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Denote by $\mathbb{Z}_p$ the ring of $p$-adic integers. Is $\mathrm{Spec}(\mathbb{Z}_p)$ of finite type over $\mathrm{Spec}(\mathbb{Z})$?

user26857
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user46578
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2 Answers2

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No; $\mathbb Z_p$ is uncountable, but any finitely-generated $\mathbb Z$-algebra is countable.

Bruno Joyal
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The obvious (and morally correct) answers is Bruno's. But, just to be amusing, I thought I'd throw in this one which uses ever so slightly more algebro-geometric language:

Suppose that $\text{Spec}(\mathbb{Z}_p)\to\text{Spec}(\mathbb{Z})$ is of finite type. Then, because both $\text{Spec}(\mathbb{Z}_p)$ and $\text{Spec}(\mathbb{Z})$ are Noetherian, we may apply Chevalley's theorem to say that the image of $\text{Spec}(\mathbb{Z}_p)$ in $\text{Spec}(\mathbb{Z})$ is constructible. But, the image is precisely $\{(0),(p)\}$ which is not constructible. Contradiction.

Alex Youcis
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