A Refresher in Galois Theory

21 Jan 2019

• algebraic number theory

In this post, we review the basic results in Galois theory.

Elementary field theory

Let $F$ be a field, and consider the additive subgroup $G$ of $F$ generated by $1_F$. This is the image of the ring homomorphism $\varphi:\bb Z\to F$ sending $1$ to $1_F$. Since $F$ has no zero divisors, there are only two possibilities:

1. $G\cong\bb Z$. Then we can check that $\varphi$ extends to an injective field homomorphism $\tilde\varphi:\bb Q\to F$ given by

   $$\tilde\varphi\left(\frac mn\right)=\frac{m1_F}{n1_F},$$

   so $F$ contains a copy of $\bb Q$. In this case we say that $F$ has characteristic zero.

2. $G\cong\bb Z/p\bb Z$ for some prime $p$. Then $\varphi$ factors into a ring isomorphism $\bb Z/p\bb Z\to G$, so $F$ contains a copy of $\bb F_p$, and we say that $F$ has characteristic $p$. Note that in this case, $px=0$ for every $x\in F$.

If $F,K$ are fields with $F\subseteq K$, we say that $K$ is an extension field (or extension) of $F$. In this case we also say that $K/F$ is an extension. Then $K$ is also a vector space over $F$; the dimension of this vector space is called the degree of the extension $K/F$, written as $[K:F]$. We say that the extension $K/F$ is finite if its degree is finite.

Proposition: If $F\subseteq K\subseteq L$ are fields, then $[L:F]=[L:K][K:F]$. In particular, if $L/K$ and $K/F$ are finite extensions, then so is $L/F$. $\qed$

Algebraic field extensions

Let $F$ be a field, and let $\alpha$ be an algebraic number over $F$, ie. an element of some field extension of $F$ which is a root of some nonzero polynomial over $F$. Then the polynomials over $F$ with $\alpha$ as a root form an ideal $I\subseteq F[X]$, which is easily seen to be prime, hence maximal (since $F[X]$ is a PID, see previous post). Note that there is a unique monic polynomial $f\in F[X]$, called the minimal polynomial of $\alpha$, such that $I=(f)$. Then $f$ is irreducible in $F[X]$, and it is a nonzero polynomial of minimal degree in $I$.

Consider the evaluation homomorphism $F[X]\to F[\alpha]$ that maps each polynomial to its value at $\alpha$. This map has kernel $I$ as defined above, so

$$F[\alpha]\cong F[X]/I.$$

Since $I$ is a maximal ideal, we immediately have:

Proposition: For any $\alpha$ algebraic over $F$, the ring $F[\alpha]$ is a field. $\qed$

Let $n$ be the degree of the minimal polynomial $f$ of $\alpha$. By minimality of $f$, the elements

$$1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$$

are linearly independent over $F$. Moreover, by the division algorithm, every element of $F[\alpha]$ is a $F$-linear combination of the above set. Hence $F[\alpha]$ is an $F$-vector space of dimension $n$.

Separable field extensions

Conversely, for any irreducible polynomial $f$ over $F$, we can construct the field $L=F[T]/(f)$, in which $F$ can be embedded as the image of constant polynomials in $F[T]$. Now if we write $\alpha=T+(f)\in L$, then $L=F[\alpha]$ and $f(\alpha)=0$. Hence there is always a finite extension of $F$ in which $f(X)$ is the product of a linear factor $X-\alpha$, and other irreducible factors of lower degree. By iteratively constructing field extensions in the same manner, we have:

Proposition: For any irreducible polynomial $f\in F[X]$, there is a finite extension $K/F$ such that $f$ splits (ie. is a product of linear factors) over $K$. $\qed$

Any field over which $f$ splits is called a splitting field for $f$. Note that every nonconstant polynomial over $F$, being a product of irreducible polynomials, has a splitting field.

A polynomial over $F$ is said to have a multiple root if it has a repeated linear factor over some splitting field. A polynomial with only simple roots (no multiple roots) over any splitting field is called separable.

