some definitions

We work over a fixed algebraically closed field $k$.And $A^{n}$ is an affine n-space over $k$,i.e. $A^{n}=k^{n}$.

An affine variety is an irreducible algebraic set in $A^{n}$.An open subset of an affine variety is called a quasi-affine variety.

My question:

Let $Y$ be a quasi-affine variety,is its closure $\bar Y$ an affine variety?

Here is my argument: Since Y is irreducible,so is $\bar Y$.Also,$\bar Y$ is closed,hence $\bar Y$ is an affine variety.

Is my argument correct? Am I missing something?

Any help will be greatly appreciated,thanks!

  • $\begingroup$ If $Y$ is a quasi-affine variety, then it's the open set of some affine variety $X$ so $\overline Y$ is just $X$ right? $\endgroup$ – Ravi Oct 31 '17 at 3:59
  • $\begingroup$ This looks OK to me. You may want to say a few words about why $Y$ irreducible implies the closure is irreducible. $\endgroup$ – Alfred Yerger Oct 31 '17 at 4:00
  • $\begingroup$ @AlfredYerger I know the reason on why $\bar Y$ is irreducible,just check by definition.I just want to determine whether my argument is true or not ? $\endgroup$ – Jiabin Du Oct 31 '17 at 4:04
  • $\begingroup$ Well, a variety is a closed and irreducible subset. You've shown it's closed, and you've shown it's irreducible. $\endgroup$ – Alfred Yerger Oct 31 '17 at 4:05
  • $\begingroup$ @Ravi Yes.Since any non-empty open set of an irreducible topological space is dense and irreducible. $\endgroup$ – Jiabin Du Oct 31 '17 at 4:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.