# Let K/F be a finite extension, given a polynomial in K[x] find another so that their product is in F[x]

Let $K$ be a finite extension of a field $F$, and let $f(x)$ be in $K[x]$. Prove that there is a nonzero polynomial $g(x)$ in $K[x]$ such that $f(x)g(x)$ is in $F[x]$.

Should I do this by induction on the degree of $f(x)$?

Obviously if $n=0$, then $g(x)=1/f(x)$

Let $f(x) = a_nx^n+...a_1x+a_0$ then I know that there exists a h(x) so that $(f(x)-a_nx^n)h(x)$ is in $F[x]$. I want now to find a $g(x)=h(x)+i(x)$ so that $f(x)g(x)$ is in $F[x]$. Thus I need to find an $i(x)$ so that $a_nx^nh(x)+i(x)f(x)$ is in $F[x]$. I feel like this is wrong because I have no control over the degrees of $h(x)$.

Any suggestions?

• Is this homework? Hint: focus on the case of irreducible $f(x)$ in $K[x]$ and think about minimal polynomials of the same number algebraic over different fields.
– KCd
Commented Mar 17, 2012 at 19:18
• I removed the tag ring-theory ; I believe it is unappropriate. Commented Mar 17, 2012 at 21:20
• @KCd Yes it is. I updated the tags. Commented Mar 17, 2012 at 21:48
• @Steven-Owen : I posted a solution of a slightly general version using field theory only . If you may take a look Commented Apr 26, 2017 at 14:24

It becomes easy if you know about integrality over rings. (Here and in the following, "ring" always means "commutative ring with $1$".)

Since $K$ is a finite extension of $F$, we see that $k\in K$ is integral over $F$ for every $k\in K$. Thus, $k\in K\left[x\right]$ is integral over $F\left[x\right]$ for every $k\in K$. Hence, $kx^i\in K\left[x\right]$ is integral over $F\left[x\right]$ for every $k\in K$ and $i\in\mathbb N$ (since $k$ and $x^i$ are both integral over $F\left[x\right]$, and the product of two integral elements is integral). Hence, $f\left(x\right) \in K\left[x\right]$ is integral over $F\left[x\right]$ (since $f\left(x\right)$ is a sum of elements of the form $kx^i$ for $k\in K$ and $i\in\mathbb N$, and since the sum of integral elements is integral). Now, all we need to prove is the following fact:

(1) If $A\subseteq B$ is a ring extension, and $u\in B$ is integral over $A$, then there exists a nonzero $v\in B$ such that $uv\in A$.

Proof of (1). Let $n$ be the smallest positive integer such that there exists a monic polynomial $P\in A\left[Y\right]$ of degree $n$ satisfying $P\left(u\right) = 0$. (Such an $n$ exists since $u$ is integral over $A$.) Write the polynomial $P$ in the form $P\left(Y\right) = \sum\limits_{i = 0}^{n - 1}a_iY^i + Y^n$ with all $a_i$ lying in $A$. Then, set $v = \sum\limits_{i = 1}^{n - 1}a_iu^{i - 1} + u^{n - 1}$. Then, $uv = \sum\limits_{i = 1}^{n - 1}a_iu^{i} + u^{n} = \underbrace{\sum\limits_{i = 0}^{n - 1}a_iu^{i} + u^{n}}_{ = P\left(u\right) =0} - a_0 = - a_0 \in A$. Also, $v\neq 0$ follows from the minimality of $n$. Thus, (1) is proven.

[I have copied some of this from one of my AoPS posts.]

