TITLE: If $f(x)$ is irreducible in $k[x]$, why is it also irreducible in $k(t)[x]$, for $t$ an indeterminate?
QUESTION [5 upvotes]: 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.

REPLY [9 votes]: 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.