We can detect whether a polynomial is separable by computing its formal derivative: the formal derivative of $f(X)=\sum a_nX^n$ is

$$f'(X)=\sum na_nX^{n-1}.$$

It is easy to check that the formal derivative satisfies the product rule

$$(fg)'=f'g+fg',$$

which implies the following:

Proposition: A nonconstant polynomial $f\in F[X]$ is separable if and only if $(f,f’)=(1)$.

Proof: Take an extension of $F$ where $f$ and $f’$ both split. For any linear factor $X-\alpha$ of $f$, write $f(X)=(X-\alpha)g(X)$. Then

$$f'(X)=g(X)+(X-\alpha)g'(X),$$

so $X-\alpha$ divides $f’(X)$ if and only if $(X-\alpha)^2$ divides $f(X)$. $\qed$

Field embeddings in $\bb C$

From now on, we consider subfields $K$ of $\bb C$. By the Fundamental Theorem of Algebra, $\bb C$ is a splitting field for any polynomial over $K$.

Corollary: Every irreducible polynomial $f\in K[X]$ is separable.

Proof: Since $(f)$ is a maximal ideal and $\deg f’=\deg f-1$, $f’\not\in(f)$ implies $(f,f’)=(1)$. $\qed$

Hence every irreducible $f\in K[X]$ has $\deg f$ many distinct roots in $\bb C$.

Let $L\subseteq\bb C$ be a finite extension of $K$. Then for any $\alpha\in L$, the elements

$$1,\alpha,\alpha^2,\ldots,\alpha^{[L:K]}$$

are linearly dependent over $K$; hence $\alpha$ is an algebraic number over $K$, so it has a monic irreducible minimal polynomial $f\in K[X]$. The roots of $f$ in $\bb C$ are called the conjugates of $\alpha$ over $K$.

We are interested in the different embeddings (injective field homomorphisms) of $K$ and $L$ into $\bb C$. It turns out that the structure of these embeddings are quite rigidly determined.

Proposition: Every embedding of $K$ in $\bb C$ extends to exactly $[L:K]$ many embeddings of $L$ in $\bb C$.

Proof: We proceed by induction on $[L:K]$. If $[L:K]=1$ then $L=K$, and there is nothing to show.

Otherwise, take any $\alpha\in L\backslash K$, and let $f$ be its minimal polynomial over $K$. For any embedding $\sigma:K\to\bb C$, note that the isomorphism $\sigma:K\to\sigma K$ extends to an isomorphism $\tilde\sigma:K[X]\to\sigma K[X]$. Hence for any root $\beta$ of $\tilde\sigma f$, we have

$$K[\alpha]\cong K[X]/(f)\cong\sigma K[X]/(\tilde\sigma f)\cong\sigma K[\beta].$$

Thus $\sigma$ can be extended to an embedding of $K[\alpha]$ in $\bb C$ sending $\alpha$ to $\beta$. Conversely, any embedding of $K[\alpha]$ extending $\sigma$ sends $\alpha$ to a root of $\tilde\sigma f$, and is determined by this choice of root. Since there are $\deg\tilde\sigma f=\deg f$ choices for $\beta$, there are exactly $\deg f$ extensions of $\sigma$ to $K[\alpha]$. By induction hypothesis, each of these embeddings of $K[\alpha]$ extends to $[L:K[\alpha]]$ embeddings of $L$, so the total number of embeddings of $L$ extending $\sigma$ is

$$[L:K[\alpha]]\deg f=[L:K[\alpha]][K[\alpha]:K]=[L:K],$$

as desired. $\qed$

Corollary: There are exactly $[L:K]$ embeddings of $L$ in $\bb C$ which fix $K$ pointwise. $\qed$

The next result states that a finite field extension in $\bb C$ can always be generated by only one element.

Primitive Element Theorem: $L=K[\alpha]$ for some $\alpha\in L$.

Proof: We proceed by induction on $[L:K]$. If $[L:K]=1$ then $L=K$, and there is nothing to show.

