizer \(G_s\) in \(G\): \(|O_s|=[G:G_s]=|G|/|G_s|\).

Because the orbits partition \(S\), if \(S\) is finite and \(s_1,\ldots,s_k\) are representatives of the distinct orbits, then \(|S|=\sum_{i=1}^k |O_{s_i}|=\sum_{i=1}^k [G:G_{s_i}]\).

Example 16.5. Let \(G\) be the group of rigid rotational symmetries of a standard cube, and pick a face \(f\). The orbit of \(f\) has size \(6\). The stabilizer consists of the \(4\) rotations around the axis through the center of \(f\), namely the rotations by \(0,90,180,270\) degrees. Hence \(|G|=|O_f||G_f|=6\cdot 4=24\).

17. Conjugation Actions and the Class Equation

17.1 Actions on Subsets

If \(G\) acts on a set \(S\), it naturally induces an action on the power set \(\mathcal{P}(S)\), the set of all subsets of \(S\). For \(U \subseteq S\) and \(g \in G\), define \(gU=\{gu : u \in U\}\).

If we restrict this action to the set of subgroups of \(G\), then the conjugate of a subgroup \(H\), namely \(gHg^{-1}\), is again a subgroup.

Definition 17.1. Let \(H\) be a subgroup of \(G\). The normalizer of \(H\) in \(G\), denoted \(N(H)\), is the stabilizer of \(H\) under the conjugation action: \(N(H)=\{g \in G : gHg^{-1}=H\}\).

The subgroup \(H\) is normal in \(N(H)\), and \(N(H)\) is the largest subgroup of \(G\) containing \(H\) as a normal subgroup. In particular, \(H\) is normal in \(G\) if and only if \(N(H)=G\).

Proposition 17.1. The number of subgroups of \(G\) conjugate to \(H\) is equal to \([G:N(H)]\).

17.2 The Class Equation

Now consider the action of \(G\) on itself by conjugation.

Definition 17.2. The orbits of the conjugation action are called conjugacy classes. The conjugacy class containing \(x\) is \(C_x=\{gxg^{-1} : g \in G\}\).

Definition 17.3. The stabilizer of \(x \in G\) under conjugation is called the centralizer of \(x\) in \(G\), denoted \(Z(x)\): \(Z(x)=\{g \in G : gxg^{-1}=x\}=\{g \in G : gx=xg\}\).

By the orbit-stabilizer formula, the size of a conjugacy class is the index of its centralizer: \(|C_x|=[G:Z(x)]=|G|/|Z(x)|\).

Definition 17.4. The center of a group \(G\), denoted \(Z(G)\), is the set of elements that commute with every element of \(G\): \(Z(G)=\{z \in G : zx=xz \text{ for all } x \in G\}\).

An element \(x \in G\) lies in \(Z(G)\) if and only if \(Z(x)=G\). Equivalently, \([G:Z(x)]=1\), or \(|C_x|=1\).

Theorem 17.5 (Class Equation). Let \(C_1,C_2,\ldots,C_k\) be the conjugacy classes of \(G\) having size strictly greater than \(1\), and choose a representative \(x_i \in C_i\) for each. Then \(|G|=|Z(G)|+\sum_{i=1}^k |C_i|=|Z(G)|+\sum_{i=1}^k [G:Z(x_i)]\).

Proposition 17.6. Let \(G\) be a nontrivial finite \(p\)-group, i.e. \(|G|=p^n\). Then \(|Z(G)|>1\).

Proof. By the class equation, \(|G|=|Z(G)|+\sum_{i=1}^k [G:Z(x_i)]\). Since \(|G|=p^n\), each index \([G:Z(x_i)]\) divides \(p^n\). The conjugacy classes appearing in the sum have size greater than \(1\), so \([G:Z(x_i)]\ne 1\) and each such index is divisible by \(p\). Therefore \(p\) divides \(|Z(G)|\), so \(|Z(G)|>1\).

18. Permutation Representations and Finite Rotation Groups

18.1 Permutation Representations

There is an exact correspondence between group actions on a set \(S\) and homomorphisms from \(G\) to the symmetric group of \(S\). Let \(\operatorname{Perm}(S)\) denote the group of all bijections from \(S\) to itself. If \(|S|=n\), then \(\operatorname{Perm}(S)\cong S_n\).

Theorem 18.1. An action of \(G\) on \(S\) defines a homomorphism \(\rho:G\to \operatorname{Perm}(S)\). Conversely, any homomorphism \(\rho:G\to \operatorname{Perm}(S)\) defines an action of \(G\) on \(S\).

Proof. Given an action, define \(m_g:S\to S\) by \(m_g(s)=g\cdot s\) for each \(g\in G\). The action axioms imply \(m_1=\operatorname{id}_S\) and \(m_{gh}(s)=(gh)\cdot s=g\cdot(h\cdot s)=m_g(m_h(s))\).

Thus the map \(\rho:G\to \operatorname{Perm}(S)\) given by \(\rho(g)=m_g\) is a homomorphism.

Conversely, given a homomorphism \(\rho\), define \(g\cdot s=\rho(g)(s)\). The homomorphism properties immediately verify the two action axioms.

The kernel of the permutation representation is \(\ker\rho=\{g\in G:g\cdot s=s\ \text{for every }s\in S\}=\bigcap_{s\in S}G_s\).

Corollary 18.2 (Cayley’s Theorem). Every finite group of order \(n\) is isomorphic to a subgroup of \(S_n\).

Proof. Let \(G\) act on itself by left multiplication, with \(S=G\) and \(m_g(x)=g\cdot x\). If \(g\in\ker\rho\), then \(m_g(1)=1\), so \(g\cdot 1=1\) and hence \(g=1\). Therefore \(\rho\) is injective, and \(G\cong\operatorname{Im}\rho\leq S_n\).

18.2 Finite Subgroups of the Rotation Group \(\operatorname{SO}(3)\)

Theorem 18.3. Every finite subgroup of \(\operatorname{SO}(3)\) is isomorphic to one of the following groups:

Proof outline. Let \(G\subseteq\operatorname{SO}(3)\) be finite of order \(N\). If \(N=1\), then \(G=C_1\). Assume \(N>1\). Every nonidentity rotation fixes exactly two points on the unit sphere \(S^2\), called its poles.

