Show the nilradical of $F_p[x] \otimes_{F_p[x^p]}F_p[x]$ is principal

Question

Here are my thoughts:

Suppose $F_p[x]$ is flat over $F_p[x^p]$. Then applying $- \otimes_{F_p[x^p]}F_p[x]$ to the natural injection $F_p[x^p] \to F_p[x]$ shows that the map $F_p[x] \to F_p[x] \otimes_{F_p[x^p]}F_p[x]$ given by $f \to f \otimes 1$ is injective.

So now if $a \in F_p[x] \otimes_{F_p[x^p]}F_p[x]$ is in the nilardical, then $a^p$ is also in the nilradical. But we can easily see that $a^p$ is a sum of elements of form $f(x^p) \otimes g(x^p)$ and each such sum has the form $h \otimes 1$ so then $h$ is in the nilradical of $F_p[x]$ by the above injection, so $a^p=0$.

I am not sure how to proceed from there, or even if $F_p[x]$ is actually flat over $F_p[x^p]$.

Any thoughts?

EDIT: OK, Hurkyl's idea kills it quickly and easily, but I'd still like to see if the above idea pans out.

Answer 1

I would start by computing the tensor product. Letting $y = x^p$ and $R = F[y]$, we have

$$ F[x^p] = R \qquad \qquad F[x] = R[t] / (t^p - y) $$

and so

$$ F[x] \otimes_R F[x] \cong R[s,t] / (t^p - y, s^p - y) \cong F[x,s]/(x^p - s^p) $$