If $f(x)$ is irreducible in $k[x]$, why is it also irreducible in $k(t)[x]$, for $t$ an indeterminate?

I've been thinking on this a few days, but I'm stuck.

Let $k$ be a field, $f(x)$ irreducible in $k[x]$. Why is $f(x)$ also irreducible in $k(t)[x]$, for $t$ an indeterminate?

I write $f(x)=c_0+c_1x+\cdots+c_nx^n$ for $c_i\in k$. Suppose that $f(x)$ is reducible in $k(t)[x]$, so $$f(x)=(a_0(t)+a_1(t)x+\cdots+a_r(t)x^r)(b_0(t)+b_1(t)x+\cdots+b_s(t)x^s)$$ for $r,s>0$. Since $c_0=a_0(t)b_0(t)$, $a_0(t)$ and $b_0(t)\in k$. Same goes for $a_r(t)$ and $b_s(t)$.

I tried something like letting $b_j(t)$ be the least index such that $b_j(t)\in k$. Then $$c_j=a_0b_j+a_1b_{j-1}+\cdots \text{ and } c_j-a_0b_j=a_1b_{j-1}+\cdots$$ where $c_j-a_ob_j\in k$, and the right hand side is the sum of polynomials in $k(t)$ of degree at least $1$. I don't see if there is a contradiction to be had. What's the right approach? Thanks.

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If it's no trouble, would the downvoter please explain the downvote? – yunone Jul 30 '12 at 2:58

If $k$ is infinite you do not need Gauss's lemma (in fact you hardly need any tools at all). Let us start the same way as you started: suppose $f$ admitted a nontrivial factorization in $k(t)[x]$: $$f(x) = (a_0(t) + ... + a_n(t) x^n)(b_0(t) + ... + b_m(t) x^m).$$

Pick $c \in k$ such that the denominators of each $a_i(t)$ and $b_i(t)$ are nonzero when $t = c$ and such that the numerators of $a_n(t), b_n(t)$ are also nonzero (this is always possible iff $k$ is infinite). Then

$$f(x) = (a_0(c) + ... + a_n(c) x^n)(b_0(c) + ... + b_m(c) x^m)$$

is a nontrivial factorization of $f$ in $k[x]$.

(The takeaway lesson here is that a polynomial in $k(t)[x]$ really lies in $k[t][x]$ localized away from finitely many elements of $k[t]$.)

If $k$ is finite you can do something similar but sneakier: pass to the algebraic closure $\bar{k}$ to find $c$ as above. By choosing $c$ more carefully you can deduce that either the $a_i$ are all constant multiples of each other (resp. the $b_i$) or $f$ admits a nontrivial factorization in $\bar{k}[x]$ with an irreducible factor having a Galois orbit larger than the degree of $f$, which is a contradiction.

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:) ............. – paul garrett Jul 30 '12 at 1:05
Thanks Qiaochu, I like this approach. – yunone Jul 30 '12 at 2:56

Another vierpoint: if $f(x)$ were reducible in $k(t)[x]$, by Gauss' lemma it would be reducible in $k[t][x]$. Thinking of $k(x)[t]$, and perhaps invoking Gauss again, a factorization in $k[t][x]$ could only involve elements of degree $0$ in $k(x)[t]$, that is, elements in $k(x)$. (The relevant special case of Gauss' lemma can be reproven directly here, of course.)

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Note that $k[x]$ is a UFD, and so $f$ irreducible in $k[x] \implies (f)$ is a prime ideal. Thus $k[x]/(f)$ is an integral domain. Adjoining $t$ gives us $(k[x]/(f))[t]\cong k[t][x]/(f)$ and this is still an integral domain. Localizing at the set $k[t]$ gives $k(t)[x]/(f)$, which is still an integral domain. Thus $(f)$ is a prime ideal of $k(t)[x]$, so $f$ is irreducible in $k(t)[x]$.

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Localizing $k[t]$ at $\{1,t,t^2,\dots\}$ does not give $k(t)$... – Bruno Joyal Jul 29 '12 at 22:06
@Bruno Good point. – Alex Becker Jul 29 '12 at 22:15
Possibly worth thinking how to make the intuition into a genuine proof... If it needed reiterating: physical intuition is often correct, but the translation into "language" is both a non-trivial task, and also may reveal otherwise-tacit assumptions, and so on. That is, in reality, "proof" is iterative, not all-or-nothing, etc. – paul garrett Jul 30 '12 at 1:08
Dear Alex, I think that you want to localize at $k[t] \setminus \{0\}$. Regards, – Matt E Jul 30 '12 at 3:06
@paulgarrett: Dear Paul, This seems like a genuine proof as written to me (modulo the minor typo I pointed out). Is there something I'm missing? Best wishes, – Matt E Jul 30 '12 at 3:07

By Gauss's lemma, $f$ is irreducible in $k(t)[x]$ if and only if it is irreducible in $k[t][x]$. Moreover, you should prove that $$k[x][t]/(f) \cong (k[x]/(f)) [t].$$

The latter is a domain, so $(f)$ is irreducible in $k[x][t]=k[t][x]$, hence also in $k(t)[x]$.

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Thanks Bruno. ${}$ – yunone Jul 30 '12 at 2:55