Let $K$ be a field with characteristic $p>0$ and $M=K(X,Y)$ the field of rational functions in 2 variables over $K$. We consider the subfield $L=K(X^p,Y^p)\subset M$.

Show that $[M:L]=p^2$.

I guess I need the property that $[K(x):K(x^n)]=n$, which we showed already. But I actually do not know how to use it here. I can not work well with that field in 2 variables..

  • $\begingroup$ Perhaps you are expected to use a similar proof technique to the proof of $[K(x):K(x^p)]=p$, rather than to use this result. $\endgroup$ – ajotatxe Dec 10 '14 at 15:15

Recall that for some field $J$ so that $L \subset J \subset M$ you have that the degree of the extension $L \subset M$ is the product of the degrees of the extensions $L \subset J $ and $ J \subset M$.

Use this for example with $J=K(X^p,Y)$, applying the result you know twice.

  • $\begingroup$ Thanks for the answer! But one problem: The statement $[K(x):K(x^n)]=n$ is true if x is transcendental.(I forget to say that above). How we can use it for our case? Because polynomials with degree $\geq 1$ are transcendental over a given field K? $\endgroup$ – Epsilondelta Dec 10 '14 at 16:06
  • $\begingroup$ Note that $J$ can be seen as $K(X^p)(Y)$ and also as $K(Y)(X^p)$; and $K(X^p)$ and $K(Y)$ are fields. $\endgroup$ – quid Dec 10 '14 at 16:39
  • $\begingroup$ Okay, so we can write $K(X^P)(Y^p)\subset K(X^p)(Y) \subset K(X)(Y)$. This is what you want to say? Fine :) .But still..can I say now that $Y^p$ is transcendental oder $Y$? If yes, why? $\endgroup$ – Epsilondelta Dec 10 '14 at 16:45
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    $\begingroup$ This is not true in general. It can happen when the characteristic of $K$ is $0$ (and I think but not sure also if the characteristic of $K$ is not $p$); in char 0 case the extension is definitely separable and the primitive element theorem implies that the extension is generated by a single element. However, if the characteristic of $K$ is $p$, which could well be the case in your context, then consider $f^p$ and recall freshman's dream en.wikipedia.org/wiki/Freshman%27s_dream to see that $f^p$ in $L$ and thus $\deg f = p$. (Perhaps better ask new q if there are further things). $\endgroup$ – quid Dec 10 '14 at 19:02
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    $\begingroup$ Sorry there was some misunderstanding. It is true for characteeristic p. I also said so in my comment. Note the part "However, if the characteristic is p..." I only had forgotten that this assumption was stated in your original question as for the original question it is not really necessary; it would work for any field. But for the follow up it is need and Without it is not true. And, this is the first part of my comment. $\endgroup$ – quid Dec 10 '14 at 22:24

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