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:
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.)
> 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:
> 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.
Seems fine to me. The extension [K:k] is of degree p^2. If u is in K then u^p is in k. Hence u is of degree p (or 1).
>> 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."
>> [...]
>> Proof:
>> [...]
>Seems fine to me.
That's a relief. Although I still couldn't see anything wrong with the proof, I was getting more and more worried, because it doesn't actually use any Galois theory!
>The extension [K:k] is of degree p^2. >If u is in K then u^p is in k. >Hence u is of degree p (or 1).
On that last point: the first draft of my post said that [k(u):k] = p or 1, but then I began to worry that it might not be true, or at least wasn't as "obvious" as I'd imagined, and I didn't need it, so I left it out, giving [k(u):k] <= p < p^2 instead. Now that I'm obliged to think about it, I think I've adapted a solution to an earlier exercise (no. 75) to prove it. I was going to post this, too - because I was worried that I was making unnecessarily heavy weather of something that might perhaps, after all, be "obvious" - but I see there's a standard result that has it as a special case: From D. J. H. Garling, /Galois Theory/ (1986), p. 88:
"Theorem 10.8 Suppose that char K = p > 0 and that
f(x) = g(x^p) = a_0 + a_1x^p + ... + x^{np}
is monic; then f is irreducible in K[x] if and only if g is irreducible in K[x], and not all of the coefficients a_i are pth powers of elements of K."
> >> 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."
> >> [...]
> >> Proof:
> >> [...]
> >Seems fine to me.
> That's a relief. Although I still couldn't see anything wrong with > the proof, I was getting more and more worried, because it doesn't > actually use any Galois theory!
> >The extension [K:k] is of degree p^2. > >If u is in K then u^p is in k. > >Hence u is of degree p (or 1).
> On that last point: the first draft of my post said that [k(u):k] = > p or 1, but then I began to worry that it might not be true, or at > least wasn't as "obvious" as I'd imagined, and I didn't need it, so > I left it out, giving [k(u):k] <= p < p^2 instead. Now that I'm > obliged to think about it, I think I've adapted a solution to an > earlier exercise (no. 75) to prove it. I was going to post this, > too - because I was worried that I was making unnecessarily heavy > weather of something that might perhaps, after all, be "obvious" - > but I see there's a standard result that has it as a special case: > From D. J. H. Garling, /Galois Theory/ (1986), p. 88:
> "Theorem 10.8 Suppose that char K = p > 0 and that
> f(x) = g(x^p) = a_0 + a_1x^p + ... + x^{np}
> is monic; then f is irreducible in K[x] if and only if g is > irreducible in K[x], and not all of the coefficients a_i are pth > powers of elements of K."
On Thu, 24 Jul 2008 07:49:35 -0700 (PDT), Mariano Suárez-Alvarez
<mariano.suarezalva...@gmail.com> wrote: >You should worry less... :-)
You caught me just as I was busy worrying that perhaps I should post another followup, explaining that I didn't mean to suggest that that general result was needed to prove the special case - and then detailing my proof, and worrying some more about the different proof of a slightly more general result that I found in another book ... It's sunny outside. I'm going shopping. No more maths. :-)