Let \(P\subseteq S^2\) be the set of all poles of all nonidentity elements of \(G\). The group \(G\) acts on \(P\): if \(p\) is a pole of \(g\), then \(x\cdot p\) is a pole of \(xgx^{-1}\).

Consider the incidence set \(\mathcal{I}=\{(g,p):1\ne g\in G,\ p\text{ is a pole of }g\}\).

Counting by rotations gives \(|\mathcal{I}|=2(|G|-1)=2(N-1)\), while counting by poles gives \(|\mathcal{I}|=\sum_{p\in P}(|G_p|-1)\).

Suppose that the action on \(P\) has orbits \(O_1,\ldots,O_k\). Choose \(p_i\in O_i\), and set \(r_i=|G_{p_i}|\). Orbit-stabilizer gives \(|O_i|=N/r_i\), so \(2-\frac{2}{N}=\sum_{i=1}^k\left(1-\frac{1}{r_i}\right)\).

Because every \(r_i\geq2\), the right-hand side shows that \(k\leq3\); the case \(k=1\) is impossible.

Case \(k=2\). The equation becomes \(\frac{2}{N}=\frac{1}{r_1}+\frac{1}{r_2}\).

Since \(r_i\leq N\), it follows that \(r_1=r_2=N\). All rotations share one axis, and \(G\cong C_N\).

Case \(k=3\). Assume \(2\leq r_1\leq r_2\leq r_3\). Then \(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=1+\frac{2}{N}\).

We must have \(r_1=2\) and \(r_2\leq3\). If \(r_2=2\), then \(N=2r_3\), yielding the dihedral group \(D_{r_3}\). If \(r_2=3\), then \(r_3<6\), and the remaining possibilities are \((r_1,r_2,r_3;N)=(2,3,3;12)\), \((2,3,4;24)\), and \((2,3,5;60)\), corresponding respectively to \(T\), \(O\), and \(I\).

19. Symmetric Polynomials and Discriminants

19.1 Symmetric Polynomials

The roots of a polynomial are closely related to its coefficients through symmetric polynomials.

Let \(R\) be a ring and consider \(R[u_1,\ldots,u_n]\). The symmetric group \(S_n\) acts on this ring by permuting variables. For \(\sigma\in S_n\) and \(P\in R[u_1,\ldots,u_n]\), define \(\sigma\bigl(P(u_1,\ldots,u_n)\bigr)=P(u_{\sigma(1)},\ldots,u_{\sigma(n)})\).

Definition 19.1. A polynomial \(P\in R[u_1,\ldots,u_n]\) is symmetric if \(\sigma(P)=P\) for every \(\sigma\in S_n\).

The set of symmetric polynomials is a subring of \(R[u_1,\ldots,u_n]\). The elementary symmetric polynomials are \(S_1=u_1+\cdots+u_n\), \(S_2=\sum_{i<j}u_i u_j\), \(S_3=\sum_{i<j<k}u_i u_j u_k\), \(\ldots\), and \(S_n=u_1u_2\cdots u_n\).

For the monic polynomial \(f(x)=(x-u_1)(x-u_2)\cdots(x-u_n)\), expansion gives \(f(x)=x^n-S_1x^{n-1}+S_2x^{n-2}+\cdots+(-1)^nS_n\).

Theorem 19.1 (Fundamental Theorem of Symmetric Polynomials). Every symmetric polynomial in \(R[u_1,\ldots,u_n]\) can be written uniquely as a polynomial in \(S_1,\ldots,S_n\) with coefficients in \(R\).

Proof. Order monomials lexicographically. Let \(P\) be a nonzero symmetric polynomial with leading term \(cu_1^{a_1}\cdots u_n^{a_n}\). Symmetry implies \(a_1\geq a_2\geq\cdots\geq a_n\). Consider \(g=cS_1^{a_1-a_2}S_2^{a_2-a_3}\cdots S_n^{a_n}\).

Its leading term is exactly \(cu_1^{a_1}\cdots u_n^{a_n}\). Hence \(P-g\) is symmetric and has strictly smaller leading term. Repeating this process terminates and proves existence.

For uniqueness, if two polynomials in \(S_1,\ldots,S_n\) represented the same symmetric polynomial, their difference would be nonzero but would have a maximal leading monomial that cannot cancel. This is impossible.

Remark. The elementary symmetric polynomials generate the invariant ring \(R[u_1,\ldots,u_n]^{S_n}\).

Corollary 19.2. Suppose \(f(x)=x^n-a_1x^{n-1}+\cdots\pm a_n\) has coefficients in a field \(F\) and splits in an extension \(K\) with roots \(\alpha_1,\ldots,\alpha_n\). If \(g\in F[u_1,\ldots,u_n]^{S_n}\), then \(g(\alpha_1,\ldots,\alpha_n)\in F\).

Proof. By Theorem 19.1, \(g=G(S_1,\ldots,S_n)\) for some polynomial \(G\) over \(F\). Evaluating at the roots expresses each \(S_i(\alpha_1,\ldots,\alpha_n)\) as a coefficient of \(f\), hence as an element of \(F\).

19.2 Discriminant

Let \(f(x)=(x-u_1)\cdots(x-u_n)\). The discriminant detects whether \(f\) has repeated roots.

Definition 19.2. Define \(\Delta=\prod_{1\leq i<j\leq n}(u_i-u_j)\) and \(D=\Delta^2\).

The element \(D\) is the discriminant of \(f\). A transposition sends \(\Delta\) to \(-\Delta\), so \(D\) is symmetric in the roots. By Corollary 19.2, it can be expressed in terms of the coefficients of \(f\).

Example 19.3 (Degree 2). For \(f(x)=x^2-S_1x+S_2\), we have \(D=(u_1-u_2)^2=S_1^2-4S_2\). If \(f(x)=ax^2+bx+c\), the normalized discriminant is \(D=\frac{b^2-4ac}{a^2}\).

Example 19.4 (Degree 3). For \(f(x)=x^3+px+q\), we have \(D=-4p^3-27q^2\).

20. Fixed Fields

20.1 The Fixed Field

Definition 20.1. Let \(K\) be a field, and let \(H\) be a group of automorphisms of \(K\). The fixed field of \(H\) is \(K^H=\{a\in K:\sigma(a)=a\ \text{for every }\sigma\in H\}\).

