Galois Extensions

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We are now ready to define what a Galois extension is, but before that we examine the following proposition.

Proposition 1 Consider an algebraic extension \(\mathbb{L}\) and an inclusion \(u:\mathbb{L}\rightarrow \overline{\mathbb{K}}\).

1. If \(u(\mathbb{L})\subseteq \mathbb{L}\), then \(u\) is a \(\mathbb{K}\)-automorphism of \(\mathbb{L}\).
2. There exists a \(\mathbb{K}\)-automorphism of \(\overline{\mathbb{K}}\) extending \(u\).

Proof

1. For any \(x\in E\), let \(f\) be the minimal polynomial of \(x\). Let \(\Phi\) be the set of roots of \(f\) in \(\mathbb{L}\); then \(\Phi\) is finite. Moreover, if \(\alpha\in\Phi\) then

   \[0=u(0)=u(f(\alpha))=f(u(\alpha))\]

   so \(u(\Phi)\subseteq\Phi\) holds. But \(u\) is not the zero map, hence it is injective (§Fields, ⁋Proposition 2), and therefore \(u\) is a bijection from \(\Phi\) to \(\Phi\). Thus \(x\in\Phi=u(\Phi)\subseteq u(E)\), and from this we obtain \(u(E)=E\).

2. Since \(\overline{\mathbb{K}}\) is an algebraic closure of both \(u(\mathbb{L})\) and \(\mathbb{L}\), the desired result follows from the universal property of §Algebraic Closures, ⁋Theorem 5.

Our goal is to examine all algebraic extensions of a fixed field \(\mathbb{K}\), or more precisely, to consider equivalence classes of algebraic extensions.

Definition 2 For two algebraic extensions \(\mathbb{L}\), \(\mathbb{M}\) of a field \(\mathbb{K}\), we say they are conjugate if there exists a \(\mathbb{K}\)-automorphism \(u:\overline{\mathbb{K}}\rightarrow \overline{\mathbb{K}}\) such that \(u(\mathbb{L})=\mathbb{M}\). In particular, two elements \(x,y\in\overline{\mathbb{K}}\) are conjugate if there exists a \(\mathbb{K}\)-automorphism \(u: \overline{\mathbb{K}}\rightarrow \overline{\mathbb{K}}\) with \(u(x)=y\).

By Proposition 1, if \(\mathbb{M}\) and \(\mathbb{L}\) are isomorphic extensions of \(\mathbb{K}\), then they are conjugate extensions, and by definition conjugate extensions are isomorphic. Moreover, the following holds.

Proposition 3 Fix two elements \(x,y\) of \(\overline{\mathbb{K}}\). The following are equivalent.

1. \(x\) and \(y\) are conjugate elements.
2. There exists a \(\mathbb{K}\)-isomorphism \(v: \mathbb{K}(x) \rightarrow \mathbb{K}(y)\) satisfying \(v(x)=y\).
3. \(x\) and \(y\) have the same minimal polynomial.

Proof

First assume the first condition. Let \(f\) be the minimal polynomial of \(x\); then

\[f(y)=f(u(x))=u(f(x))=u(0)=0\]

so the minimal polynomial of \(y\) divides \(f\). By the same logic the minimal polynomial of \(x\) divides that of \(y\), and therefore they are equal.

On the other hand, if \(x\) and \(y\) have the same minimal polynomial \(f\), then by the first isomorphism theorem

\[\mathbb{K}(x)\cong \mathbb{K}[\x]/(f)\cong \mathbb{K}(y)\]

so it is obvious that the third condition implies the second. Finally, assuming the second condition, Proposition 1 yields a \(\mathbb{K}\)-isomorphism \(u:\overline{\mathbb{K}}\rightarrow\overline{\mathbb{K}}\) extending \(v\), and therefore \(x\) and \(y\) are conjugate.

From this, if an algebraic element \(x\in \overline{\mathbb{K}}\) of degree \(n\) is given, we know that elements conjugate to \(x\) must be roots of the minimal polynomial of \(x\), and hence there are at most \(n\) such elements. Moreover, having fewer than \(n\) elements conjugate to \(x\) is exactly equivalent to the minimal polynomial of \(x\) not being separable. That is, denoting the group of \(\mathbb{K}\)-automorphisms of \(\overline{\mathbb{K}}\) by \(\Aut_\mathbb{K}(\overline{\mathbb{K}})\), and letting \(\Aut_{\mathbb{K}}\overline{\mathbb{K}}\) act on \(\overline{\mathbb{K}}\) in the obvious way, the set of elements fixed by this action \(\overline{\mathbb{K}}^{\Aut_{\mathbb{K}}(\overline{\mathbb{K}})}\) is exactly \(\mathbb{K}^{p^{-\infty}}\).

