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4 votes
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If the Frobenius endomorphism of a characteristic $p$ ring is epimorphic, is it surjective?

MO question 19282 is about integral epimorphisms of commutative rings, and a counterexample is given to surjectivity. What about the case of the Frobenius endomorphism of a commutative, characteristic ...
Matthieu Romagny's user avatar
3 votes
0 answers
192 views

How can I prove this stronger version of Fedder's Criterion?

I was reading Fedder's original paper which proved what is now known as "Fedder's criterion". I noticed that the abstract stated something which is a priori stronger than what is proved in ...
Anon's user avatar
  • 317
1 vote
0 answers
82 views

Relative 1 form of Frobenius morphism

Let $X$ be a connected smooth algebraic variety over a field $k$ of characteristic $p > 0$. We can consider the Frobenius morphism associated to $X \mapsto Spec(k)$. I'd like to show $\Omega^{1}_{X/...
Analyse300's user avatar
1 vote
0 answers
55 views

If $ n \in \mathbb{N} $, then does the Reynolds operator of $ \mathbb{G}_{m}^{n} $ commute with the Frobenius endomorphism?

If $ n \in \mathbb{N} $, then $ \mathbb{G}_{m}^{n} $ is linearly reductive. Let $ \beta: \mathbb{G}_{m}^{n} \to \operatorname{GL}(\mathbf{V}) $ where $ \mathbf{V} $ is a vector space over an ...
Schemer1's user avatar
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