Galois Theory, Part 1: The Fundamental Theorem of Galois Theory

October 19th, 2017

Introduction

Beginning with a polynomial $f(x)$, there exists a finite extension of $F$ which contains the roots of $f(x)$. Galois THeory aims to relate the group of permutations fo the roots of $f$ to the algebraic structure of its splitting field. In a similar way to representation theory, we study an object by how it acts on another.

Definition: An isomorphism $\sigma$ of $K$ with itself is called an automorphism of $K$. The collection of automorphism $K$ is denoted $Aut(K)$.

Definition: If $F$ is a subset of $K$ (like a subfield), then an automorphism $\sigma$ is said to fix $F$ if it fixes every element of $F$.

Note that any field has at least one automorphism: the identity map, called the trivial automorphism.

Note that the prime subfield is generated by $1$, and since any automorphism sends $1$ to $1$, any automorphism of a field fixes its prime subfield. For example, $\mathbb{Q}$ and $\mathbb{F}_p$ have only the trivial automorphism.

Definition: Let $K/F$ be an extension of fields. Then, $Aut(K/F)$ is the collection of automorphisms of $K$ which fix $F$.

Note that the above discussion gives us that $Aut(K) = Aut(K/F)$, if $F$ is the prime subfield. Note that under composition, there is a group structure on automorphisms.

Proposition 1

$Aut(K)$ is a group under composition and $Aut(K/F)$ is a subgroup.

Proposition 2

Let $K/F$ be a field extension, and $\alpha \in K$ algebraic over $F$. Then for any $\sigma \in Aut(K/F)$, $\sigma\alpha$ is a root of the minimal polynomial for $\alpha$. In other words, $Aut(K/F)$ permutes the roots of irreducible polynomials.

Suppose that $\alpha$ satisfies the equation:

$$ \alpha^n + c_{n-1}\alpha^{n-1} + \dots + c_1\alpha + c_0 = 0 $$

Where $c_i \in F$. Then apply the automorphism $\sigma$ to obtain:

$$ (\sigma\alpha)^n + c_{n-1}(\sigma\alpha)^{n-1} + \dots + c_0 = 0 $$

And thus, $\sigma\alpha$ is a root of the same polynomial over $F$ as $\alpha$.

In general, if $K$ is generated over $F$ by some elements, then an automorphism is completely determined by its action on the generators.

In particular, if $K/F$ is finite, then it is finitely generated over $F$ by algebraic elements. In this case, the number of automorphisms fixing $F$ is finite, and $Aut(K/F)$ is a finite group. In this case, the automorphisms of a finite extension are permutations of the roots of a finite number of equations (though not every permutation necessarily gives an automorphism).

We have described a field associated to each extension; we now reverse the process.

Proposition 3

Let $H \leq Aut(K)$ be a subgroup of $Aut(K)$. The collection of all elements $F$ of $K$ which are fixed by $H$ is a subfield.

This follows from readily from the definition of an field isomorphism.

Note here that we do not necessarily need a subgroup, but just a subset of $K$.

Proposition 4

The above association is inclusion reversing:

It maybe should be clear here that we are heading towards a bijection of some sort. We begin by investigating the size of the automorphism group of a splitting field.

Let $F$ be a field and let $E$ be the splitting field over $F$ of $f(x)$. We know that we can extend an isomorphism $\varphi: F \rightarrow F’$ to an isomorphism $\sigma: E \rightarrow E’$, where $E’$ is the splitting field over $F’$ of $f’(x)$.

We now show that the number of such extensions is at most $[E:F]$, with equality if $f$ is separable over $F$. We proceed by induction. If $[E:F]=1$, then $E=F$ and there is only one extension (the identity).

If $[E:F]>1$, then $f(x)$ has at least one irreducible factor $p(x)$ of degree greater than $1$ which maps to $p’(x)$. Fix $\alpha$, a root of $p(x)$. Then, if $\sigma$ is any extension of $\varphi$ to $E$, then $\sigma$ restricted to $F(\alpha)$ is an isomorphism $\tau$ which maps $F(\alpha)$ to $F’(\beta)$, where $\beta$ is a root of $p’(x)$. We have the two extensions:

$$ \begin{aligned} \sigma: E &\rightarrow E' \\ \tau: F(\alpha) &\rightarrow F'(\beta) \\ \varphi: F &\rightarrow F' \end{aligned} $$

Now conversely, say $\beta$ is a root of $p’(x)$. Then we can by the same process construct such a diagram.

Counting the number of extensions $\sigma$ of $\varphi$ is now counting the number of diagrams.