It is straightforward to check that \(K^H\) is a subfield of \(K\), and \(H\) is a subgroup of \(\operatorname{Gal}(K/K^H)\).

Example 20.1. Let \(K=\mathbb{C}\) and \(H=\{\operatorname{id},\sigma\}\), where \(\sigma(a+bi)=a-bi\) is complex conjugation. Then \(\mathbb{C}^H=\mathbb{R}\) and \([\mathbb{C}:\mathbb{R}]=2=|H|\).

20.2 Orbit Polynomial

Lemma 20.2. Let \(H\) be a finite group of automorphisms of a field \(K\), let \(F=K^H\), and let \(\{\beta_1,\ldots,\beta_r\}\) be the \(H\)-orbit of \(\beta_1\in K\).

  1. The minimal polynomial of \(\beta_1\) over \(F\) is \(g(x)=(x-\beta_1)\cdots(x-\beta_r)\).
  2. The element \(\beta_1\) is algebraic over \(F\), and its degree over \(F\) equals the size of its orbit. In particular, this degree divides \(|H|\).

Proof. The coefficients of \(g\) are elementary symmetric functions of the orbit. Since elements of \(H\) permute the orbit, these coefficients are fixed by \(H\) and therefore lie in \(F\).

If \(h(x)\in F[x]\) has \(\beta_1\) as a root, then applying elements of \(H\) shows that every \(\beta_i\) is a root of \(h\). Hence \(g\mid h\), so \(g\) is the minimal polynomial.

20.3 The Fixed Field Theorem

Lemma 20.3 (Characteristic-zero setting). Let \(K/F\) be an infinite algebraic extension with \(\operatorname{char}F=0\). Then \(K\) contains elements of arbitrarily large degree over \(F\).

Proof. Build a strictly increasing chain of finite intermediate extensions \(F\subsetneq F_1\subsetneq F_2\subsetneq\cdots\).

Since \(\operatorname{char}F=0\), every finite extension \(F_i/F\) is separable, so the Primitive Element Theorem gives \(F_i=F(\gamma_i)\). As the degrees \([F_i:F]\) are unbounded, so are the degrees of the elements \(\gamma_i\).

Theorem 20.4 (Fixed Field Theorem). Let \(H\) be a finite group of automorphisms of a field \(K\), and let \(F=K^H\). Then \(K/F\) is finite and \([K:F]=|H|\).

Proof in characteristic zero. Let \(n=|H|\). By Lemma 20.2, every element of \(K\) is algebraic over \(F\) with degree at most \(n\). Lemma 20.3 therefore implies that \(K/F\) is finite. By the Primitive Element Theorem, write \(K=F(\gamma)\). An element of \(H\) that fixes \(\gamma\) fixes all of \(K\), so the stabilizer of \(\gamma\) is trivial. Its orbit has size \(n\), and Lemma 20.2 gives \([F(\gamma):F]=n\). Thus \([K:F]=|H|\).

Note. The theorem remains true without the characteristic-zero assumption; the proof above records the characteristic-zero route used in these notes.

21. Extending Isomorphisms and Splitting Fields

21.1 Isomorphism Extension

Lemma 21.1 (Isomorphism Extension Lemma). Let \(\varphi:F\to F'\) be an isomorphism of fields, and let \(f(x)\in F[x]\) be irreducible. Let \(f'(x)=\varphi(f(x))\). Suppose \(\alpha\) is a root of \(f\) in \(K\), and \(\alpha'\) is a root of \(f'\) in \(K'\). Then there exists a unique isomorphism \(\Phi:F(\alpha)\to F'(\alpha')\) such that \(\Phi|_F=\varphi\) and \(\Phi(\alpha)=\alpha'\).

21.2 Uniqueness of Splitting Fields

Theorem 21.2 (Uniqueness of Splitting Fields). Let \(\varphi:F\to F'\) be an isomorphism. Let \(f(x)\in F[x]\) and \(f'(x)=\varphi(f(x))\). If \(K\) is a splitting field of \(f\) over \(F\), and \(K'\) is a splitting field of \(f'\) over \(F'\), then \(\varphi\) extends to an isomorphism \(\Phi:K\to K'\).

Corollary. For any \(f(x)\in F[x]\), the splitting field of \(f\) is unique up to isomorphism.

21.3 Counting Automorphisms

Theorem 21.3. Let \(f(x)\in F[x]\) be separable, and let \(K\) be a splitting field of \(f\) over \(F\). Then \(|\operatorname{Aut}(K/F)|=[K:F]\).

This equality is the defining property of a Galois extension.

Proof sketch. More generally, given an isomorphism \(\varphi:F\to F'\) and corresponding splitting fields \(K\) and \(K'\) of a separable polynomial, count the extensions \(\Phi:K\to K'\) of \(\varphi\) by induction on \([K:F]\).

If \([K:F]=1\), there is nothing to prove. Otherwise choose an irreducible factor \(p(x)\) of degree \(d>1\) and a root \(\alpha\in K\). Separability gives exactly \(d\) possible images for \(\alpha\). By Lemma 21.1, each choice extends \(\varphi\) uniquely to \(F(\alpha)\). The remaining degree is \([K:F(\alpha)]=\frac{[K:F]}{d}\).

The inductive hypothesis gives \([K:F(\alpha)]\) extensions for each of the \(d\) choices, for a total of \([K:F]\).

22. Galois Extensions

22.1 Definition and Basic Lemma

Definition 22.1. A finite extension \(K/F\) is a Galois extension if \(|\operatorname{Gal}(K/F)|=[K:F]\).

Lemma 22.1.

  1. The Galois group \(G=\operatorname{Gal}(K/F)\) of a finite field extension \(K/F\) is finite, and \(|G|\) divides \([K:F]\).
  2. If \(H\) is a finite group of automorphisms of a field \(K\), then \(K/K^H\) is Galois and \(H=\operatorname{Gal}(K/K^H)\).

Proof. For part 1, \(G\) fixes \(F\), so \(F\subseteq K^G\subseteq K\). By the Fixed Field Theorem, \(|G|=[K:K^G]\mid [K:F]\).