Otherwise, let $n=[L:K]$. Take any $\alpha\in L\backslash K$, so by induction hypothesis we have $L=K[\alpha,\beta]$ for some $\beta$.

For any $\lambda\in K$, consider the field $K[\alpha+\lambda\beta]$. If this is not all of $L$, then the minimal polynomial of $\alpha+\lambda\beta$ has degree at most $[K[\alpha+\lambda\beta]:K]<[L:K]=n$, so $\alpha+\lambda\beta$ has less than $n$ conjugates over $K$. Since $L$ has $n$ embeddings in $\bb C$ fixing $K$ pointwise, there must be two such embeddings $\sigma,\tau$ sending $\alpha+\lambda\beta$ to the same conjugate. Hence

$$\begin{aligned} \sigma(\alpha)+\lambda\sigma(\beta)&=\tau(\alpha)+\lambda\tau(\beta)\\ \implies\lambda&=\frac{\sigma(\alpha)-\tau(\alpha)}{\tau(\beta)-\sigma(\beta)}. \end{aligned}$$

(Note that $\sigma(\beta)=\tau(\beta)$ implies $\sigma(\alpha)=\tau(\alpha)$, so $\sigma=\tau$ on $K[\alpha,\beta]=L$, contradiction; hence the denominator above is nonzero.)

Since there are finitely many choices for $\sigma(\alpha),\sigma(\beta),\tau(\alpha),\tau(\beta)$, there are finitely many possible values for $\lambda$. Thus $K[\alpha+\lambda\beta]=L$ for all but finitely many values of $\lambda\in K$, and we are done since $K$ has infinitely many elements. $\qed$

Normal extensions

For fields $K\subseteq L\subseteq\bb C$, we say that $L$ is normal over $K$ if $L$ is closed under taking conjugates over $K$.

Proposition: $L$ is normal over $K$ if and only if every embedding of $L$ in $\bb C$ fixing $K$ pointwise has image $L$; ie. is an automorphism of $L$.

Proof: ($\Rightarrow$) Let $\sigma$ be an embedding of $L$ in $\bb C$ fixing $K$ pointwise. For any $\alpha\in L$ with minimal polynomial $f$ over $K$, $\sigma$ sends $\alpha$ to a root of the polynomial $\tilde\sigma f=f$, ie. to one of its conjugates. Hence $\sigma L\subseteq L$. Now $\sigma$ is a surjective homomorphism between two $K$-vector spaces of the same dimension, so it must be injective; hence $\sigma L=L$.

($\Leftarrow$) If $\alpha,\beta\in L$ are conjugates, let $f$ be their minimal polynomial over $K$; then

$$K[\alpha]\cong K[X]/(f)\cong K[\beta],$$

and this isomorphism fixes $K$. This map extends to an embedding of $L$ in $\bb C$ fixing $K$ (hence an automorphism of $L$ by hypothesis) sending $\alpha$ to $\beta$; hence $\beta\in L$. $\qed$

Corollary: $L$ is normal over $K$ if and only if $L$ has exactly $[L:K]$ automorphisms fixing $K$ pointwise. $\qed$

Corollary: If $K\subseteq L\subseteq M\subseteq\bb C$ are fields and $M/K$ is normal, then $M/L$ is normal.

Proof: Note that every embedding of $M$ in $\bb C$ fixing $L$ pointwise also fixes $K$ pointwise, and hence is an automorphism of $M$. $\qed$

From the definition, to check if $L$ is normal over $K$, we have to check that the conjugate of every element over $K$ is in $L$. In practice, it suffices to check this for the generating elements:

Proposition: If $L=K[\alpha_1,\ldots,\alpha_n]$ and $L$ contains all conjugates of all $\alpha_k$, then $L$ is normal over $K$.