Galois Extensions

Definition 4 A field extension \(\mathbb{L}/\mathbb{K}\) is called a quasi-Galois extension or normal extension if \(\mathbb{L}/\mathbb{K}\) is algebraic and any irreducible polynomial \(f\in \mathbb{K}[\x]\) having a root in \(\mathbb{L}\) splits into a product of linear factors in \(\mathbb{L}[\x]\).

Thus a quasi-Galois extension is essentially nothing but another name for a splitting field.

Proposition 5 For an algebraic extension \(\mathbb{L}/\mathbb{K}\), the following are all equivalent.

1. \(\mathbb{L}/\mathbb{K}\) is quasi-Galois.
2. For any \(x\in \mathbb{L}\), the conjugates of \(x\) (in \(\overline{\mathbb{K}}\)) all belong to \(\mathbb{L}\).
3. Any \(\mathbb{K}\)-automorphism of \(\overline{\mathbb{K}}\) sends \(\mathbb{L}\) to \(\mathbb{L}\).
4. Any \(\mathbb{K}\)-homomorphism from \(\mathbb{L}\) to \(\overline{\mathbb{K}}\) lands in \(\mathbb{L}\).
5. \(\mathbb{L}\) is the splitting field of some family of non-constant polynomials \((f_i\in \mathbb{K}[\x])\).

Proof

The equivalence of the third and fourth conditions follows from Proposition 1. On the other hand, a quasi-Galois extension can be viewed as the splitting field of the minimal polynomials of its elements, so the last condition is implied by the first. Conversely, if the last condition holds, then by the same logic as in Proposition 1 any \(\mathbb{K}\)-automorphism of \(\overline{\mathbb{K}}\) sends roots of \(f_i\) to roots of \(f_i\), and hence sends \(\mathbb{L}\) to \(\mathbb{L}\). Thus the third condition holds. Also, the third condition trivially implies the second by definition. Therefore

\[(1)\implies (5)\implies (3)\iff (4)\implies (2)\]

so it suffices to show \((2)\implies (1)\). To this end, let \(f\in \mathbb{K}[\x]\) be a (monic) irreducible polynomial having a root in \(\mathbb{L}\). Then since \(\overline{\mathbb{K}}\) is algebraically closed, \(f\) factors in \(\overline{\mathbb{K}}\) as

\[f(\x)=\prod_{i=1}^d (\x- a_i), \qquad a_i\in \overline{\mathbb{K}}.\]

Now the \(a_i\) are all conjugate, and hence by assumption they must all belong to \(\mathbb{L}\). That is, \(f\) splits into a product of linear factors in \(\mathbb{L}[\x]\).

From this the following holds.

Corollary 6 The following hold.

1. An algebraic extension \(\mathbb{L}/\mathbb{K}\) is quasi-Galois if and only if every conjugate of \(\mathbb{L}\) is itself.
2. For an algebraic extension \(\mathbb{K}\subseteq \mathbb{L}\subseteq \mathbb{M}\), if \(\mathbb{M}/\mathbb{K}\) is quasi-Galois then so is \(\mathbb{L}/\mathbb{K}\).
3. Let a quasi-Galois extension \(\mathbb{M}/\mathbb{K}\) and its subextension \(\mathbb{L}/\mathbb{K}\) be given. Then for any \(\mathbb{K}\)-homomorphism \(u: \mathbb{L}\rightarrow \overline{\mathbb{K}}\) we have \(u(\mathbb{L})\subseteq \mathbb{M}\), and there exists a \(\mathbb{K}\)-automorphism \(v\) of \(\mathbb{M}\) extending this.
4. For any field extension \(\mathbb{K}'/\mathbb{K}\) and quasi-Galois extension \(\mathbb{L}/\mathbb{K}\), the composite \(\mathbb{K}'(\mathbb{L})\) is quasi-Galois over \(\mathbb{K}'\).