For part 2, the Fixed Field Theorem gives \([K:K^H]=|H|\). The inclusion \(H\leq\operatorname{Gal}(K/K^H)\) and part 1 force equality.

22.2 Characterization of Galois Extensions

Theorem 22.2. Let \(K/F\) be a finite extension, let \(G=\operatorname{Gal}(K/F)\), and let \(F'=K^G\). The following are equivalent:

  1. \(K/F\) is Galois, i.e. \(|G|=[K:F]\).
  2. \(F=K^G\).
  3. \(K\) is the splitting field over \(F\) of a separable polynomial \(f\in F[x]\).

Proof note. The equivalence of 1 and 2 follows from Lemma 22.1; the equivalence with 3 follows from Theorem 21.3.

For \((1)\Rightarrow(3)\), we pick a primitive element \(\gamma\) for \(K\) over \(F\). Let \(f\) be the minimal polynomial of \(\gamma\) over \(F\). Then the degree of \(f(x)\) is exactly \(n\).

Let \(\gamma_1,\ldots,\gamma_r\) be the distinct roots of \(f(x)\) that lie in \(K\). Any \(F\)-automorphism \(\sigma\in G\) is completely determined by its action on \(\sigma(\gamma)\).

Furthermore, for each \(\gamma_i\in K\), there is exactly one \(F\)-automorphism mapping \(\gamma\mapsto\gamma_i\). Thus, the order of \(G\) is exactly \(r\), the number of distinct roots. Therefore \(r=n\). Since \(f(x)\) has degree \(n\), \(r=n\) iff \(f(x)\) splits completely into distinct linear factors in \(K\).

22.3 Consequences and Example

Corollary 22.3.

  1. In characteristic zero, every finite extension \(K/F\) is contained in a Galois extension of \(F\).
  2. If \(K/F\) is Galois and \(F\subseteq L\subseteq K\), then \(K/L\) is Galois, and \(\operatorname{Gal}(K/L)\) is a subgroup of \(\operatorname{Gal}(K/F)\).

Proof idea. For part 1, let \(\gamma\) be a primitive element for \(K/F\), let \(f\) be its minimal polynomial, and take the splitting field \(K'\) of \(f\). Since \(f\) is separable in characteristic zero, \(K'/F\) is Galois and \(K'\supseteq K\).

For part 2, if \(K\) is the splitting field over \(F\) of a separable polynomial \(f\), then \(f\in L[x]\) and \(K\) remains its splitting field over \(L\).

Example. Let \(K=\mathbb{Q}(\sqrt2,\sqrt3)\) over \(F=\mathbb{Q}\). Then \(K\) is the splitting field of \(f(x)=(x^2-2)(x^2-3)\), so \(K/F\) is Galois. Moreover, \(|\operatorname{Gal}(K/\mathbb{Q})|=4\) and \(\operatorname{Gal}(K/\mathbb{Q})\cong C_2\times C_2\). Its automorphisms are determined by the independent sign choices \(\sqrt2\mapsto \pm\sqrt2\) and \(\sqrt3\mapsto \pm\sqrt3\).

23. Fundamental Theorem of Galois Theory

Let \(K/F\) be a Galois extension with Galois group \(G=\operatorname{Gal}(K/F)\).

For any subgroup \(H\leq G\), associate the intermediate field \(K^H\).

For any intermediate field \(L\), where \(F\subseteq L\subseteq K\), associate the subgroup \(\operatorname{Gal}(K/L)\leq G\).

Theorem 23.1. The Galois correspondence says the maps \(H\mapsto K^H\) and \(L\mapsto \operatorname{Gal}(K/L)\) are mutually inverse, inclusion-reversing bijections between the set of subgroups of \(G\) and the set of intermediate fields of \(K/F\). In particular, \(H'\leq H\) iff \(K^H\subseteq K^{H'}\).

For any subgroup \(H\) and its associated intermediate field \(L=K^H\), we have \([K:L]=|H|\) and \([L:F]=[G:H]\).

An intermediate field \(L\) is a Galois extension of \(F\) iff its associated subgroup \(H=\operatorname{Gal}(K/L)\) is a normal subgroup of \(G\). In this case, the Galois group of \(L/F\) is isomorphic to the quotient group: \(\operatorname{Gal}(L/F)\cong G/H\).

Proof notes. Let \(\Phi(H)=K^H\) and \(\Psi(L)=\operatorname{Gal}(K/L)\), where \(H\leq G\) and \(F\subseteq L\subseteq K\).

By Corollary 22.3(b), the extension \(K/L\) is Galois. Hence Theorem 22.2 gives \(K^{\operatorname{Gal}(K/L)}=L\).

Conversely, let \(H\leq G\). Since \(H\) is a finite group of automorphisms of \(K\), Lemma 22.1(b) gives \(\operatorname{Gal}(K/K^H)=H\).

Thus \(\Phi\) and \(\Psi\) are mutually inverse bijections. For inclusion reversal, this is an easy exercise.

For degrees, let \(L=K^H\). By the Fixed Field Theorem, \([K:L]=[K:K^H]=|H|\). The tower law and the Galois hypothesis \([K:F]=|G|\) then give \([L:F]=[K:F]/[K:L]=|G|/|H|=[G:H]\).

For normality, let \(L\) be an intermediate field and set \(H=\operatorname{Gal}(K/L)\). For \(\sigma\in G\), compute the subgroup corresponding to \(\sigma(L)\): \(\operatorname{Gal}(K/\sigma(L))=\sigma H\sigma^{-1}\). Thus \(\sigma(L)=L\) for every \(\sigma\in G\) iff \(\sigma H\sigma^{-1}=H\) for every \(\sigma\in G\). In other words, \(H\triangleleft G\) iff every element of \(G\) preserves \(L\).

Assume first that \(H\triangleleft G\). Then every \(\sigma\in G\) restricts to an \(F\)-automorphism of \(L\), so restriction defines a homomorphism \(\rho:G\to\operatorname{Gal}(L/F)\) by \(\rho(\sigma)=\sigma|_L\). Its kernel is exactly \(\operatorname{Gal}(K/L)=H\). The image satisfies \(\operatorname{Im}\rho\leq \operatorname{Gal}(L/F)\).

