1
$\begingroup$

Let $F/k$ be an algebraic extension. Let $S(F/k)$ be the set of all embeddings of F over k into algebraic closure $k^\mathrm{a} $. I'm trying to prove that the smallest normal extension of k containing F is

$E= \displaystyle\prod_{\sigma \in S(F/k)} \sigma F$ ($\Pi$ is used to denote the compositum of fields)

In the finite extension case, matter is simple as if $\tau$ is embedding of E over k, $\sigma \mapsto \tau\sigma$ is an injective mapping of $S(F/k)$ into $S(F/k)$. Since $S(F/k)$ is finite, the above mapping is bijective and $E= \displaystyle\prod_{\sigma \in S(F/k)} \sigma F=\prod{\tau\sigma F} $.

But in the infinite extension case when $S(F/k) $ might not be finite, surjective part must be added to prove that $\tau$ induces permutation on $S(F/k) $. I don't quite get an idea of how I should prove the surjectiveness.

$\endgroup$
0
$\begingroup$

The easiest way to think about this is to not care about the induced map on $S(F/k)$ but instead just look at the field map. If $\tau:E\to k^\mathrm{a}$ is an embedding over $k$, then its image is contained in $E$, and so it gives an endomorphism of $E$ over $k$. Such an endomorphism is automatically surjective: if $E$ is an algebraic extension of a field $k$ and $\tau:E\to E$ is a homomorphism over $k$, then $\tau$ is surjective.

To prove this, consider and $a\in E$. The set $S$ of elements of $K$ which have the same minimal polynomial over $k$ as $a$ is finite, and $\tau$ maps $S$ to itself. Since $\tau$ is injective, this implies $\tau$ must restrict to a bijection from $S$ to itself. In particular, since $a\in S$, $a$ is in the image of $\tau$, and since $a$ was arbitrary, $\tau$ is surjective.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.