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I'm having a problem solving this problem from dummit & foote.

prove that $F_p(x,y)/F_p(x^p,y^p)$ is not a simple extension n by explicitly exhibiting an infinite number of intermediate sub-fields

Honestly i have no idea where to start.

Moshe Levy
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  • I hope your $F_p$ denotes the finite field on $p$ elements, isnt it? – vidyarthi Jul 25 '19 at 12:20
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    Here are some intermediate fields: $\Bbb F_p(x^p,y^p,x)$, $\Bbb F_p(x^p,y^p,y)$, $\Bbb F_p(x^p,y^p,xy)$. Can you show that these are indeed intermediate, and are pairwise different? Can you guess a pattern to find infinitely many such fields?, – Hagen von Eitzen Jul 25 '19 at 12:23
  • @vidyarthi yes of course – Moshe Levy Jul 25 '19 at 12:54
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    @HagenvonEitzen but wouldn't it stop at $x^p$ or $ y^p$? and hence not infinite? – Moshe Levy Jul 25 '19 at 12:55
  • @MosheLevy I think you could keep extending the pattern, for example, by setting the next field $\mathbb{F}_p(x^p,y^p,x^py,xy^p)\ldots\mathbb{F}_p(x^p,y^p,x^ny^n)$ for $n$ approaching to infinty (a multiple of $p$)? – vidyarthi Jul 25 '19 at 13:04
  • @vidyarthi you are probably right but why are they different? could i get a hint on how to show that they are not equal? – Moshe Levy Jul 25 '19 at 13:08
  • @MosheLevy the field $\mathbb{F}_p(x^p,y^p,x^py)$ for instance consists of all the linear combinations of the elements inside the brackets, which would not have, for example the term $xy^p$, which would be present in $\mathbb{F}_p(x^p,y^p,xy^p)$ – vidyarthi Jul 25 '19 at 13:11
  • @vidyarthi thank you – Moshe Levy Jul 25 '19 at 13:35
  • I seem to have handled this question here. Thankfully I don't have a dupehammer here, so I can vote to close as a duplicate freely. See here for a more thorough discussion by people who know more than I do. – Jyrki Lahtonen Jul 27 '19 at 21:38

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