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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?

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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 Mar 17 '12 at 19:18
I removed the tag ring-theory ; I believe it is unappropriate. – Patrick Da Silva Mar 17 '12 at 21:20
@KCd Yes it is. I updated the tags. – Steven-Owen Mar 17 '12 at 21:48
up vote 1 down vote accepted

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,

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I worked out the case where the extension is Galois. Can you work out the other cases? – Patrick Da Silva Mar 17 '12 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 $\sigma(a_nx^n \ldots a_1x+a_0)=\sigma(a_n)x^n \ldots \sigma(a_1)x+\sigma(a_0)$? – Steven-Owen Mar 17 '12 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 $\sigma$. – Patrick Da Silva Mar 17 '12 at 19:26
Awesome. So in the case where the extension is finite but not Galois, I don't want to just consider some Galois extension G/F that has K as an intermediate field and proceed by the same method because then I would not be guaranteed that $w(x)/f(x)$ is in $K[x]$. Also, I don't think you need the assumption that the extension is Galois to know there are only finitely many automorphisms of K that fix F. This comes just from knowing that K/F is a finite extension and it must be contained in some larger extension that is Galois. – Steven-Owen Mar 17 '12 at 20:28
Yes, but then again you wanted a suggestion and I gave you one. Perhaps you can work the rest out by yourself or wait for someone else to work it out for you, because my Galois theory is rusty ; I gave you my thoughts about this question in $5$ minutes but didn't think it through very much. – Patrick Da Silva Mar 17 '12 at 20:33

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.]

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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? – Patrick Da Silva Mar 18 '12 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.) – darij grinberg Mar 18 '12 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! – Steven-Owen Mar 21 '12 at 3:02
I wish I could upvote this amazingly elegant and perfectly written answer more: congratulations, Darij! – Georges Elencwajg Mar 27 '12 at 22:57

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