## Galois Theory: The Fundamental Theorem

After the preliminaries of field automorphisms and the conditions of normality and separability, I can prove the fundamental theorem of Galois theory. The basic idea is to relate intermediate fields of L/K and subgroups of Aut(L/K). Every intermediate field F maps to Aut(L/F), regarded as a subgroup of Aut(L/K), and every subgroup H of Aut(L/K) maps to L^H.

From the previous post, those maps are inverses of each other when L/K is finite and K = L^Aut(L/K); in that case, Aut(L/L^H) = H and L^Aut(L/F) = F. The ideal case in Galois theory is when K is indeed equal to L^Aut(L/K). In particular, the fundamental theorem states that this case holds precisely when L/K is separable and normal.

First, two examples. Let K = Q and L = Q(2^(1/3)). The only root of the minimal polynomial of 2^(1/3), x^3 – 2, that is contained in L is 2^(1/3). Automorphisms fairly obviously preserve minimal polynomials, so any automorphism must be the identity on 2^(1/3). But that generates L over K, so the only automorphism is the identity. This implies that L^Aut(L/K) = L.

For the second example, let $K = \mathbb{Z}_{p}(t), L = \mathbb{Z}_{p}(\sqrt[p]{t})$. The minimal polynomial of t^(1/p) over K is x^pt = (xt^(1/p))^p, so it’s again the only root lying in L. And again, as it generates L over K, every K-automorphism of L must be the identity, so that L^Aut(L/K) = L.

Before I prove the fundamental theorem, there are two small preliminaries. One, from [L:L^Aut(L/K)] = |Aut(L/K)| it follows that L^Aut(L/K) = K iff |Aut(L/K)| = [L:K]. And two, a not necessarily finite extension is normal and separable iff it is the splitting field of a family of separable polynomials. The latter result follows from the fact that if L/K is normal then it is a splitting field of a family of polynomials, and if it is separable then they all have to be separable; conversely, a splitting field of a family of separable polynomials is normal, and generated by separable elements.

For one direction, suppose that L^Aut(L/K) = K, and let a be any element of L. Let a1 = a, a2, a3, …, a(m) be the distinct images of a under the various elements of Aut(L/K); note that I’m assuming that the field extension is finite. Then under each element g of Aut(L/K), the set of a(i)’s gets mapped to itself, because g(a(i)) = gh(a) for some h in Aut(L/K), and g(a(i)) = g(a(j)) implies that a(i) = a(j) since g is an automorphism.

Consider the polynomial f(x) = (xa1)(xa2)…(xa(m)). The polynomial is invariant under the action of Aut(L/K), since every g in it merely permutes the a(i)’s. Therefore the coefficients of the polynomial are in L^Aut(L/K) = K, and the polynomial is in K[x]. The minimal polynomial of a then divides f, so that it inherits the properties of splitting over L and having distinct roots. As this applies to any a in L, it makes L/K separable and normal.

Conversely, it’s enough to show that Aut(L/K) acts transitively on the roots of each minimal polynomial $m_{a}(x), a \in L$. In other words, it’s enough to show that if a and b have the same minimal polynomial over K, then there exists a K-automorphism g of L such that g(a) = b. This is because if a is any element of L outside K, then deg(m(a)) > 1. By normality, all roots of m(a) are in L, and by separability they’re distinct, so there exists a b != a such that g(a) = b for some g, so that a is not in L^Aut(L/K).

To see that the action of Aut(L/K) is indeed transitive, let a and b have the same minimal polynomial over K. Then K(a) = K[x]/(m(a)) = K[x]/(m(b)) = K(b), so there exists a K-isomorphism from K(a) to K(b) mapping a to b. In L, m(a) = m(b) splits, so that the isomorphism from K(a) to K(b) can be extended to an automorphism of L fixing K and sending a to b.

If L/K is separable and normal, it’s called a Galois extension, and Aut(L/K) is called the Galois group of L/K and denoted Gal(L/K). Both separability and normality are inherited from L/K to L/F, where F is an intermediate field, so we get a bijection from each intermediate field F to each subgroup of Gal(L/K), denoted Gal(L/F).

If L/K is Galois, then F/K is separable; however, as the case of L = Q(2^(1/3), w), K = Q, F = Q(2^(1/3)) readily shows, F/K need not be normal. This induces the second fundamental theorem of Galois theory, which states that F/K is normal iff Gal(L/F) is a normal subgroup of Gal(L/K), and that in that case, Gal(F/K) is the quotient group Gal(L/K)/Gal(L/F).

