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.