Claim: \(\operatorname{Im}\rho=\operatorname{Gal}(L/F)\). By part 2, \(|G/H|=[G:H]=[L:F]\). Since Lemma 22.1(a) gives \(|\operatorname{Gal}(L/F)|\leq [L:F]\), and by the First Isomorphism Theorem \(G/H\cong\operatorname{Im}\rho\), we get \(|\operatorname{Gal}(L/F)|\geq [L:F]\). Hence \(\operatorname{Gal}(L/F)\cong G/H\).

Conversely, assume that \(L/F\) is Galois. By Theorem 22.2, \(L\) is the splitting field over \(F\) of some separable polynomial \(f(x)\in F[x]\). Every \(\sigma\in G\) fixes \(F\), so it sends roots of \(f\) in \(K\) to roots of \(f\) in \(K\). Since all roots of \(f\) already lie in \(L\) and generate \(L\) over \(F\), we get \(\sigma(L)=L\) for every \(\sigma\in G\). By the previous lemma, \(H\triangleleft G\).

24. Goal of the Examples

Guiding principle. The Fundamental Theorem of Galois Theory is a dictionary: \(H\leq \operatorname{Gal}(K/F)\leftrightarrow K^H\), with inclusions reversed.

In both examples we follow the same four steps:

  1. construct the splitting field \(K\);
  2. compute \([K:F]\);
  3. describe \(G=\operatorname{Gal}(K/F)\) by generators;
  4. identify fixed fields using fixed elements and degree checks.

The Fixed-Field Test

Suppose \(K/F\) is finite Galois and \(H\leq G=\operatorname{Gal}(K/F)\).

Degree formula. \([K^H:F]=[G:H]=|G|/|H|\).

How we identify \(K^H\) in practice. If \(H\) fixes an element \(\beta\), then \(F(\beta)\subseteq K^H\). If the two fields have the same degree over \(F\), then \(F(\beta)=K^H\).

Example 1: The Splitting Field of \(x^3-2\)

Let \(f(x)=x^3-2\), \(\alpha=\sqrt[3]{2}\), and \(\omega=e^{2\pi i/3}\).

The roots are \(\alpha\), \(\alpha\omega\), and \(\alpha\omega^2\).

So the splitting field is \(K=\mathbb{Q}(\alpha,\omega)\).

Degree computation. \([\mathbb{Q}(\alpha):\mathbb{Q}]=3\), \([K:\mathbb{Q}(\alpha)]=2\), and \([K:\mathbb{Q}]=6\).

The Galois Group of \(x^3-2\)

Every automorphism is determined by the images of \(\alpha\) and \(\omega\): \(\alpha\mapsto \alpha\omega^a\) for \(a=0,1,2\), and \(\omega\mapsto \omega^{\pm1}\).

Define \(\sigma(\alpha)=\alpha\omega\), \(\sigma(\omega)=\omega\), \(\tau(\alpha)=\alpha\), and \(\tau(\omega)=\omega^2\).

Then \(\sigma^3=1\), \(\tau^2=1\), and \(\tau\sigma\tau^{-1}=\sigma^{-1}\).

Conclusion. \(\operatorname{Gal}(K/\mathbb{Q})=\langle \sigma,\tau\rangle\cong S_3\).

Subgroups and Fixed Fields for \(S_3\)

subgroup \(H\) \(\lvert H\rvert\) \([K^H:\mathbb{Q}]\) fixed field \(K^H\)
\(S_3\) 6 1 \(\mathbb{Q}\)
\(A_3=\langle\sigma\rangle\) 3 2 \(\mathbb{Q}(\omega)\)
\(\langle\tau\rangle\) 2 3 \(\mathbb{Q}(\alpha)\)
\(\langle\sigma\tau\rangle\) 2 3 \(\mathbb{Q}(\alpha\omega^2)\)
\(\langle\sigma^2\tau\rangle\) 2 3 \(\mathbb{Q}(\alpha\omega)\)
\(\{1\}\) 1 6 \(K=\mathbb{Q}(\alpha,\omega)\)

Each entry is found by fixed element plus degree formula.

A Representative Fixed-Field Check

For \(H=\langle\sigma\tau\rangle\), observe that \((\sigma\tau)(\alpha\omega^2)=\sigma(\alpha\omega)=\alpha\omega^2\).

Thus \(\mathbb{Q}(\alpha\omega^2)\subseteq K^{\langle\sigma\tau\rangle}\).

But \(\alpha\omega^2\) is a root of \(x^3-2\), so \([\mathbb{Q}(\alpha\omega^2):\mathbb{Q}]=3\).

Also \([K^{\langle\sigma\tau\rangle}:\mathbb{Q}]=[S_3:\langle\sigma\tau\rangle]=3\).

Therefore \(K^{\langle\sigma\tau\rangle}=\mathbb{Q}(\alpha\omega^2)\).

Normality in the Cubic Example

Fundamental theorem: normality part. \(H\triangleleft G\) iff \(K^H/F\) is Galois.

For \(G=S_3\), the normal subgroups are \(\{1\}\), \(A_3\), and \(S_3\).

Therefore the Galois intermediate extensions are \(K/\mathbb{Q}\), \(\mathbb{Q}(\omega)/\mathbb{Q}\), and \(\mathbb{Q}/\mathbb{Q}\).

The three cubic fields \(\mathbb{Q}(\alpha)\), \(\mathbb{Q}(\alpha\omega)\), and \(\mathbb{Q}(\alpha\omega^2)\) are not Galois over \(\mathbb{Q}\).

Example 2: The Splitting Field of \(x^4-2\)

Let \(f(x)=x^4-2\) and \(\alpha=\sqrt[4]{2}\).

The roots are \(\alpha\), \(-\alpha\), \(i\alpha\), and \(-i\alpha\).

Hence \(K=\mathbb{Q}(\alpha,i)\).

Degree computation. \([\mathbb{Q}(\alpha):\mathbb{Q}]=4\), \([K:\mathbb{Q}(\alpha)]=2\), and \([K:\mathbb{Q}]=8\).

The Galois Group of \(x^4-2\)

Define automorphisms by \(r(\alpha)=i\alpha\), \(r(i)=i\), \(s(\alpha)=\alpha\), and \(s(i)=-i\).