Proof: Let $\sigma$ be an embedding of $L$ in $\bb C$ fixing $K$ pointwise. Now any $\alpha\in L$ can be written as

$$\alpha=f(\alpha_1,\ldots,\alpha_n)$$

for some $f\in K[X_1,\ldots,X_n]$. Then

$$\sigma\alpha=f(\sigma\alpha_1,\ldots,\sigma\alpha_n),$$

so $\sigma L\subseteq L$, which implies $\sigma L=L$ as in the previous proof. $\qed$

Corollary: If $L/K$ is a finite extension, then there exists a finite extension $M/L$ which is normal over $K$. Any such $M$ is also normal over $L$.

Proof: By the primitive element theorem, $L=K[\alpha]$ for some $\alpha$. If $\alpha_1,\ldots,\alpha_n$ are the conjugates of $\alpha$, let

$$M=K[\alpha_1,\ldots,\alpha_n],$$

so by the previous proposition, $M$ is normal over $K$. The second part follows from a previous corollary. $\qed$

Galois groups

For fields $K\subseteq L\subseteq\bb C$, the Galois group $\Gal(L/K)$ of $L$ over $K$ is the group of automorphisms of $L$ fixing $K$ pointwise, with composition as the group operation.

Proposition: $\lvert\Gal(L/K)\rvert\leq[L:K]$, with equality if and only if $L/K$ is normal. $\qed$

For a subgroup $H\leq\Gal(L/K)$, the fixed field of $H$ is

$$L^H\coloneqq\{\alpha\in L\,:\,\sigma(\alpha)=\alpha\text{ for all }\sigma\in H\}.$$

It is easy to check that this is in fact a subfield of $L$.

Proposition: Let $L/K$ be normal, and let $G=\Gal(L/K)$. Then for any $H\leq G$,

$$L^H=K\iff H=G.$$

Proof: ($\Leftarrow$) Note that

$$\begin{aligned} [L:K]&=|\Gal(L/K)|\\ &\leq|\Gal(L/L^G)|\\ &\leq[L:L^G]\leq[L:K], \end{aligned}$$

so equality holds throughout, hence $[L^G:K]=1$ implies $L^G=K$.

($\Rightarrow$) We have $L=K[\alpha]$ for some $\alpha$. Consider the polynomial

$$f(X)=\prod_{\sigma\in H}(X-\sigma\alpha).$$

Now for any $\tau\in H$, we have

$$\tilde\tau f(X)=\prod_{\sigma\in H}(X-\tau\sigma\alpha)=f(X),$$

so $\tau$ fixes all coefficients of $f$ for any $\tau\in H$. Hence $f\in L^H[X]=K[X]$, so

$$|H|=\deg f=[L:K]=|G|$$

implies $H=G$. $\qed$

The Galois correspondence

For fields $K\subseteq L\subseteq\bb C$ with $L/K$ normal, set $G=\Gal(L/K)$. Then we have maps between subfields of $L$ containing $K$ and subgroups of $G$, defined as follows:

$$\begin{aligned} \{K\subseteq F\subseteq L\}&\Leftrightarrow\{H\leq G\}\\ F\qquad&\mapsto\Gal(L/F)\\ L^H\qquad&\leftarrow\quad H \end{aligned}$$

Fundamental Theorem of Galois Theory: The mappings above are inverses of each other.

Proof: For any field $K\subseteq F\subseteq L$, $L/K$ normal implies $L/F$ normal, so the previous proposition gives $G^{\Gal(L/F)}=F$.

Now for any subgroup $H\leq G$, clearly we have $H\leq\Gal(L/L^H)$. But by the previous proposition, $L^H$ is not fixed by any proper subgroup of $\Gal(L/L^H)$. Hence $\Gal(L/L^H)=H$. $\qed$

Hence we get a one-to-one correspondence between subfields of $L$ containing $K$ and subgroups of $G$, which we may call the Galois correspondence.

Proposition: Let $K,L$ be as above. If $F\leftrightarrow H$ under the Galois correspondence, then $F/K$ is normal if and only if $H$ is a normal subgroup of $G$. In this case, there is an isomorphism

$$G/H\to\Gal(F/K),$$

obtained by restricting automorphisms to $F$.