Proof

1. By Proposition 5, \(\mathbb{L}/\mathbb{K}\) being quasi-Galois is equivalent to every \(\mathbb{K}\)-automorphism of \(\overline{\mathbb{K}}\) sending \(\mathbb{L}\) to \(\mathbb{L}\).
2. Suppose \(\mathbb{M}/\mathbb{K}\) is quasi-Galois. Then since \(\overline{\mathbb{K}}\) is also an algebraic closure of \(\mathbb{L}\), it suffices by Proposition 5 to show that for any \(\mathbb{L}\)-automorphism \(u: \overline{\mathbb{K}}\rightarrow\overline{\mathbb{K}}\) we have \(u(\mathbb{M})=\mathbb{M}\). But \(\mathbb{M}\) is a quasi-Galois extension of \(\mathbb{K}\), and \(u\) is an \(\mathbb{L}\)-automorphism, so it is automatically a \(\mathbb{K}\)-automorphism as well. From this we know that \(u\) satisfies the desired condition.
3. From Proposition 1 we know that there exists a \(\mathbb{K}\)-automorphism \(v:\overline{\mathbb{K}}\rightarrow\overline{\mathbb{K}}\) extending \(u\). Then, since \(\mathbb{M}\) is quasi-Galois by assumption, its restriction to \(\mathbb{M}\) must satisfy \(v(\mathbb{M})=\mathbb{M}\), and hence the desired claim holds.
4. If \(\mathbb{L}\) is the splitting field of \(f_i\in \mathbb{K}[\x]\), then \(\mathbb{L}'\) is the splitting field of \(f_i\in \mathbb{K}'[\x]\).

As can be seen from the proof of the above corollary, the most important property characterizing a quasi-Galois extension \(\mathbb{L}/\mathbb{K}\) is that any \(\mathbb{K}\)-automorphism sends \(\mathbb{L}\) to \(\mathbb{L}\). The following proposition is also obvious from this fact.

Proposition 7 Let quasi-Galois extensions \(\mathbb{L}_i\) inside an algebraic closure \(\overline{\mathbb{K}}\) of \(\mathbb{K}\) be given. Then both \(\bigcap \mathbb{L}_i\) and \(\mathbb{K}(\bigcup \mathbb{L}_i)\) are quasi-Galois.

Proof

Both extensions are subextensions of \(\overline{\mathbb{K}}\), hence algebraic, so it suffices to verify only the third condition of Proposition 5. Let \(u\) be an arbitrary \(\mathbb{K}\)-automorphism of \(\overline{\mathbb{K}}\). Since each \(\mathbb{L}_i\) is quasi-Galois, by Proposition 5 again we have \(u(\mathbb{L}_i)=\mathbb{L}_i\).

First, since \(u\) is bijective,

\[u\left(\bigcap_i\mathbb{L}_i\right)=\bigcap_iu(\mathbb{L}_i)=\bigcap_i\mathbb{L}_i.\]

On the other hand, since \(u\) is a field automorphism fixing \(\mathbb{K}\), for any subset \(S\subseteq\overline{\mathbb{K}}\) we have \(u(\mathbb{K}(S))=\mathbb{K}(u(S))\). Applying this to \(S=\bigcup_i\mathbb{L}_i\) gives

\[u\left(\mathbb{K}\left(\bigcup_i\mathbb{L}_i\right)\right)=\mathbb{K}\left(\bigcup_iu(\mathbb{L}_i)\right)=\mathbb{K}\left(\bigcup_i\mathbb{L}_i\right)\]

so we obtain the desired result.

In particular, for any set \(S\) of elements of \(\overline{\mathbb{K}}\), we can consider the smallest quasi-Galois extension containing it. By definition, this is the extension of \(\mathbb{K}\) generated by all conjugates of each element of \(S\). We call this the quasi-Galois extension generated by \(S\).

In Definition 4, when we defined a quasi-Galois extension, we required that an irreducible polynomial \(f\) split into a product of linear factors, but we did not require these factors to be distinct. A Galois extension is obtained by adding the separability condition.

Theorem 8 Let an algebraic extension \(\mathbb{L}/\mathbb{K}\) and the group \(\Gamma\) of \(\mathbb{K}\)-automorphisms of \(\mathbb{L}\) be given. The following are all equivalent.

1. Every \(\Gamma\)-invariant element of \(\mathbb{L}\) lies in \(\mathbb{K}\).
2. \(\mathbb{L}\) is a separable quasi-Galois extension of \(\mathbb{K}\).
3. For any \(x\in \mathbb{L}\), the minimal polynomial \(f\in \mathbb{K}[\x]\) of \(x\) splits into a product of distinct linear factors in \(\mathbb{L}[\x]\).

Proof

The equivalence of the second and third conditions is obvious, so it suffices to show that these are equivalent to the first condition.