Then \(r^4=1\), \(s^2=1\), and \(srs=r^{-1}\).

Conclusion. \(\operatorname{Gal}(K/\mathbb{Q})=\langle r,s\mid r^4=s^2=1,\ srs=r^{-1}\rangle\cong D_4\).

Quadratic Fixed Fields in the Quartic Example

The subgroups of order 4 have quadratic fixed fields.

subgroup \(H\) fixed element fixed field \(K^H\)
\(\langle r\rangle\) \(i\) \(\mathbb{Q}(i)\)
\(\langle r^2,s\rangle\) \(\alpha^2=\sqrt2\) \(\mathbb{Q}(\sqrt2)\)
\(\langle r^2,rs\rangle\) \(i\alpha^2=i\sqrt2\) \(\mathbb{Q}(i\sqrt2)\)

Why the degree check works. Each subgroup has order 4, so \([K^H:\mathbb{Q}]=[D_4:H]=2\). Each field listed on the right is quadratic over \(\mathbb{Q}\).

Quartic Fixed Fields

The five subgroups of order 2 have fixed fields of degree 4.

subgroup \(H\) fixed element(s) fixed field \(K^H\)
\(\langle r^2\rangle\) \(\sqrt2,\ i\) \(\mathbb{Q}(\sqrt2,i)\)
\(\langle s\rangle\) \(\alpha\) \(\mathbb{Q}(\alpha)\)
\(\langle r^2s\rangle\) \(i\alpha\) \(\mathbb{Q}(i\alpha)\)
\(\langle rs\rangle\) \((1+i)\alpha\) \(\mathbb{Q}((1+i)\alpha)\)
\(\langle r^3s\rangle\) \((1-i)\alpha\) \(\mathbb{Q}((1-i)\alpha)\)

The first field is generated by two quadratic elements; the remaining four are generated by one quartic element.

A Representative Check in the \(D_4\) Case

For \(H=\langle rs\rangle\), compute \((rs)((1+i)\alpha)=r((1-i)\alpha)=(1-i)i\alpha=(1+i)\alpha\).

Thus \(\mathbb{Q}((1+i)\alpha)\subseteq K^{\langle rs\rangle}\).

Also \([K^{\langle rs\rangle}:\mathbb{Q}]=[D_4:\langle rs\rangle]=4\).

Since \(((1+i)\alpha)^4=-8\) and \(x^4+8\) is irreducible over \(\mathbb{Q}\), we have \([\mathbb{Q}((1+i)\alpha):\mathbb{Q}]=4\).

Therefore \(K^{\langle rs\rangle}=\mathbb{Q}((1+i)\alpha)\).

Normality in the Quartic Example

Normal subgroups give Galois subextensions. \(H\triangleleft D_4\) iff \(K^H/\mathbb{Q}\) is Galois.

Important normal subgroups and fields:

normal subgroup Galois fixed field
\(D_4\) \(\mathbb{Q}\)
\(\langle r\rangle\) \(\mathbb{Q}(i)\)
\(\langle r^2,s\rangle\) \(\mathbb{Q}(\sqrt2)\)
\(\langle r^2,rs\rangle\) \(\mathbb{Q}(i\sqrt2)\)
\(\langle r^2\rangle\) \(\mathbb{Q}(\sqrt2,i)\)
\(\{1\}\) \(K\)

The reflection subgroups of order 2 are not normal, so their quartic fixed fields are not Galois over \(\mathbb{Q}\).

Comparison of the Two Examples

\(x^3-2\) \(x^4-2\)
splitting field \(\mathbb{Q}(\sqrt[3]{2},\omega)\) \(\mathbb{Q}(\sqrt[4]{2},i)\)
degree 6 8
Galois group \(S_3\) \(D_4\)
main generators \(\sigma,\tau\) \(r,s\)
non-Galois fields three cubic fields four reflection-fixed quartic fields
Galois test normal subgroups of \(S_3\) normal subgroups of \(D_4\)

Common method. Compute \(G\), list subgroups, then translate subgroup data into field data.

Takeaways

  1. The Galois correspondence is practical: fixed fields can be found by fixed elements plus degree checks.
  2. For \(x^3-2\), the splitting field has group \(S_3\).
  3. For \(x^4-2\), the splitting field has group \(D_4\).
  4. Normal subgroups exactly identify the Galois intermediate extensions.
  5. These examples show how group theory organizes the entire lattice of intermediate fields.

25. Cubic and Quartic Equations

Goal. Understand the Galois group of an irreducible polynomial of degree 3 or 4.

25.1 Cubic Equations

Let \(F\) be a field with \(\operatorname{char}(F)\neq 2,3\). A general cubic \(x^3+a_2x^2+a_1x+a_0\) can be reduced, by the translation \(x=y-a_2/3\), to a depressed cubic \(f(y)=y^3+py+q\).

Assume first that \(f\) is irreducible over \(F\), and let its roots in the splitting field \(K\) be \(\alpha_1,\alpha_2,\alpha_3\). Then \(G=\operatorname{Gal}(K/F)\) acts transitively on the three roots, so \(G\leq S_3\) is a transitive subgroup. Hence, there are only two possibilities: \(G\cong A_3\) or \(G\cong S_3\).

The discriminant decides between them. Define \(\delta=(\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_2-\alpha_3)\) and \(D=\delta^2\).

Even permutations fix \(\delta\), while odd permutations send \(\delta\mapsto -\delta\). Therefore \(G\leq A_3\) iff \(\delta\in F\).

Since \(D=\delta^2\), this is equivalent to \(D\) being a square in \(F\).

For the depressed cubic one computes \(D=-4p^3-27q^2\).

Theorem 25.1. Let \(f(x)=x^3+px+q\in F[x]\) be irreducible, and let \(K\) be its splitting field. Then \(\operatorname{Gal}(K/F)=A_3\) if \(D\) is a square in \(F\), and \(\operatorname{Gal}(K/F)=S_3\) otherwise.

When \(G\cong S_3\), the subgroup \(A_3\triangleleft S_3\) gives the unique quadratic subfield \(K^{A_3}=F(\delta)=F(\sqrt D)\).

Examples. For \(x^3+3x+1\), \(D=-135\). For \(x^3-3x+1\), \(D=81\).

