If $L/K$ is purely inseparable and $M/K$ is arbitrary, then $|\mathrm{Spec}(L \otimes_K M)| = 1$

Question

In Ravi Vakil's book "The Rising Sea", Exercise 10.5.H (c) states that if $L/K$ is a purely inseparable extension of fields and if $M/K$ is any field extension, then $L \otimes_K M$ has precisely one prime ideal. I am wondering if it is possible to prove this by only appealing to things discussed in the book at that point. In particular, dimension theory and transcendence theory only appear later.

It is not terribly hard to see that any prime ideal of $L \otimes_K M$ is maximal: The ring extension $M \subseteq L \otimes_K M$ is integral, so if $\mathfrak{p} \trianglelefteq L \otimes_K M$ is prime, then $M \subseteq (L\otimes_K M)/\mathfrak{p}$ is an integral extension of integral domains with one ring being a field, implying that both are fields.

However, I do not really see how to use pure inseparability to show that there can be only one prime ideal. It is enough to show that the nilradical is prime, which is for example done in Bourbaki's Algebra (V, 17.2 Prop. 1, p.139), but they use transcendence bases and in general, the argument seems quite complicated to me.

Does anyone know of a more direct proof of this fact?

Answer 1

$L/K$ purely inseparable implies $\operatorname{char} K=p>0$ and for every $\ell\in L$ there exists $n$ such that $\ell^{p^{n}}\in K$.

For $x=\sum_i \ell_i\otimes m_i\in L\otimes_K M$, for $n$ sufficiently large one has $x^{p^{n}}=\sum_i \ell_i^{p^{n}}\otimes m_i^{p^{n}}\in K\otimes_K M\cong M$.

Let $\mathfrak p\in\operatorname{Spec}(L\otimes_K M)$. Since $M$ is a field and $M\subset L\otimes_K M$, we have $\mathfrak p\cap M=(0)$.

Hence $x^{p^{n}}\in\mathfrak p \iff x^{p^{n}}=0$.

If $\mathfrak p,\mathfrak q\in\operatorname{Spec}(L\otimes_K M)$, then $x^{p^{n}}\in\mathfrak p \iff x^{p^{n}}=0 \iff x^{p^{n}}\in\mathfrak q$.

Since prime ideals are radical, this implies $x\in\mathfrak p \iff x\in\mathfrak q$.

Therefore $\mathfrak p=\mathfrak q$, and hence $|\operatorname{Spec}(L\otimes_K M)|=1$.

Answer 2

The following is not really a "direct" proof to be honest, but I find it conceptually pleasing. It just uses the definition of an inseparable polynomial and goes the usual way "primitive extension $\leadsto$ finite extension $\leadsto$ algebraic extension".

First assume that $L=K(\alpha)$ is a simple extension, then $L = K[t]/\langle t^q-\kappa\rangle$ for $q$ some power of $\operatorname{char}(K)$. Then $L \otimes_K M \cong M[t]/\langle t^q-\kappa\rangle$. Since $t^q-\kappa$ has only one solution in $\overline K$, it also has only one solution in $\overline{M}$, so $\operatorname{Spec} M[t]/\langle t^q-\kappa\rangle$ is a point.

Next, if $L$ is a general finite purely inseparable extension, then it is a tower of such simple extensions. Now use $L\otimes_K M = \dots\otimes_{K(\alpha,\beta)} (K(\alpha,\beta)\otimes_{K(\alpha)} (K(\alpha)\otimes_K M))$, and that base change commutes with "taking the reduced subscheme"$^{[1]}$ $$(X \otimes_S Y)^{\mathrm{red}} = (X \otimes_S Y^{\mathrm{red}})^{\mathrm{red}} \qquad \text{assuming }X=X^{\mathrm{red}}$$ to conclude that $\operatorname{Spec}(L \otimes_K M)$ is a point.

Finally, every purely inseparable field extension is the direct limit (union) of its finite subextensions $L = \operatorname{colim}_{L'/K\text{ fin.}} L'$, so $L \otimes_K M = \operatorname{colim}_{L'/K\text{ fin.}} L' \otimes_K M$. Now use the easy fact that a direct limit of local rings is local to conclude that $L \otimes_K M$ is local. As you already note, this implies that $\operatorname{Spec}(L \otimes_K M)$ is a single point and we win.

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$[1]$ This result is [Görtz-Wedhorn's AG I, Prop 4.34], but it also easily follows from the results in The Rising Sea up to that point: Reduced subscheme is surjective closed embedding, and both of these properties are stable under base change, the first by Exercise 10.4.G.