Show that a polynomial is irreducible but not separable over a particular field

Hi, this was a problem on an exam, and I wasn't sure how to do it.

The problem is as follows:

Let $\displaystyle f(y)=y^p-x \in Z_p(x)[y]$, where $\displaystyle p$ is prime. Show that $\displaystyle f(y)$ is irreducible in $\displaystyle Z_p(x)$.

I'm lost on this one. I feel like I should assume that it's reducible, which implies that $\displaystyle f(y)$ can be factored into irreducibles of lesser degree, but somehow there aren't any products of irreducibles that give us $\displaystyle f(y)$. Unfortunately, I don't know how to show the "somehow" part :)

Hope someone can help! Thanks!

Re: Show that a polynomial is irreducible but not separable over a particular field

you do know that in a field F with char(F) = p, that:

$\displaystyle (y - a)^p = y^p - a^p$, right?

in particular, $\displaystyle \sqrt[p]{x}$ is the ONLY root of $\displaystyle y^p - x$, in any extension of $\displaystyle \mathbb{Z}_p(x)$.

so it suffices to show $\displaystyle \sqrt[p]{x} \not \in \mathbb{Z}_p(x)$.

for suppose it were. then we would have:

$\displaystyle x = \frac{(p(x))^p}{(q(x))^p}$ for some polynomials $\displaystyle p(x),q(x) \in \mathbb{Z}_p[x]$.

suppose deg(p) = k, and deg(q) = m. then deg(p(x)^{p}) = kp, and deg(q(x)^{p}) = mp.

thus we have:

1 + mp = kp

but p divides kp, and cannot divide 1 + mp. so we have our contradiction.

Re: Show that a polynomial is irreducible but not separable over a particular field

Ah, yes! I do know that, I just didn't make the connection that the pth root will be the only root. Thank you!!