# Galois group - extend homomorphism to automorphism

Let $K \subset L$ be a finite Galois extension, $M$ a field with $K \subset M \subset L$ and $G := \text{Aut}(L/K)$.

I want to show that if $\sigma \, \colon M \longrightarrow L$ is a $K$-homomorphism there exists a $\tau \in G$ with $\tau\vert_M=\sigma$.

$K$-homomorphism means that $\sigma\vert_K=\text{id}$.

$\\ \\$

$\textbf{My ideas:}$

I thought to use the Isomorphism extension theorem, which I know in the following form:

Let $\varphi \, \colon K \longrightarrow K'$ be a field isomorphism, $f \in K[X]$ a polynomial with a root $x$ in a field extension of $K$ and $x'$ a root of $\varphi(f)$ in a field extension of $K'$.

Then there exists exactly one field isomorphism $\phi \, \colon K[x] \longrightarrow K'[x']$ with $\phi(x)=x'$ and $\phi\vert_K=\varphi$.

But I think this doesn't help in this situation.

• What have you tried? Where do you struggle? Also what do you mean by $K$- homomorphism? I'm unfamiliar with this notation; does that mean it fixes $K$? – DanZimm May 25 '14 at 0:30
• Are the extensions finite? – Tom Oldfield May 25 '14 at 9:13
• @user148364 I'm fairly sure that there are infinite Galois extensions, although I don't know how they're categorised. – Tom Oldfield May 25 '14 at 9:38

## 2 Answers

Suppose all degrees are finite.

We first prove that there are at most $\lvert M : K \rvert$ $K$-homomorphisms $M \to L$. Proceed by induction on $\lvert M : K \rvert$; suppose we know that there are $\le \lvert N : K \rvert$ $K$-homomorphisms $N \to L$. Let $\alpha \in L \setminus N$, with minimal polynomial $g \in N[X]$ of degree $k > 1$ over $N$. Then $\lvert N(\alpha) : N \rvert = k$, and any $\sigma : N \to L$ can be extended to at most $k$ $K$-homomorphisms $N(\alpha) \to L$, as $\alpha$ can only be mapped in one of the $k$ roots of $g$. Hence there are at most $\lvert N(\alpha) : N \rvert \cdot \lvert N : K \rvert = \lvert N(\alpha) : K \rvert$ $K$-homomorphisms $N(\alpha) \to L$.

Now consider the restriction of the elements $\tau \in Aut(L/K)$ to $M$. (Of course these are not necessarily automorphisms of $M$, but rather $K$-homomorphisms $M \to L$.) The restrictions of $\tau, \tau'$ coincide on $M$ iff $\tau Aut(L/M) = \tau' Aut(L/M)$. So by Lagrange there are $$\frac{\lvert Aut(L/K) \rvert}{\lvert Aut(L/M) \rvert} = \frac{\lvert L:K \rvert}{\lvert L : M \rvert} = \lvert M : K \rvert$$ such restrictions.

So there are exactly $\lvert M : K \rvert$ $K$-homomorphisms $M \to L$, and each of these is the restriction to $M$ of an element of $Aut(L/K)$.

• This is really nice, it is very reminiscent of the fact that $L/K$ is Galois iff $[L:K] = \#Aut(L/K)$, and I would never have thought to solve this by counting! – Tom Oldfield May 25 '14 at 19:01
• @TomOldfield, thanks! Actually, one can probably concoct a proof not dissimilar from yours from just the first part. – Andreas Caranti May 26 '14 at 8:32

You can use the key theorem you state, if $L/M$ is finite then $L = M(x_1,\dots,x_n)$ for some $x_i$ in $L$. Let $M_0 = M$ and $M_{i+1} = M_i(x_{i+1})$.

The crucial part of this question is that $L/K$ is Galois, which for finite extensions is equivalent to being normal and separable. This means that the minimal polynomial of each of the $x_i/K$ splits completely into linear factors in $L$. Thus the minimal polynomial of each of the $x_i$ over $M_{i-1}$ also splits completely in $L$.

So, if we let $f_i(X)$ be the minimal polynomial for $x_i$/$M_{i-1}$, then for example $f_1(X)$ $\sigma(f_1)$ splits completely in $L$*. Thus we can extend $\sigma$ to some $\sigma_1:M_1\rightarrow L$ by the theorem, because we have shown that we have a root of $\sigma(f_1)$ in $L$. We can repeat this process iteratively until we have extended $\sigma$ up through each extension to $L$.

As pointed out in the comments, my original argument for this was invalid. We can prove this by taking $g_i(X)\in K[X]$ to be the minimal polynomial of $x_i$ over $K$, $f_i|g_i$, $g_i$ splits completely in $L$ and is fixed by $\sigma$ since it lies in $K[X]$. Thus $\sigma(f_i)|g_i$ also, so it too splits completely in $L$.

• @user148364 For fields, any homomorphism is injective. (This is true since given any homomorphism of rings, the kernel is an ideal and the only ideals in a field are ${0}$ and the field itself, for example. There are other ways to show this too.) So $\sigma$ gives an isomorphism onto its image $\sigma(K)$ in $L$, and $L$ is a field extension of $\sigma(K)$ as required in the statement of the theorem. The theorem is almost always used when you have a homomorphism from one field into another one, since you can always apply what I have mentioned here. – Tom Oldfield May 25 '14 at 10:49
• @user148364 Yes, I did, sorry, I automatically use $K$ and $L$ for fields without thinking. To see that the image is all of $L$, we know $M(x_1,\dots,x_n)=L$ $\tau$ is a homomorphism from $L$ to $L$. It is injective (since all field homomorphisms are) and since $L$ is a finite dimensional $K$ vector space (this is what it means for the extension to be finite) $\tau$ is a linear map from $L$ to $L$ (thinking of $L$ as a $K$ vector space). For linear maps on vector spaces of the same dimension, being injective is equivalent to being surjective so the image of $\tau$ is all of $L$. – Tom Oldfield May 25 '14 at 11:53
• @user148364 No problem, happy to help! – Tom Oldfield May 25 '14 at 12:00
• @TomOldfield, could you please clarify the sentence $f_1(X)$ splits completely in $L$ and so $\sigma(f_1)$ does also? It is clear that if $\sigma(f_1)$ has a root in $L$, then it splits completely in $L$. But why does it have to have a root in $L$? – Andreas Caranti May 25 '14 at 17:24
• @AndreasCaranti Because if $f_1(X) = (X-\lambda_1)\dots(X-\lambda_m)$, $\sigma(f_1)(X)=(X-\sigma(\lambda_1))\dots(X-\sigma(\lambda_m))$. In fact, in general, if $f(X) \in K[X]$ has $\alpha$ as a root and $\sigma:K\rightarrow K'$ is a homomorphism, $\sigma(f)(X) \in k'[X]$ has $\sigma(\alpha)$ as a root, since $\sigma(f)(\sigma(\alpha)) = \sigma(f(\alpha))=\sigma(0)$. – Tom Oldfield May 25 '14 at 18:03