0
$\begingroup$

Let $k$ be a field, $f(x)\in k[x]$, and let $F$ be the splitting field of $f(x)$ over $k$. Let $k\subseteq K$ be an extension such that $f(x)$ splits as a product of linear factors over $K$. Prove that there is a homomorphism $F\to K$ extending the identity on $k$.

This is a question from Aluffi's Algebra Chapter 0.

It looks to me as if here, $K$ is just the algebraic closure of $F$. Thus, a homomorphism would just be the inclusion. Is this all there is to it or am I missing something?

$\endgroup$
  • $\begingroup$ $K$ is just any extension of $k$ in which the polynomial splits. It need not be the algebraic closure of $F$. For example, you could have $k=\mathbb{Q}$, $f(x) = x^2-2$, and $K=\mathbb{Q}[\sqrt[4]{2}]$. $K$ is not the algebraic closure of $F$. Or it could be a field that does not obviosuly include $\mathbb{Q}[\sqrt{2}]$. For example, it could be $\mathbb{Q}[x]/(x^4-10x^2+1,x^2-3)$. $\endgroup$ – Arturo Magidin Apr 11 at 22:48
  • $\begingroup$ @ArturoMagidin I see your point. So it is not necessary that $F \subseteq K$, correct? How else can I construct a homomorphism, or in other words, where can I map the roots of $f$ to in $K$? $\endgroup$ – blanchey Apr 11 at 23:02
  • $\begingroup$ Well, the polynomial $f(x)$ is supposed to split in $K$, so there are some natural candidates... $\endgroup$ – Arturo Magidin Apr 12 at 1:43

Your Answer

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

Browse other questions tagged or ask your own question.