To extend $\varphi$ to $\tau$ is to count the number of distinct roots $\beta$ of $p’(x)$. Since $p(x)$ and $p’(x)$ both have degree $[F(\alpha):F]$, the number of extensions of $\varphi$ to $\tau$ is at most $[F(\alpha):F]$, with equality if the roots are distinct.

Now, since $E$ is the splitting field of $f$ over $F(\alpha)$ and $E’$ is the splitting field of $f’$ over $F’(\beta)$, and by hypothesis $[E:F(\alpha)] < [E:F]$, we apply the induction hypothesis to say that the number of extensions of $\tau$ to $\sigma$ is at most $[E:F(\alpha)]$, with equality if $f$ has distinct roots.

Finally, since $[E:F] = [E: F(\alpha)][F(\alpha):F]$, it follows that the number of extensions of $\varphi$ to $\sigma$ is at most $[E:F]$, with equality if $f(x)$ has distinct roots.

In particular, when $F=F’$ and $\varphi$ is the identity map, the isomorphisms $\sigma$ are exactly the automorphisms of $E$ fixing $F$.

Corollary 1

Let $E$ be the splitting field over $F$ of the polynomial $f(x) \in F[x]$. Then:

$$ \vert Aut(E/F) \vert \leq [E:F] $$

With equality if $f(x)$ is separable over $F$.

Therefore, the splitting field of a separable polynomial is exactly the “bijective” correspondence we are looking for, in which $[E:F] = \vert Aut(E/F)\vert$.

Definition: Let $K/F$ be a finite extension. Then $K$ is said to be Galois over $F$ and $K/F$ is a Galois extension if $\vert Aut(E/F)\vert = [K:F]$. The group of automorphisms is called the Galois group of $K/F$, denoted $Gal(K/F)$.

Corollary 2

If $K$ is the splitting field over $F$ of a separable polynomial $f(x)$ then $K/F$ is Galois.

We will see that the converse is also true.

Note also that this tells us that the splitting field of any polynomial over $\mathbb{Q}$ is Galois, since the splitting field of a polynomial is the same as the one obtained by removing multiple factors, which is separable.

Definition: If $f(x)$ is a separable polynomial over $F$, then the Galois group of $f$ over $F$ is the Galois group of the spliting field of $f(x)$ over $F$.

The Fundamental Theorem of Galois Theory

Definition: A character of a group $G$ with values in a field $L$ is a homomorphism from $G$ to the multiplicative group $L^\times$.

Definition: The characters $\chi_1, \chi_2, \dots, \chi_n$ are linearly independent if they are linearly independent functions on $G$.

Theorem 1

If $\chi_1, \chi_2, \dots, \chi_n$ are distinct characters of $G$, then they are linearly independent.

Now, consider an injective homomorphism $\sigma$ of a field $K$ into a field $L$, which is called an embedding of $K$ into $L$. In particular, $\sigma$ can be viewed as a character of $K^\times$ with values in $L$.

Corollary 3

If $\sigma_1, \dots, \sigma_n$ are distinct embeddings of $K$ into $L$, then they are linearly independent as functions on $K$. In particular, the distinct automorphisms of a field $K$ are linearly independent as functions on $K$.

Theorem 2

Let $G = \sigma_1, \dots \sigma_n$ be a subgroup of automorphisms of a field $K$ and let $F$ be its fixed field. Then:

$$ [K:F] = n = \vert G \vert $$

This proof will be omitted; it follows from analyzing systems of equations.

Corollary 4

Let $K/F$ be any finite extension. Then:

$$ \vert Aut(K/F) \vert \leq [K:F] $$

With equality iff $F$ is the fixed field of $Aut(K/F)$. This tells us that $K/F$ is Galois iff $F$ is the fixed field of $Aut(K/F)$.

To prove this, let $F_1$ be the fixed field of $Aut(K/F)$. In other words:

$$ F \subseteq F_1 \subseteq K $$

By Theorem $2$, we have:

$$ [K:F_1] = \vert Aut(K/F) \vert $$

Hence, we have:

$$ [K:F] = \vert Aut(K/F) \vert [F_1: F] $$

And this proves the corollary.

Corollary 5

Let $G$ be a finite subgroup of automorphisms of a field $K$ and let $F$ be its fixed field. Then every automorphism of $K$ fixing $F$ is contained in $G$, i.e.:

$$ Aut(K/F) = G $$

Therefore, $K/F$ is Galois, with Galois group $G$.

Note that by definition $G \leq Aut(K/F)$. But by the theorem we have $\vert G \vert = [K:F]$. By the previous corollary we have $\vert Aut(K/F) \vert \leq [K:F] = \vert G \vert$. This gives:

$$ [K:F] \leq \vert Aut(K/F) \vert \leq [K:F] $$

And therefore, if we have a subgroup of automorphisms, then $K$ is a Galois extension over its fixed field.