Proof: Every embedding of $F$ in $\bb C$ extends to an embedding of $L$ in $\bb C$, which must be a member of $G$. Hence $F/K$ is normal if and only if $\sigma F=F$ for all $\sigma\in G$. By the Galois correspondence, this holds if and only if $\Gal(L/\sigma F)=\Gal(L/F)=H$ for all $\sigma\in G$.

On the other hand, for any $\tau\in G$, we have $\begin{aligned} \tau\in\Gal(L/\sigma F)&\iff\tau\sigma x=\sigma x\text{ for all }x\in F\\ &\iff\sigma^{-1}\tau\sigma x=x\text{ for all }x\in F\\ &\iff\sigma^{-1}\tau\sigma\in\Gal(L/F)=H\\ &\iff\tau\in\sigma H\sigma^{-1}. \end{aligned}$

Hence $F/K$ is normal if and only if $\sigma H\sigma^{-1}=H$ for all $\sigma\in G$, ie. $H$ is a normal subgroup of $G$.

For the second part, consider the group homomorphism $G\to\Gal(F/K)$ given by restricting automorphisms of $L$ to $F$. (This is well-defined since such restrictions are embeddings of $F$ in $\bb C$, and hence are automorphisms of $F$ since $F/K$ is normal.) By definition, $H$ lies in the kernel of this homomorphism, so we have an embedding

$$G/H\hookrightarrow\Gal(F/K).$$

But

$$|G/H|=\frac{|G|}{|H|}=\frac{[L:K]}{[L:F]}=[F:K]=|\Gal(F/K)|,$$

so the above embedding is an isomorphism. $\qed$

Field compositions

Given two fields $L,L’\subseteq\bb C$, the composite (or compositum) of $L$ and $L’$, written $LL’$, is the intersection of all subfields of $\bb C$ containing both $L$ and $L’$. This is clearly the smallest subfield of $\bb C$ containing both $L$ and $L’$.

Proposition: Let $K,L,L’$ $L=K[\alpha]$ for some $\alpha\in L$. Then $LL’=L’[\alpha]$, and the minimal polynomial of $\alpha$ over $L’$ divides the minimal polynomial of $\alpha$ over $K$.

Proof: Note that $L’\subseteq LL’$ and $\alpha\in LL’$ implies $L’[\alpha]\subseteq LL’$. Conversely, $L=K[\alpha]\subseteq L’[\alpha]$ and $L’\subseteq L’[\alpha]$ implies $LL’\subseteq L’[\alpha]$.

For the second part, note that the minimal polynomial of $\alpha$ over $K$ has coefficients in $K\subseteq L$ and has $\alpha$ as a root. $\qed$

Corollary: Let $K=L\cap L’$. Then $[LL’:L’]\leq[L:K]$. $\qed$

Proposition: Let $L/K$ be normal and $E/K$ be any extension. Then $EL/E$ is normal, and there is an embedding

$$\Gal(EL/E)\hookrightarrow\Gal(L/K)$$

obtained by restricting automorphisms to $L$. Moreover, this embedding is an isomorphism if and only if $E\cap L=K$.

Proof: Let $L=K[\alpha]$ for some $\alpha\in L$. Then $EL=E[\alpha]$, and conjugates of $\alpha$ over $E$ are also conjugates of $\alpha$ over $K$, all of which are in $L$. Hence $EL/E$ is normal.

The given homomorphism is well-defined since such restrictions are embeddings of $L$ in $\bb C$, and hence are automorphisms of $L$ since $L/K$ is normal. Also note that the kernel is trivial, since any embedding of $EL$ in $\bb C$ fixing $E$ and $L$ must fix all of $EL$.

Now $\Gal(EL/E)$ has fixed field $E$, so the image of $\Gal(EL/E)$ under this homomorphism has fixed field $E\cap L$. Thus the image is $\Gal(L/E\cap L)$ by the Galois correspondence, so the homomorphism is surjective if and only if $E\cap L=K$. $\qed$