25.2 Quartic Equations

Still assume the quartic \(f\) is irreducible and \(\operatorname{char}F\neq 2,3\).

After a translation, \(f\) also has a depressed form \(f(x)=x^4+px^2+qx+r\).

Let its roots be \(\alpha_1,\alpha_2,\alpha_3,\alpha_4\), and let \(K\) be the splitting field.

The Galois group acts transitively on \(\alpha_1,\alpha_2,\alpha_3,\alpha_4\), so it is a transitive subgroup of \(S_4\). Up to conjugacy, the possibilities are \(S_4\), \(A_4\), \(D_4\), \(C_4\), and \(V_4=D_2\).

Here \(V_4=\{1,(12)(34),(13)(24),(14)(23)\}\).

25.3 The Resolvent Cubic

Group the four roots into their pairings: \(\beta_1=\alpha_1\alpha_2+\alpha_3\alpha_4\), \(\beta_2=\alpha_1\alpha_3+\alpha_2\alpha_4\), and \(\beta_3=\alpha_1\alpha_4+\alpha_2\alpha_3\).

Every permutation of the roots permutes the three numbers \(\beta_1,\beta_2,\beta_3\).

Moreover, the permutations that fix all three \(\beta_i\) are exactly the elements of \(V_4\). Therefore the action on the \(\beta_i\) gives a homomorphism \(G\to S_3\) with kernel \(G\cap V_4\).

The polynomial \(g(x)=(x-\beta_1)(x-\beta_2)(x-\beta_3)\) has coefficients in \(F\) because its coefficients are symmetric functions in the \(\alpha_i\).

One can verify the resolvent cubic is \(g(x)=x^3-px^2-4rx+(4pr-q^2)\).

Let \(L=F(\beta_1,\beta_2,\beta_3)\) be the splitting field of \(g(x)\).

By the Galois correspondence and the kernel computation, \(\operatorname{Gal}(L/F)\cong G/(G\cap V_4)\).

25.4 A Quartic Galois Group Test

The behavior of the resolvent cubic \(g(x)\) determines the following possibilities for the Galois group \(G\) of an irreducible quartic:

behavior of the resolvent cubic \(g(x)\) \(\operatorname{Gal}(L/F)\) \(G\leq S_4\)
irreducible, discriminant non-square \(S_3\) \(S_4\)
irreducible, discriminant square \(A_3\) \(A_4\)
splits completely over \(F\) \(1\) \(V_4\)
one root in \(F\) and an irreducible quadratic factor \(C_2\) \(D_4\) or \(C_4\)

One standard additional test is to look over the discriminant field \(F(\sqrt{\Delta_f})\):

26. Roots of Unity and Kummer Extension

26.1 Roots of Unity

Let \(\zeta_n\) be a primitive \(n\)-th root of unity, and let \(K=\mathbb{Q}(\zeta_n)\). This is the splitting field of \(x^n-1\) over \(\mathbb{Q}\).

Let \(\Phi_n\) be the cyclotomic polynomial. It is an irreducible factor of \(x^n-1\) of degree \(\varphi(n)\). By the irreducibility of \(\Phi_n(x)\), \([K:\mathbb{Q}]=\deg\Phi_n=\varphi(n)=|(\mathbb{Z}/n\mathbb{Z})^\times|\).

An automorphism \(\sigma\in\operatorname{Gal}(K/\mathbb{Q})\) is determined by \(\sigma(\zeta_n)=\zeta_n^a\). Since automorphisms preserve multiplicative order, \(\sigma(\zeta_n)\) must be a primitive \(n\)-th root of unity. Hence \(\sigma(\zeta_n)=\zeta_n^a\) for some \(a\) relatively prime to \(n\).

This gives an injective homomorphism \(\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\hookrightarrow(\mathbb{Z}/n\mathbb{Z})^\times\), defined by \(\sigma\mapsto a\pmod n\).

Both groups have order \(\varphi(n)\), so the map is an isomorphism.

Corollary 26.1. \(\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^\times\).

In particular, cyclotomic extensions of \(\mathbb{Q}\) are Abelian Galois extensions.

Remark 26.1. The above fact is only one direction of a much deeper story. The Kronecker–Weber theorem says that every finite Abelian Galois extension of \(\mathbb{Q}\) is contained in some cyclotomic extension.

26.2 Kummer Extension

Let \(F\) be a field of characteristic zero and suppose that \(F\) already contains a primitive \(p\)-th root of unity \(\zeta_p\), where \(p\) is prime. For \(a\in F^\times\), an extension of the form \(F(\sqrt[p]{a})/F\) is called a Kummer extension.

Since all solutions of \(x^p-a\) are of the form \(\sqrt[p]{a}\zeta_p^i\), and \(\zeta_p\in F\), the field \(F(\sqrt[p]{a})\) is a splitting field of \(x^p-a\).

Theorem 26.2.

  1. If \(a\in F^\times\) and \(\alpha^p=a\), then \(F(\alpha)\) is the splitting field of \(x^p-a\) over \(F\), and \(\operatorname{Gal}(F(\alpha)/F)\) is cyclic of order dividing \(p\).
  2. Conversely, if \(K/F\) is a cyclic Galois extension of degree \(p\), then there exists \(a\in F^\times\) such that \(K=F(\sqrt[p]{a})\).

Proof.

For part 1, we have seen that \(F(\alpha)\) is the splitting field of \(x^p-a\). Let \(G=\operatorname{Gal}(F(\alpha)/F)\). If \(\tau\in G\), then \(\tau(\alpha)\) is another root of \(x^p-a\).

Therefore \(\chi:G\to\langle\zeta_p\rangle\), defined by \(\tau\mapsto\tau(\alpha)/\alpha\), is an injective homomorphism. Thus \(G\) is isomorphic to a subgroup of the cyclic group \(\langle\zeta_p\rangle\), so \(G\) is cyclic and its order divides \(p\).

For part 2, suppose conversely that \(K/F\) is cyclic of degree \(p\), and write \(G=\langle\sigma\rangle\). Regard \(K\) as an \(F\)-vector space, and regard \(\sigma\) as an \(F\)-linear operator. Since \(\sigma^p=\operatorname{id}\), the operator \(\sigma\) is annihilated by \(x^p-1\).

