The Stacks project

Lemma 33.36.6. Let $p > 0$ be a prime number. Let $S$ be a scheme in characteristic $p$. Let $X$ be a scheme over $S$. Then the relative frobenius $F_{X/S} : X \to X^{(p)}$ is a universal homeomorphism, is integral, and induces purely inseparable residue field extensions.

Proof. By Lemma 33.36.3 the morphisms $F_ X : X \to X$ and the base change $h : X^{(p)} \to X$ of $F_ S$ are universal homeomorphisms. Since $h \circ F_{X/S} = F_ X$ we conclude that $F_{X/S}$ is a universal homeomorphism (Morphisms, Lemma 29.45.8). By Morphisms, Lemmas 29.45.5 and 29.10.2 we conclude that $F_{X/S}$ has the other properties as well. $\square$

Comments (2)

Comment #7878 by Shogōki on

While homeomorphisms have "2 out of 3" property, it's unclear if universal homeomorphism also has the property. (In fact, the one being used in your second sentence is the only unclear one, because one needs to look at base changes that doesn't come from the "most downstairs base"? I tried to see this, and easily reduced to the case where the composition is identity, but then got stuck. Maybe this is already discussed somewhere in the SP?)

In any case, it is still true that F_{X/S} is a universal homeomorphism: WLOG one may assume both X and S are affine, then one can check condition (2) in [0CNF] holds true directly.

Comment #8149 by on

OK, Shizhang Li added the universal homeomorphism has 2-out-of-3 property and then I added a link to that (new) lemma here. If you want your name added to the contributors, please give first+last name.

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  • 6 comment(s) on Section 33.36: Frobenii

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