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I'm trying to find a simple counterexample that proves that, given a ring A with prime characteristic, the Frobenius endomorphism is not surjective in general. Are there any elegant examples?

Wild Feather
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You can think about $\mathbb{F}_p(X)$. Then $X$ is not in the image of the Frobenius endomorphism.

ChocoSavour
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  • Thank you, but I was looking for a simpler example, it seems that this one is the only one that I can find. Maybe there isn't a simpler one? – Wild Feather Jul 06 '20 at 20:53
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    @WildFeather Maybe you should have a look on the "perfect field", in the particular case where the characteristic of the field is equal to $p > 0$. It's closely related to the surjectivity of the Frobenius endomorphism. By the way, as any finite field is perfect, I think $\mathbb{F}_p(X)$ is a simple example of non perfect field. As you can't choose a finite field, it seems natural to add a transcendental element and obtain $\mathbb{F}_p(X)$, which is clearly non perfect. I don't know what you mean by "simpler one", but it seems very natural to look at $\mathbb{F}_p(X)$. – ChocoSavour Jul 06 '20 at 21:32