2
$\begingroup$

Let $E/L,\ L/F$ be fields extensions and $a\in E$ algebraic over $F$. Then,

$$\text{Irr}_{(a,L)}(x)|\text{Irr}_{(a,F)}(x)$$ where $\text{Irr}_{(a,L)}(x),\text{Irr}_{(a,F)}(x)$ are the irreducible polynomials of $a\in E$ over $L,F$ respectively.

Proof. If we take the evaluation epimorphisms in $a$: $\text{ev}_a:L[x]\longrightarrow L[a],\ f(x)\mapsto f(a)$ and $\text{ev*}_a:F[x]\longrightarrow F[a],\ g(x)\mapsto g(a)$ then $\ker \text{ev}_a=\langle \text{Irr}_{(a,L)}(x) \rangle$ and $\ker \text{ev*}_a=\langle \text{Irr}_{(a,F)}(x) \rangle$.

Τhen, if $f(x)\in \ker \text{ev*}_a \implies f(a)=0_F$ with $f(x) \in F[x]$. But, $F[x] \subseteq L[x]$ so $f(a)=0_F$ with $f(x) \in L[x] \implies f(x) \in \ker \text{ev}_a$. So, $$\ker \text{ev*}_a\subseteq \ker \text{ev*}_a \iff \langle \text{Irr}_{(a,L)}(x) \rangle \subseteq \langle \text{Irr}_{(a,F)}(x) \rangle \iff \text{Irr}_{(a,L)}(x)|\text{Irr}_{(a,F)}(x).$$

Is this proof completely correct?

$\endgroup$
  • 1
    $\begingroup$ I think you have reversed the inclusion relationship here as $ev^{\star}_a$ is just $ev_a$ restricted to $F[x]$. So you should have $Ker(ev^{\star}_a)\subset Ker(ev_a)$. $Ker(ev^{\star}_a)$ is an ideal of $F[x]$ not $L[x]$. I guess you are treating this a module but you are treating this as an ideal. So you should have $Ker(ev^{\star}_a)L[x]\subset Ker(ev_a)$. $\endgroup$ – user45765 Aug 5 '17 at 0:41
  • $\begingroup$ Thank you for your comment. You were right, I fixed it. $\endgroup$ – Chris Aug 5 '17 at 19:26
  • $\begingroup$ Please do not have $\LaTeX$-only titles - they make it impossible to right click > open in new tab. $\endgroup$ – anon Aug 6 '17 at 16:47
  • $\begingroup$ @anon Ok, fixed! Thank you. $\endgroup$ – Chris Aug 7 '17 at 2:01

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.