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In many books I have seen that the function field is defined for integral schemes. However, many times they refer in further chapters for a function field of the irreducible component of scheme (not necessarily reduced). For example, here in the answer we have only the assumption that the variety is irreducible, while in Liu's book a variety is not assumed to be reduced. So, my question is: do we always assume in such cases that we induce the reduce structure on the irreducible components? Or in closed irreducible subsets in general?

T. Wildwolf
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  • You can always define the ring of rational functions by localizing with respect to all elements that are not zero divisors. – Sasha Dec 17 '22 at 17:47
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    The linked question deals with dimension, which is a topological property that does not depend on the scheme structure. So picking the reduced structure is okay. If you have questions on material that's not from the linked question, please add more references: right now your question is somewhat vague. – KReiser Dec 17 '22 at 17:53
  • I will try my best to find other examples, but don't we need a reduced scheme to prove the result in the answer of the link? I mean, he even uses the notion of function field for irreducible varieties and not necessarily integral schemes... – T. Wildwolf Dec 17 '22 at 17:56
  • So @KReiser , can you answer my question in the previous comment, please? – T. Wildwolf Dec 17 '22 at 20:10
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    As I detailed in my previous comment, the dimension does not depend on the scheme structure. So one may fix the linked answer by saying "since dimension does not depend on the scheme structure, we may take $V$ to be reduced" then applying the argument verbatim. – KReiser Dec 17 '22 at 20:11
  • Ok I got it but in the answer he used an argument which involves the function field, and function fields are defined at least for integral schemes, so it was not so clear. So, for proving topological properties, we can have the reduced scheme structure for an irreducible closed subscheme without loss of generality, right? – T. Wildwolf Dec 17 '22 at 20:20
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    Yes, you're correct, there was a little bit missing from the linked answer, but as I mentioned before it's a very quick and short fix. And also as I've said before, topological properties are independent of the scheme structure, so it is alright to pick the reduced structure if you are only interested in topological properties. – KReiser Dec 17 '22 at 20:30

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