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Let $p$ be a prime. The p-adic numbers $\mathbb{Q}_p$ are an algebra under the $p$-adic integers $\mathbb{Z}_p$ via the localization $\mathbb{Z}_p\to \mathbb{Z}_p[\frac{1}{p}]=\mathbb{Q}_p$. Is $\mathbb{Q}_p$ a $\mathbb{Z}_p$-algebra of finite type? Geometrically the map above corresponds to the open immersion $\operatorname{Spec}(\mathbb{Q}_p)\to \operatorname{Spec}(\mathbb{Z}_p)$ and I thought that an open immersion is almost never of finite type and only locally of finite type.

user228077
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1 Answers1

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Any nonzero element in $\mathbb{Q}_p$ can be represented in a unique way as $p^ku$, where $u\in\mathbb{Z}_p$ is invertible (in $\mathbb{Z}_p$) and $k$ is an integer.

If $x_1,\dots,x_n$ are elements $\mathbb{Q}_p$, we can write them as $p^my_i$, where $y_i\in \mathbb{Z}_p$, by taking the “common denominator”. Assume $m<0$, so the submodule $M$ they generate properly contains $\mathbb{Z}_p$. Then $p^{m-1}\notin M$.

This shows that $\mathbb{Q}_p$ is not a finitely generated module over $\mathbb{Z}_p$. However, it is finitely generated as algebra, because just adding $p^{-1}$ suffices to generate it.

egreg
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  • Thank for your answer egreg. Why isn't $\mathbb{Q}_p=\mathbb{Z}_p[X]/(pX-1)$ and thus of finite type? – user228077 Apr 07 '15 at 23:36
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    The question was pretty clear: Is $\mathbb{Q}_p$ a $\mathbb{Z}_p$-algebra of finite type? and the answer is Yes. Most of your answer has nothing to do with the question. – user26857 Apr 08 '15 at 07:21
  • @user26857 Sorry for the confusion about the previous comment (if you had a proper nickname… ;-)). Since “finite type” can have different meanings, I showed both possibilities. – egreg Apr 08 '15 at 07:28