Show that $[\mathbb{Z}_p(x,y):\mathbb{Z}_p(x^p,y^p)]=p^2$ and that $\mathbb{Z}_p(x,y)$ is not a simple extension of $\mathbb{Z}_p(x^p,y^p)$

Question

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.

Answer 1

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.

Answer 2

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 $p^2$.