Corollary 6

If $G_1 \ne G_2$ are distinct finite subgroups of automorphisms of a field $K$, then their fixed fields are also distinct.

If the fixed fields $F_1 = F_2$, then by definition $F_1$ is fixed by $G_2$. But then $G_2 \neq G_1$, and similarly $G_1 \leq G_2$ adn thus the two groups are equal.

The corollaries above tell us that taking fixed field for distinct finite subgroups of $Aut(K)$ gives distinct subfields of $K$ over which $K$ is Galois. The degrees of the extensions are given by the orders of the subgroups.

The next result completely characterizes Galois extensions.

Theorem 3

The extension $K/F$ is Galois iff $K$ is the splitting field of some separable polynomial over $F$. If this is the case then every irreducible polynomial with coefficients in $F$ which has a root in $K$ is separable and has all its roots in $K$ ($K/F$ is in particular separable).

We showed earlier that the splitting field of a separable polynomial is Galois. We now show, essentially, the converse.

Let $G=Gal(K/F)$ and let $\alpha \in K$ be a root of $p(x)$, an irreducible polynomial in $F[x]$ which has a root in $K$. Consider the elements:

$$ \alpha, \sigma_2(\alpha), \dots, \sigma_n(\alpha) \in K $$

Where $\sigma_i$ represent the elements of the Galois group. Of this list, denote the distinct elements by:

$$ \alpha, \alpha_2, \dots, \alpha_r $$

If $\tau \in G$ then since $G$ is a group applying $\tau$ to the first list just permutes it. In particular, teh following polynomial has coefficients which are fixed by all the elements of $G$:

$$ f(x) = (x-\alpha)(x-\alpha_2)\dots(x-\alpha_r) $$

The coefficients thus lie in the fixed field of $G$. However, note that $K/F$ is Galois iff $F$ is the fixed field of $Aut(K/F)$, so the fixed field of $G$ is exactly $F$. Thus, $f(x) \in F[x]$.

Since $p(x)$ is irreducible and has $\alpha$ as a root, $p(x)$ is the minimal polynomial for $\alpha$ over $F$, and it follows that $p(x)$ divides $f(x)$ in $F[x]$. So we have:

$$ p(x) = f(x) $$

This shows that $p(x)$ is separable and all its roots lie in $K$.

To complete the proof, suppose $K/F$ is Galois and let $\omega_1, \dots, \omega_n$ be a basis for $K/F$. let $p_i(x)$ be the minimal polynomial for $\omega_i$. Then $p_i(x)$ is separable and has all its roots in $K$. Let $g(x)$ be the polynomial obtained by removing multiple factors in this product. Then the splitting field of the two polynomials is the same and this field is $K$. Hence, $K$ is the splitting field of the separable polynomial $g(x)$.

Definition: Let $K/F$ be a Galois extension. If $\alpha \in K$ then the elements $\sigma\alpha$ for $\sigma \in Gal(K/F)$ are called the Galois conjugates of $\alpha$ over $F$. If $E$ is a subfield of $K$ containing $F$, the field $\sigma(E)$ is called the conjugate field of $E$ over $F$.

The proof of Theorem 3 shows that in a Galois extension $K/F$, if we have $\alpha \in K$ which is a root of a minimal polynomial over $F$, then the other roots are precisely the distinct conjugates of $\alpha$ under the Galois group of $K/F$.

The theorem also says that $K$ is not Galois over $F$ if we can find an irreducible polynomial over $F$ which has a root in $K$ but not all its roots in $K$. Now we have four characterizations of Galois extensions $K/F$:

Theorem (Fundamental Theorem of Galois Theory)

Let $K/F$ be a Galois extension and let $G=Gal(K/F)$. Then there is a bijection between subfields:

$$ F \subseteq E \subseteq K $$

And subgroups of the Galois group:

$$ 1 \subseteq H \subseteq G $$

In particular, the correspondence identifies $E$ to the elements of $G$ which fix $E$. Conversely, it identifies $H$ with the fixed field of $H$.

We will number these points 1 through 5 and prove each separately.

Part 1

Given any subgroup $H$ of $G$, we saw that there is a unique fixed field $E=K_H$. The correspondence is thus injective from subgroups to subfields. We now need to see that it is surjective, i.e. we can find a subgroup of the Galois group which fixes any subfield.

Now, if $K$ is the splitting field of a separable polynomial $f(x) \in F[x]$ then it is an element of $E[x]$ for any subfield $F \subseteq E \subseteq K$. Thus, $K$ is also the splitting field of $f$ over $E$, and therefore $K/E$ is Galois. Thus, $E$ is the fixed field of $Aut(K/E) \leq G$. This shows that indeed our correspondence is bijective. Concretely, the automorphisms fixing $E$ are precisely $Aut(K/E)$ since $K/E$ is Galois.