First assume the first condition. For any \(x\in \mathbb{L}\) and its minimal polynomial \(f\in \mathbb{K}[\x]\), we must show that \(f\) splits into distinct linear factors in \(\mathbb{L}[\x]\). To this end, let \(S\) be the set of all roots of \(f\) in \(\mathbb{L}\), and define a new polynomial

\[g(\x)=\prod_{a\in S}(\x-a)\]

then \(g\) is an element of \(\mathbb{L}[\x]\) and for any \(\sigma\in\Gamma\),

\[(\sigma\cdot g)(\x)=\prod_{a\in S}(\x-\sigma(a))=\prod_{a\in S}(\x-a)\]

so the coefficients of \(g\) are unchanged by \(\sigma\), and therefore by the assumption of the first condition we have \(g\in\mathbb{K}[\x]\). Now since \(g(x)=0\), by §Algebraic Extensions, ⁋Theorem 15 we know that \(g\) divides \(f\), and considering their degrees we must have \(g=f\). That is, the third condition holds.

Conversely, assume the third condition and show the first. If \(x\in\mathbb{L}\) does not belong to \(\mathbb{K}\), we must show that there exists \(\sigma\in\Gamma\) sending \(x\) to another element. Let \(f\) be the minimal polynomial of \(x\); then from \(x\not\in\mathbb{K}\) we know that \(f\) has degree at least 2, and by assumption

\[f(\x)=\prod_{a\in S}(\x-a), \qquad \text{$S$ the set of conjugates of $x$ in $\overline{\mathbb{K}}$}\]

can be so decomposed, and on the other hand since \(\mathbb{L}/\mathbb{K}\) is quasi-Galois there exists a \(\mathbb{K}\)-automorphism \(u\) of \(\overline{\mathbb{K}}\) sending \(x\) to some other \(a\in S\), which by Proposition 5 is a \(\mathbb{K}\)-automorphism of \(\mathbb{L}\). From this we obtain the desired result.

We may now define the following.

Definition 9 An algebraic extension \(\mathbb{L}/\mathbb{K}\) is said to be Galois if \(\mathbb{L}/\mathbb{K}\) satisfies the conditions of Theorem 8.

Then from the result of Proposition 7 and the results on separable extensions we obtain the following two propositions.

Proposition 10 Let Galois extensions \(\mathbb{L}_i\) inside an algebraic closure \(\overline{\mathbb{K}}\) of \(\mathbb{K}\) be given. Then both \(\bigcap \mathbb{L}_i\) and \(\mathbb{K}(\bigcup \mathbb{L}_i)\) are Galois.

Proof

By the second condition of Theorem 8, a Galois extension is the same as a separable quasi-Galois extension. That both extensions are quasi-Galois was shown in Proposition 7, so we only need to check separability.

First, \(\bigcap_i\mathbb{L}_i\) is a subextension of a separable extension \(\mathbb{L}_{i_0}/\mathbb{K}\), hence it is separable. (§Separable Extensions, ⁋Proposition 15) On the other hand, the elements of \(\bigcup_i\mathbb{L}_i\) are each elements of separable extensions, so they are all separable over \(\mathbb{K}\) (first result of §Separable Extensions, ⁋Proposition 12), and therefore \(\mathbb{K}(\bigcup_i\mathbb{L}_i)\) is an extension generated by separable elements, hence is separable by the second result of the same proposition.

Proposition 11 Let a Galois extension \(\mathbb{L}/\mathbb{K}\) and a finite-degree subextension \(\mathbb{M}/\mathbb{K}\) be given. Then there exists a subextension \(\mathbb{N}/\mathbb{K}\) of \(\mathbb{L}/\mathbb{K}\) containing \(\mathbb{M}\) that is Galois of finite degree.

Proof

Since \(\mathbb{M}/\mathbb{K}\) is of finite degree, we can write \(\mathbb{M}=\mathbb{K}(x_1,\ldots,x_n)\). Let \(S\) be the set of all conjugates (in \(\overline{\mathbb{K}}\)) of \(x_1,\ldots,x_n\). Each \(x_i\) is algebraic, so it has only finitely many conjugates (Proposition 3), and since \(\mathbb{L}/\mathbb{K}\) is quasi-Galois, by the second condition of Proposition 5 they all belong to \(\mathbb{L}\). That is, \(S\) is a finite subset of \(\mathbb{L}\), and \(\mathbb{N}=\mathbb{K}(S)\) is a subextension of finite degree containing \(\mathbb{M}\), since it is generated by finitely many algebraic elements.

