# Irreducibility of the polynomial $f (x) = x^{2p} − x ^p + t$

If $k$ is a ﬁeld of characteristic $p > 0$ and $f (x) = x^{2p} − x ^p + t \in k(t)[x]$, how can we show that $f (x)$ is an irreducible polynomial in $k(t)[x]$ and that $f (x)$ is inseparable?

If $f(x)$ is irreducible then $D_xf(x) = 0 \implies( f, D_xf) = ( f, 0) = f$. Hence, if $f(x)$ is irreducible then $( f, D_xf) \neq 1 \implies f(x)$ is inseparable.

• Homework problem? If it is, people will not go beyond giving you a hint. Mar 4 '13 at 15:34
• @Lubin no sir it is not a homework . a hint will do.
– jim
Mar 4 '13 at 15:38
• I think that $f'=0$ suffices to show it is inseparable Mar 4 '13 at 15:45

Here’s a hint: notice that $k(t)=k(\tau)$ when $\tau=1/t$. If $\rho$ is a root of $f$, what polynomial in $k(\tau)[x]$ does $1/\rho$ satisfy?

• sir i have posted an answer could you tell me is it right?
– jim
Mar 5 '13 at 3:05

$K$ is a field $\implies K[t]$ is a U.F.D.

By gauss lemma: $f(x)$ is irreducible in $K(t)[x] \iff f(x)$ is irreducible in $K[t][x]$

then $K[t][x]= K[x][t]$

$f(x)$ is irreducible in $K[x][t]$ being a linear polynomial in $K[x][t]$

hence $f(x)$ is irreducible in $K(t)[x]$

• Nice answer. In Gauss Lemma, $f$ should be primitive.
– user18119
Mar 5 '13 at 20:14
• Very nice indeed, and much quicker than what I had in mind! Mar 8 '13 at 20:17