0

Let $X$ be a scheme.

  • if the underlying space of $X$ is the disjount union of spaces $S_1$ and $S_2$, is $X$ necessarily a disjoint union of schemes with underlying spaces $S_1$ and $S_2$?
  • if the underlying space of $X$ is the direct product of spaces $S_1$ and $S_2$, is $X$ necessarily a direct product of schemes with underlying spaces $S_1$ and $S_2$?
  • if the ring of global functions on $X$ is the direct product of commutative unital rings $R_1$ and $R_2$, is $X$ necessarily a disjoint union of schemes with rings of global functions $R_1$ and $R_2$?
  • if the ring of global functions on $X$ is the tensor product (over $\mathbb{Z}$) of commutative unital rings $R_1$ and $R_2$, is $X$ necessarily a direct product of schemes with rings of global functions $R_1$ and $R_2$?

More generally, what about finite (co)limits?

I believe that the answer to the first question is yes; the ringed space $(S_1\sqcup S_2, \mathcal{O}_X|_{S_1}\oplus \mathcal{O}_X|_{S_2})$ is isomorphic to $X$ as a ringed space.

  • You are right about the first one (provided by disjoint union you really mean the coproduct in the respective category). What are your thoughts about the other questions? – asdq May 04 '19 at 14:40
  • 1
    @asdq I think the rest should fail. –  May 04 '19 at 14:44
  • note that for affine schemes, the third bullet point is true (https://math.stackexchange.com/q/501056/631975). –  May 04 '19 at 15:31
  • If $X$ is a scheme, let $|X|$ be the underlying topological space. It is very rare that $|X\times Y| = |X|\times |Y|$, which one can see by the definition of the product of schemes (consider the product topology on $\Bbb A^1\times\Bbb A^1$ versus the topology on $\Bbb A^2$) - it may even be true that if $|X|=|S_1|\times |S_2|$, one of $S_1$ or $S_2$ must be simple enough to make your claim true. – KReiser May 04 '19 at 18:17

0 Answers0