We now show that \(\mathbb{N}/\mathbb{K}\) is Galois. First, any \(\mathbb{K}\)-automorphism \(u\) of \(\overline{\mathbb{K}}\) sends conjugates to conjugates, so \(u(S)=S\), and therefore \(u(\mathbb{N})=\mathbb{K}(u(S))=\mathbb{N}\), whence by Proposition 5 we see that \(\mathbb{N}/\mathbb{K}\) is quasi-Galois. On the other hand, \(\mathbb{N}\) is a subextension of the separable extension \(\mathbb{L}/\mathbb{K}\), so it is separable (§Separable Extensions, ⁋Proposition 15), and therefore by the second condition of Theorem 8 it is Galois.

Galois Groups

As we have already seen, when dealing with a Galois extension \(\mathbb{L}/\mathbb{K}\), what is used importantly is the collection of \(\mathbb{K}\)-automorphisms of \(\mathbb{L}\).

Definition 12 For a Galois extension \(\mathbb{L}/\mathbb{K}\), the group of \(\mathbb{K}\)-automorphisms of \(\mathbb{L}\) is called the Galois group and is denoted \(\Gal(\mathbb{L}/\mathbb{K})\).

In particular, let us fix a field \(\mathbb{K}\) and consider its algebraic closure \(\overline{\mathbb{K}}\). For a separable polynomial \(f\) in \(\mathbb{K}[\x]\) and the set \(A\) of roots of \(f\), we have already seen that \(\mathbb{L}=\mathbb{K}(A)\) is a Galois extension of \(\mathbb{K}\). But since \(\mathbb{L}\) is generated by \(A\), any \(\sigma\in \Gal(\mathbb{L}/\mathbb{K})\) is completely determined by its values on \(A\), and from this an injective group homomorphism

\[\Gal(\mathbb{L}/\mathbb{K})\rightarrow S_A\]

is induced. In general this homomorphism need not be surjective. That is, among the roots of an arbitrary separable polynomial \(f\) there may be ones that are not conjugate to each other, and by Proposition 3 this is equivalent to these two roots \(x,y\) having different minimal polynomials. On the other hand, if \(x,y\) are roots of \(f\), then by §Algebraic Extensions, ⁋Theorem 15 their minimal polynomials each divide \(f\), and therefore \(x\) and \(y\) are roots of different irreducible factors of \(f\). From this we can further determine what the image of the above injective homomorphism looks like.

On the other hand, for a Galois extension \(\mathbb{L}/\mathbb{K}\) and another Galois extension \(\mathbb{M}/\mathbb{K}\) that is a subextension of \(\mathbb{L}\), we obtain the following result from Proposition 1.

Proposition 13 In the above situation, the restriction homomorphism

\[\Gal(\mathbb{L}/\mathbb{K})\rightarrow\Gal(\mathbb{M}/\mathbb{K});\qquad \sigma\mapsto \sigma\vert_\mathbb{M}\]

is surjective.

Proof

First let us verify that this map is well defined. For any \(\sigma\in\Gal(\mathbb{L}/\mathbb{K})\), by Proposition 1 there exists a \(\mathbb{K}\)-automorphism \(v\) of \(\overline{\mathbb{K}}\) extending \(\sigma\). Since \(\mathbb{M}/\mathbb{K}\) is quasi-Galois, by Proposition 5 we have \(v(\mathbb{M})=\mathbb{M}\), and therefore \(\sigma\vert_\mathbb{M}=v\vert_\mathbb{M}\) is a \(\mathbb{K}\)-automorphism of \(\mathbb{M}\), that is, an element of \(\Gal(\mathbb{M}/\mathbb{K})\).

We now show surjectivity. Given any \(\tau\in\Gal(\mathbb{M}/\mathbb{K})\), regarding \(\tau\) as a \(\mathbb{K}\)-homomorphism from \(\mathbb{M}\) to \(\overline{\mathbb{K}}\) and applying Proposition 1 again, we obtain a \(\mathbb{K}\)-automorphism \(u\) of \(\overline{\mathbb{K}}\) extending \(\tau\). Then since \(\mathbb{L}/\mathbb{K}\) is quasi-Galois we have \(u(\mathbb{L})=\mathbb{L}\), and \(\sigma=u\vert_\mathbb{L}\) is an element of \(\Gal(\mathbb{L}/\mathbb{K})\) with \(\sigma\vert_\mathbb{M}=u\vert_\mathbb{M}=\tau\).