• I removed the tag ring-theory, because I believe it was inappropriate. Yet your argument uses integrality over rings, which is something usually one does not cover when beginning Galois theory (or maybe I am judging, but I have faith in what I'm saying). Do you think you can do a purely field/Galois theory proof by inspiring yourself from your integrality argument? Commented Mar 18, 2012 at 23:00
• In my opinion, the fact that my proof does not depend on Galois theory and fields is more of a feature than a bug; the focus on fields is antiquated and precludes one from exploiting the flexibility of rings. (Natural and useful constructions such as direct sums, quotients, polynomial rings, tensor products etc. of rings are not possible with fields.) Commented Mar 18, 2012 at 23:19
• This is exactly why I had put the ring theory tag on there as well. My course begins with rings and it seemed as though this problem could be solved using those methods, and here we have it. Thanks Darij! Commented Mar 21, 2012 at 3:02
• I wish I could upvote this amazingly elegant and perfectly written answer more: congratulations, Darij! Commented Mar 27, 2012 at 22:57

Assume that the extension is Galois. By assumption, there are only finitely many automorphisms of $K$ that fix $F$. The polynomial $$\omega(x) = \prod_{\sigma \in \mathrm{Aut}(K/F)} \sigma(f(x))$$ is a polynomial such that $f(x) \, | \, \omega(x)$ and $\omega$ is fixed by every automorphism of $K/F$, because $\mathrm{Aut}(K/F)$ is a group, so that when one tries to apply an automorphism on $\omega(x)$, the conjugate factors get permuted... hence all the coefficients of $\omega$ lie in the fixed field $F$. You can choose $g(x) = \omega(x) / f(x)$.

Hope that helps,

• I worked out the case where the extension is Galois. Can you work out the other cases? Commented Mar 17, 2012 at 19:00
• Hello Patrick, Thanks so much for the help. Just to make sure when you say $\sigma(f(x))$ does this act like $\sigma(a_nx^n \ldots a_1x+a_0)=\sigma(a_n)x^n \ldots \sigma(a_1)x+\sigma(a_0)$? Commented Mar 17, 2012 at 19:23
• @jake : Yes, exactly. Automorphisms go through addition/multiplication, so you just apply those rules to apply the automorphisms to $f(x)$. Although it's true that a priori $\sigma : K \to K$, my writing suggests that you can naturally extend this function to $\sigma : K[x] \to K[x]$ by letting $\sigma(x) = x$. When I write $\sigma(f(x))$ I suppose you understand this natural way of extending $\sigma$. Commented Mar 17, 2012 at 19:26
• I think the place where you used the Galois property was in knowing that $w(x)$ is in $F[x]$ because it is fixed by group, this is not something that is true in general for galois groups of non-Galois extensions, right? Commented Mar 17, 2012 at 20:34
• @jake : Don't need to be sorry, your questions were fine and I didn't feel overwhelmed or bothered or anything like that. Feel free to keep asking questions ; although at the moment it is better if you continue asking questions in chat if you have to. Commented Mar 17, 2012 at 20:38

A proof using field theory and assuming only that $K/F$ is an algebraic extension .

Let $K^a$ be an algebraic closure of $K$ , $f(x)\in K[x]$ splits in $K^a[x]$ as $f(x)=(x-a_1)...(x-a_n)$

where $a_1,..,a_n \in K^a$ ($a_i$'s are not necessarily distinct) . Each $a_i$ is

algebraic over $K$ and $K$ is algebraic over $F$ , so each $a_i$ is algebraic over $F$ ; let $g_{a_i}(x)=Irr(a_i , F)$

be the irreducible polynomial of $a_i$ over $F$ . Then $h(x)=\prod_{i=1}^n g_{a_i}(x) \in F[x]$ is a non-zero

polynomial . Moreover , $x-a_i|g_{a_i}(x)$ in $K^a[x]$ , so $f(x)|h(x)$ in $K^a[x]$ . Now $f(x),h(x) \in K[x]$ ;

by division algorithm , $\exists q(x) , r(x)\in K[x]$ such that $h(x)=f(x)q(x)+r(x)$ where either

$r(x)=0$ or $\deg r(x)< \deg f(x)$ . But then , as $f(x),h(x),q(x),r(x) \in K^a[x]$ and $f(x)|h(x)$ in

$K^a[x]$ , so $f(x)|r(x)$ in $K^a[x]$ , then $\deg r(x) <\deg f(x)$ is not possible , hence $r(x)=0$ . Hence

$h(x)=f(x)q(x)$ , where $h(x) \in F[x]$ is a non-zero polynomial and $f(x) , q(x) \in K[x]$ ,

which is what we wanted to prove .