The Galois correspondence is evidently inclusion reversing.

Part 2

If $E=K_H$ is the fixed field of $H$ (which is Galois), then by Theorem 2 $[K:E] =\vert H \vert$, and similarly $[K:F] = \vert G \vert$. Taking the quotient gives $[E:F] = [G:H]$.

Part 3

Since $E$ is the fixed field of a subgroup $H \leq G$, by Corollary 5, $K/E$ is Galois with Galois group $Gal(K/E) = H$.

Part 4
Lemma

Let $E$ be the fixed field of a subgroup $H$. Then $\sigma$ is an embedding of $E$ iff it is the restriction of some automorphism $\sigma \in G$ to $E$.

Let $E = K_H$ be the fixed field of the subgroup $H$. Then every $\sigma \in G$, when restricted to $E$, gives an embedding of $E$ with a subfield $\sigma(E)$ of $K$. We shall show that these are indeed the only embeddings of $E$.

Conversely, let $\tau: E \rightarrow \tau(E) \subseteq \overline{F}$ be any embedding of $E$ (into a fixed algebraic closure $\overline{F}$ containing $K$) which fixes $F$. Then, if $\alpha \in E$ has minimal polynomial $m_\alpha$ over $F$ then $\tau(\alpha)$ is another root of $m_\alpha(x)$ and so $K$ contains $\tau(\alpha)$ as well. Thus, $\tau(E)\subseteq K$.

As above, $K$ is the splitting field of $f(x)$ over $E$ and so it is also the splitting field of $\tau f(x) =f(x)$ (since $\tau$ fixes $F$) over $\tau(E)$.

So, we can extend $\tau$ to an isomorphism $\sigma$ from $K$ to $K$. Since $\sigma$ fixes $F$, what we have just shown is that every embedding $\tau$ of $E$ fixing $F$ can be extended to an automorphism $\sigma$ of $K$ fixing $F$. In other words, every embedding of $E$ is the action of some $\sigma \in G$.

Proof

Now, two automorphisms $\sigma, \sigma’ \in G$ restrict to the same embedding of $E$ iff $\sigma^{-1}\sigma’$ is the identity on $E$. But then $\sigma^{-1}\sigma’ \in H$ since the automorphisms of $K$ which fix $E$ are exactly $H$. Another way of saying this is that $\sigma’ \in \sigma H$.

What we have just shown is that distinct embeddings of $E$ are in bijection with cosets $\sigma H$ of $H$ in $G$. In particular, this gives us that:

$$ \vert \text{Emb} (E/F) \vert = [G:H] = [E:F] $$

Where $\text{Emb}$ denotes the set of embeddings of $E$ into a fixed algebraic closure of $F$. Note that $\text{Emb}(E/F)$ contains the automorphisms $Aut(E/F)$, since any automorphism admits to an embedding by our lemma.

The extension $E/F$ is Galois iff $\vert Aut(E/F) \vert = [E:F]$. By the equality above, this is the case iff each embedding of $E$ is an automorphism of $E$, i.e. $\sigma(E) = E$.

Now note that if $\sigma\alpha \in \sigma(E)$, then:

$$ (\sigma h \sigma^{-1})(\sigma \alpha) = \sigma(h \alpha) = \sigma \alpha $$

For any $\alpha \in E$, since $H$ fixes $E$. Thus $\sigma H \sigma^{-1}$ fixes $\sigma(E)$. The group fixing $\sigma(E)$ has order equal to $[K:\sigma(E)] = [K:E] = \vert H \vert H$, so indeed $\sigma H \sigma^{-1}$ is precisely the group fixing $\sigma(E) = E$.

Because the Galois correspondence is a bijection, $\sigma H \sigma^{-1} = H$ and hence $H$ is normal. Thus, $E$ is Galois over $F$ iff $H$ is normal in $G$.

Furthermore, this proof shows that the group of cosets $G/H$ is identified with the group of automorphisms of the Galois extension $E/F$. Thus, $G/H \cong Gal(E/F)$.

Part 5

Suppose $H_1$ is the subgroup of elements fixing $E_1$ and $H_2$ the subgroup of elements fixing $E_2$. Then any element in $H_1 \cap H_2$ fixes both $E_1$ and $E_2$ and hence fixes the composite. Conversely, if an automorphism $\sigma$ fixes the composite $E_1E_2$, then in particular $\sigma \in H_1 \cap H_2$. Similarly, the intersection $E_1 \cap E_2$ corresponds to the subgroup generated by $H_1, H_2$, and this proves the final part.

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