Exercise 93 in Joseph Rotman's /Galois Theory/ (2nd ed. 1998):

 "Show that Z_p(x, y) is a finite extension of its subfield
 Z_p(x^p, y^p), but it is not a simple extension."

My head still swims when I do abstract algebra (although I'm
slowly becoming more confident and more able to check my own
work). The following proof seems vaguely OK, but I'm not very
confident that I haven't written gibberish somewhere, and/or
made the proof too complicated.  Also, from the placement of
the exercise in the text (after a series of applications of
the Fundamental Theorem of Galois Theory), I thought we were
expected to use Steinitz's theorem (on simple extensions being
those with finitely many intermediate fields) - but I didn't
use it.

Someone please deliver either a pat on the head or a smack on
the backside, so that I know whether I've been a good boy or
a bad boy!

Proof:

Z_p[x, y] is a ring of characteristic p, therefore we have
(f(x, y) + g(x, y))^p = f(x, y)^p + g(x, y)^p, for all f(x, y),
g(x, y) in Z_p[x, y].  Therefore, the set {f(x, y) : f(x, y)^p
= f(x^p, y^p)} is a subring of Z_p[x, y] containing Z_p and x
and y, therefore it is the whole of Z_p[x, y].

Therefore, for all f(x, y)/g(x, y) in Z_p(x, y), we have:

 f(x, y)/g(x, y) = f(x, y)g(x, y)^{p-1}/g(x^p, y^p)

therefore Z_p(x, y) = Z_p(x^p, y^p)[x, y]. This clearly has the
finite basis x^iy^j (i, j = 0, 1, ..., p-1) over Z_p(x^p, y^p).
Therefore, [Z_p(x, y) : Z_p(x^p, y^p)] = p^2.

On the other hand, by what has just been proved, every element
t of Z_p(x, y) satisfies a polynomial equation of the form
t^p - c = 0, where c is in Z_p(x^p, y^p), therefore:

 [Z_p(x^p, y^p)(t) : Z_p(x^p, y^p)] <= p

therefore Z_p(x, y) =/= Z_p(x^p, y^p)(t).

Q.E.D.

(My head's still swimming, so I'd better post this, to see if
I've made an ass of myself.)