This polynomial splits over \(F\) into distinct linear factors, so \(\sigma\) is diagonalizable and its eigenvalues are \(p\)-th roots of unity. Because \(\sigma\neq\operatorname{id}\) and \(p\) is prime, at least one eigenvalue is \(\zeta_p^m\) with \(m\not\equiv0\pmod p\). Replacing the generator \(\sigma\) by a suitable power, we may choose a nonzero \(v\in K\) such that \(\sigma(v)=\zeta_pv\).

Then \(\sigma(v^p)=(\sigma(v))^p=\zeta_p^pv^p=v^p\), so \(v^p\) is fixed by \(G\). Since \(K^G=F\), we have \(v^p=a\in F\). Moreover, \(v\notin F\), and its Galois orbit is \(v,\zeta_pv,\zeta_p^2v,\ldots,\zeta_p^{p-1}v\).

Thus \([F(v):F]=p\). Since \([K:F]=p\), it follows that \(K=F(v)=F(\sqrt[p]{a})\).

Remark 26.2 (due to Lagrange). The eigenvector \(v\) in the proof of Theorem 26.2 can be constructed explicitly. For \(\alpha\in K\), define \((\alpha,\zeta_p):=\alpha+\zeta_p^{-1}\sigma(\alpha)+\zeta_p^{-2}\sigma^2(\alpha)+\cdots+\zeta_p^{-(p-1)}\sigma^{p-1}(\alpha)\). Then \(\sigma((\alpha,\zeta_p))=\zeta_p(\alpha,\zeta_p)\).

Proposition 26.3. Let \(F\) be a field of characteristic zero containing a primitive \(p\)-th root of unity \(\zeta_p\), with \(p\) prime. For \(b\in F^\times\), the polynomial \(x^p-b\) is either irreducible over \(F\), or it splits completely over \(F\).

Proof. Let \(\beta\) be a root of \(x^p-b\). If \(\beta\in F\), then all roots \(\beta,\zeta_p\beta,\ldots,\zeta_p^{p-1}\beta\) belong to \(F\), so the polynomial splits completely over \(F\).

Suppose now that \(\beta\notin F\). Since \(\zeta_p\in F\), the field \(K=F(\beta)\) already contains all roots of \(x^p-b\), so \(K/F\) is the splitting field. Any nontrivial automorphism \(\tau\in\operatorname{Gal}(K/F)\) sends \(\beta\) to another root, say \(\tau(\beta)=\zeta_p^m\beta\) for \(m\not\equiv0\pmod p\).

The powers of \(\tau\) send \(\beta\) to \(\beta,\zeta_p^m\beta,\ldots,\zeta_p^{(p-1)m}\beta\). Because \(p\) is prime and \(m\not\equiv0\pmod p\), these are distinct roots. Hence the Galois orbit of \(\beta\) is all of the roots of \(x^p-b\). By Lemma 20.2, \(x^p-b\) is the minimal polynomial of \(\beta\) and is irreducible over \(F\).

Kummer theory applies cleanly to prime-degree radicals. The next lemma says that this is not a serious restriction.

Lemma 26.4. Let \(F\) be a field of characteristic zero and suppose \(E=F(\alpha)\) with \(\alpha^n=a\in F\). Then \(E/F\) can be obtained through a tower of prime radical adjunctions.

Proof. Write \(n=p_1p_2\cdots p_r\) as a product of primes, with repetition allowed. Define \(\alpha_i=\alpha^{p_{i+1}\cdots p_r}\) for \(1\leq i\leq r\), and set \(\alpha_0=a\). We have \(\alpha_i^{p_i}=\alpha_{i-1}\) for each \(i\). Therefore \(F\subseteq F(\alpha_1)\subseteq F(\alpha_2)\subseteq\cdots\subseteq F(\alpha_r)=E\) is a tower in which each step adjoins a \(p_i\)-th root.

Remark 26.3. A radical adjunction need not be Galois over the base field. Kummer theory says that, after the necessary roots of unity have been adjoined, each individual prime radical adjunction becomes a cyclic Galois extension of prime degree. Thus a tower of radical adjunctions becomes, after adjoining roots of unity and passing to a suitable normal closure, a tower whose successive Galois groups are Abelian.

27. Solvability

Definition 27.1. Let \(f(x)\in F[x]\), and let \(K\) be its splitting field over \(F\). We say that \(f(x)\) is solvable by radicals if there exists a field \(E\supseteq K\) and a tower \(F=E_0\subseteq E_1\subseteq E_2\subseteq\cdots\subseteq E_r=E\) such that each step has the form \(E_{i+1}=E_i(\sqrt[n_i]{a_i})\) for some \(a_i\in E_i\).

To state the Galois-theoretic criterion, we need one group-theoretic definition.

Definition 27.2. A finite group \(G\) is called solvable if there is a chain of subgroups \(\{1\}=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_r=G\) such that each quotient \(G_i/G_{i-1}\) is Abelian.

(()) () refinement (H_0H_1H_s=G) s.t. (H_i/H_{i-1}) is cyclic.

Examples.

The group (A_5) is simple, meaning that it has no nontrivial normal subgroups. Also, (A_5S_5).

Theorem 27.1. (Galois criterion for radicals) Let (F) be a field of char (=0), let (f(x)F[x]), and let (K) be the splitting field of (f(x)) over (F). Then (f(x)) is solvable by radicals iff. ((K/F)) is a solvable gp.

Pf. (Sketch) “()” follows from Theorem 26.2(1) and Corollary 26.1. “()” follows from. Theorem 26.2.(2).

Ex 27.2. (f(x)=x^5-6x+3Q[x]).

  1. Eisenstein’s criterion (f(x)) irred. (G) transitive () five-cycle in (G).

  2. “Complex conjugates gives a transposition”. (f’(x)=5x^4-6). It has two real critical pts at (x=(6/5)^{1/4}). (f(x)) has three real roots. (f(x)) has two conjugate complex roots.

  3. 5-cycle + transposition generates the full (S_5).

Rmk 27.1

  1. If a deg (n) polynomial (f(x)) is “generic enough”, then the Galois group of (f(x)) is (S_n).

  2. “Insolvability for polynomial; if deg ()” refers to generic polynomials. Some special polynomials can be solved by radicals.