Suppose K is a finite field extension of $\mathbb{Q}$. Let K ⊆ L be a Galois field extension and K ⊆ K′ be a finite field extension. Show that K′ ⊆ K′L is a Galois field extension and $$\text{Gal}(K′L \ / \ K′) \cong \text{Gal}(L\ / \ L ∩ K′)$$

I have seen this stated in the theorem that allows us to conclude that a polynomial is solvable by radicals if and only if its Galois group is solvable, however, I'm not able to formulate a proof of this intermediate result.


I will use the notation $K'/K$ for $K\subseteq K'$.

First show that $K'L/K'$ is a normal extension: we know that $L$ is the splitting field of some polynomials $f_i$ with coefficients in $K$. I claim that $K'L$ is the splitting field over $K'$ of the same polynomials.

Let $E=K'L$. The field $E$ is the splitting field of the $f_i$ if the following hold:

  1. The $f_i$ decompose into linear factors over $E$.

  2. If $E/F/K'$ is a tower of field extensions such that the $f_i$ decompose into linear factors over $F$, then $E=F$.

Part 1 is easy. For part 2, use that $K'L$ is the smallest field containing both $K'$ and $L$.

Since we are over $\mathbb{Q}$, everything is separable. Hence the extension $K'L/K$ is a Galois extension.

Next define the restriction map

$$\mathrm{Aut}(K'L/K')\to\mathrm{Aut}(L/L\cap K')\colon \ \sigma\mapsto\sigma|_L.$$

We want to show that this is an isomorphism. To do this, we need to show that each automorphism $\tau$ of $L/L\cap K'$ extends uniquely to an automorphism $\sigma$ of $K'L$ that fixes $K'$.

Write $K'=K(\alpha_1,\ldots,\alpha_n)$ with algebraic $\alpha_i\in K'$. Then $K'L=L(\alpha_1,\ldots\alpha_n).$ Now we can extend $\tau$ uniquely by setting $\alpha_i\mapsto \alpha_i$.

In general, let $L/K$ be a normal finite field extension, let $\sigma\colon K\to K$ be an autormorphism. We want to find a field extension $L'/K$ and an homomorphism $\sigma'\colon L\to L'$ such that $\sigma'|_K = \sigma$ (we say that $\sigma'$ extends $\sigma$).

The plan is to do it stepwise, so we break the extension into simple extensions


with the $\alpha_i\in L$ being algebraic over $K$. We start by extending $\sigma$ to $\sigma_1\colon K(\alpha_1)\to L_1$ for some field $L_1/K$ and we continue by choosing $\sigma_1$ as the new starting point, so that we arrive inductively at $\sigma'\colon L\to L'$ .

Hence we may assume that $L=K(\alpha)$ is a simple field extension.

Let $f$ be the minimal polynomial of $\alpha$, and let $f^\sigma$ be the polynomial obtained by $f$ after applying $\sigma$ to the coefficients. Choose some element $\alpha'$ of the algebraic closure of $K$ such that $f^\sigma(\alpha)=0$. Then we can define $\sigma'\colon K(\alpha)\to K(\alpha')$ by setting $\alpha\mapsto\alpha'$.

To see that this is well-defined, note that this map corresponds to the map $K[X]/(f)\to K[X]/(f^\sigma)$ that sends $X$ to $X$ and an element $\lambda$ in $K$ to $\sigma(\lambda)$.

In our special case, we see that at every step of the construction, we have $f^{\sigma}=f$, so that we don't actually have to change fields, and at every step we can just map $\alpha_i$ to $\alpha_i$. Hence in our case we end up with an automorphism of the big field, which is $K'L$.

  • 1
    $\begingroup$ I would like to add that the last step uses a more general procedure to extend field homomorphisms of field extensions. A good reference for this is in Bosch, Algebra. Maybe I will outline the general construction if it is needed. $\endgroup$ May 25 '16 at 8:41
  • $\begingroup$ I don't have access to that text, can you outline the general construction? $\endgroup$
    – user319128
    May 25 '16 at 8:58
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    $\begingroup$ I edited the question to explain what I mean in the last step. In field theory, there are many similar constructions to this one (also more general ones, etc.). See for example these questions: math.stackexchange.com/questions/897660/…, math.stackexchange.com/questions/1370880/… $\endgroup$ May 25 '16 at 9:53

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