Is an affine formal scheme quasi-compact?

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It is well known that an affine scheme $X=mathrmSpec(A)$ is quasi compact. In analogy, what can we say about an affine formal scheme $mathrmSpf(A)$ (here $A$ should be an adic ring and $mathrmSpf(A)$ is defined to be the set of all open primes in $A$ with respect to the adic topology)? Is it also quasi-compact? If the answer is no, what additional conditions can guarantee the quasi-compactness?




algebraic-geometry soft-question schemes affine-schemes




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    Let $I$ be an ideal of definition. Then the underlying topological spaces of $Spf(A)$ and $Spec(A/I)$ are homeomorphic to each other.
    – Rieux
    Jul 14 at 21:45










  • So you mean it is still quasicompact right？
    – Hang
    Jul 14 at 22:07










  • Yes. This follows from what I said.
    – Rieux
    Jul 14 at 22:45














up vote
0
down vote

favorite












It is well known that an affine scheme $X=mathrmSpec(A)$ is quasi compact. In analogy, what can we say about an affine formal scheme $mathrmSpf(A)$ (here $A$ should be an adic ring and $mathrmSpf(A)$ is defined to be the set of all open primes in $A$ with respect to the adic topology)? Is it also quasi-compact? If the answer is no, what additional conditions can guarantee the quasi-compactness?




algebraic-geometry soft-question schemes affine-schemes




share|cite|improve this question















  • 2




    Let $I$ be an ideal of definition. Then the underlying topological spaces of $Spf(A)$ and $Spec(A/I)$ are homeomorphic to each other.
    – Rieux
    Jul 14 at 21:45










  • So you mean it is still quasicompact right？
    – Hang
    Jul 14 at 22:07










  • Yes. This follows from what I said.
    – Rieux
    Jul 14 at 22:45












up vote
0
down vote

favorite









up vote
0
down vote

favorite











It is well known that an affine scheme $X=mathrmSpec(A)$ is quasi compact. In analogy, what can we say about an affine formal scheme $mathrmSpf(A)$ (here $A$ should be an adic ring and $mathrmSpf(A)$ is defined to be the set of all open primes in $A$ with respect to the adic topology)? Is it also quasi-compact? If the answer is no, what additional conditions can guarantee the quasi-compactness?




algebraic-geometry soft-question schemes affine-schemes




share|cite|improve this question











It is well known that an affine scheme $X=mathrmSpec(A)$ is quasi compact. In analogy, what can we say about an affine formal scheme $mathrmSpf(A)$ (here $A$ should be an adic ring and $mathrmSpf(A)$ is defined to be the set of all open primes in $A$ with respect to the adic topology)? Is it also quasi-compact? If the answer is no, what additional conditions can guarantee the quasi-compactness?





algebraic-geometry soft-question schemes affine-schemes





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 14 at 18:26









Hang

395214




395214







  • 2




    Let $I$ be an ideal of definition. Then the underlying topological spaces of $Spf(A)$ and $Spec(A/I)$ are homeomorphic to each other.
    – Rieux
    Jul 14 at 21:45










  • So you mean it is still quasicompact right？
    – Hang
    Jul 14 at 22:07










  • Yes. This follows from what I said.
    – Rieux
    Jul 14 at 22:45












  • 2




    Let $I$ be an ideal of definition. Then the underlying topological spaces of $Spf(A)$ and $Spec(A/I)$ are homeomorphic to each other.
    – Rieux
    Jul 14 at 21:45










  • So you mean it is still quasicompact right？
    – Hang
    Jul 14 at 22:07










  • Yes. This follows from what I said.
    – Rieux
    Jul 14 at 22:45







2




2




Let $I$ be an ideal of definition. Then the underlying topological spaces of $Spf(A)$ and $Spec(A/I)$ are homeomorphic to each other.
– Rieux
Jul 14 at 21:45




Let $I$ be an ideal of definition. Then the underlying topological spaces of $Spf(A)$ and $Spec(A/I)$ are homeomorphic to each other.
– Rieux
Jul 14 at 21:45












So you mean it is still quasicompact right？
– Hang
Jul 14 at 22:07




So you mean it is still quasicompact right？
– Hang
Jul 14 at 22:07












Yes. This follows from what I said.
– Rieux
Jul 14 at 22:45




Yes. This follows from what I said.
– Rieux
Jul 14 at 22:45















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