For one direction, suppose that F/K is normal. Every automorphism of L sends a to an element with minimal polynomial m(a), so the normality of F implies that if a is in F, then so is g(a) for every g in Gal(L/K). In that case, g can be regarded as a homomorphism from F to itself, i.e. an automorphism of F.

Hence there is a group homomorphism from Gal(L/K) to Gal(F/K) defined by restriction to F. The kernel of the homomorphism is $\{g \in textrm{Gal}(L/K): \forall a \in F, g(a) = a\} = \textrm{Gal}(L/F)$, making Gal(L/F) a normal subgroup since kernels are normal. The homomorphism is surjective by a counting argument, so Gal(F/K) = Gal(L/K)/Gal(L/F).

For the other, if E and F are two intermediate extensions such that Gal(L/E) = H and Gal(L/F) = J, then there exists g in Gal(L/K) with g(E) = F iff there exists g in Gal(L/K) with gHg^(-1) = J. To see why, suppose that g(E) = F; then for every h in H and a in F, $ghg^{-1}(a) = g(h(g^{-1}(a))) = g(g^{-1}(a)) = a$ since $g^{-1}(a) \in E$, and by a counting argument J = gHg^(-1). Conversely, if gHg^(-1) = J, then for every a in E and h in H, $ghg^{-1}(g(a)) = gh(a) = g(a)$ so J fixes g(E) and g(E) = F.

Thence, if Gal(L/F) is normal, then for every g in Gal(L/K), gGal(L/F)g^(-1) = Gal(L/F). Thus g(F) = F, or else Gal(L/F) is not normal because different intermediate fields have different Galois groups. But if F is not normal, there exists some a in F that shares a minimal polynomial with some b in L but not in F. By the same argument I used to prove separable normal extensions satisfy L^Aut(L/K) = K, there exists some g in Gal(L/K) such that g(a) = b, which is a contradiction.

To summarize, the fundamental theorem of Galois theory states that the following are equivalent for a finite extension L/K:

1. L/K is separable and normal, i.e. Galois;

2. L/K is the splitting field of a separable polynomial;

3. L^Aut(L/K) = K;

4. |Aut(L/K)| = [L:K].

For an intermediate field F of a Galois extension L/K, the following are equivalent:

1. F/K is normal;

2. F/K is Galois;

3. Gal(L/F) is normal in Gal(L/K), in which case the quotient group is Gal(F/K);

4. For every g in Gal(L/K), g(F) = F.

Among other things, if L/K is Galois, then every polynomial over L that is invariant under the action of Gal(L/K) has coefficients in K. In algebraic number theory, that’s the easiest way of proving that the norm and trace of an element do indeed belong to the ground field (i.e. Q, classically).

Next time, I’ll be more concrete and compute a few Galois groups, as well as point out a few general computationsl tricks.

### 18 Responses to Galois Theory: The Fundamental Theorem

1. muppt says:

Galois correspondence, me like

2. SLC says:

Time for Mr. Levy to leave off bashing Israel and American politicians for a while and aim his big guns at Conservapedia, particularly their inane entries on mathematics.

3. Tyler DiPietro says:

Time for Mr. Levy to leave off bashing Israel and American politicians for a while and aim his big guns at Conservapedia, particularly their inane entries on mathematics.

I actually think it would be more fun to go over there and write insidious paraodies to undermine the site itself. But your idea is cool too!

4. Alon Levy says:

I’m trying to ignore it on the grounds that it shouldn’t be taken seriously, but I’ll see what I can come up with.

5. muppt says:

wanna overthrow the Wikipedia ruling class leaded by the evil libertarian burgeois, Ayn Rand objectivistic rich capitalist Jimmy Wales, then sign up at

http://www.wikipediareview.com

6. muppt says:

you should also read this shocking essay that exposes zionists’s iron grip on the heart of wikipedia

http://robertlindsay.blogspot.com/2006/04/wikipedia-ziopedia-or-judeopedia.html

in fact, there’s actully a website called ziopedia

7. muppt says:

and the real ziopedia is actully an anti-zionist website, not a pro-zionist website.

8. Tyler DiPietro says:

Yes, the Jews control everything, even the blogosphere. Lindsay Beyerstein? Erza Klein? ALON FUCKING LEVY? SHIT, RUN!

9. […] Theory: Examples of Galois Groups Armed with the fundamental theorem of Galois theory, the aspiring mathematician can now go and compute Galois groups of some easy […]

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