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Question: Let $x,y$ be independent indeterminates over $\mathbb{Z}_p$, $K=\mathbb{Z}_p(x,y)$ and $F=\mathbb{Z}_p(x^p,y^p)$. Show that $[K:F]=p^2$ and that $K$ is not a simple extension of $F$.

This question (at least the second part) is asked and answered here: $F(x,y)$ over $F$ is not simple, but I thought this was an interesting way of approaching it, by first showing $[K:F]=p^2$. I figured that the minimal polynomial of $x^p$ over $K$ has degree $p$ and same with $y^p$ (I suppose we can say this since the indeterminates are independent of each other), thus we get $[K:F]=p^2$. But, I know that a simple extension has degree of the minimal polynomial, so, and I am not quite sure about this, is the degree of the minimal polynomial $p$, and so we get the contradiction? I am just getting a bit lost, but I figured it was a worthwhile way of showing this.. Any help is greatly appreciated! Thank you.

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    $\begingroup$ I don't see how your argument uses the base field. If you replace $\mathbb{Z}_p$ with $\mathbb{Q}$, then you'd have $[K:F]=p^2$ and the extension is simple. (Also note your linked problem doesn't apply here, as in that case $x,y$ were a transcendence basis for $F(x,y)$ over $F$ while in your case the extension is finite) $\endgroup$ Oct 13, 2021 at 0:06
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    $\begingroup$ Related. $\endgroup$ Oct 15, 2021 at 5:04
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    $\begingroup$ It is easy to find several threads where this extension is discussed. Many (most) of those simply state that this extension is not simple, and also that there are infinitely many intermediate extensions (those two are related). That is how standard an example this is! $\endgroup$ Oct 15, 2021 at 5:09

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Let $K=\mathbb{Z}_p(x^p,y^p)$, $F=\mathbb{Z}_p(x,y^p)$, $L=\mathbb{Z}_p(x,y)$ then $L=F(y)$, $K(x)=F $. Observe that, $p(t) = t^p-x^p$ is a polynomial in $K[t]$. Because, $(x^p)$ is a prime ideal in $K$, $x^p \in (x^p)$ but $x^{p} \notin (x^p)^2$. Then, the Einsenstein Criterion for Irreduciblity holds and $p(t)$ is an irreducible polinomial over $K[t]$ and $p(x)=0$. Hence, $p$ is the minimal polinomial for $F/K$ and $[F:K]=p$. Analogously, using the same argument $q(t)=t^p-y^p$ is an irreducible polynomial over $F[t]$, then $q(t)$ is the minimal polynomial for $L/F$.

$$[L:K]=[L:F][F:K]=p^2$$

It's well-known fact that field extension $L/K$ is simple if only if there exist finitely many subfields. Consider the family of subfields $K(f(y^p)x+y)$ with $f\in \mathbb{Z_p}[x]$. Suppose that $f_1\neq f_2$ and $K(y+f_1(y^p)x)=K(y+f_2(y^p)x)$.

$$x= \frac{(y+f_1(y^p)x)-(y+f_2(y^p)x)}{f_1(y^p)-f_2(y^p)} \in K(y+f_1(x^p)x)$$ Then $y = (y+f_1(y^p)x) - f_1(y^p)x \in K(y+f_1(y^p)x)$. Consequently $K(y+f_1(y^p)x)=K(x,y)$. But consider the polinomial $h(t)=t^p-(y^p+f_1(y^p)^px^p)$. Observe that $h(y+f_1(y^p)x)=0$ then $$[K(y+f_1(y^p)x):K] \leq p.$$ This is a contradiction, hence $K(y+f_1(y^p)x) \neq K(y+f_2(y^p)x)$. Finally, $\{K(y+f(y^p)x)\}_{f \in \mathbb{Z}_p[x]}$ is infinite family of intermediate subfields, therefore $L$ cannot be a simple extension.

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Here is an alternate way to see this result. Within any characteristic $p$ field $L$, we have the Frobenius endomorphism $x\mapsto x^p$, and the image of this is naturally a sub field $L^p$ of $L$.

Then by construction, any element of $L$ satisfies a polynomial equation of degree at most $p$ over $L^p$, just take $x^p-a^p$ for any $a$ in $L$.

So now if the extension of $L/L^p$ is of degree more than $p$, it can’t have a primitive element, since all possible degrees of minimal polynomials are either $p$ or $1$ depending on whether the element is a power of $p$.

Your example is precisely this, for the field $\mathbb{Z}_p(x,y)$, your subfield is the subfield of $p$th powers. To see this, just note that the Frobenius map is replacing $f(x,y)$ by $f(x^p,y^p)$, since it’s a homomorphism. But then we are done, since you already computed the degree of this to be $p^2$.

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