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Let $S$ be a finite set of maximal ideals in $ O_K$, where $O_K$ is the ring of integers of some number field $K$. Define $A= O_K[S^{-1}]$.

Let $X$ be an arbitrary $A$-scheme. Consider the scheme $X_A=X\times_{\mathbb Z} A$ as an $A$-scheme via the second projection.

Is $X_A$ the disjoint union of at most $[K:\mathbb Q]$ copies of $X$?

Edit: I rewrote the question to make it more clear.

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    $\begingroup$ Question 1: This is just abstract nonsense: if a morphism factors through a monomorphism, then pulling back (or, base change, if you like) along that monomorphism doesn't do anything. $\endgroup$
    – Zhen Lin
    Commented Jul 27, 2014 at 16:17
  • $\begingroup$ Dear Zhen Lin, thanks for pointing this out. $\endgroup$
    – Jinbi Wang
    Commented Jul 27, 2014 at 16:23

1 Answer 1

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The answers to question 2 are "no". Take $R = A = \mathbb{Z}[x]/(x^2)$, and let $X = \operatorname{Spec} A$. There is no ring isomorphism between $A \otimes A$ and $A \oplus A$.

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