Take a Noetherian affine scheme. Can it be covered by finitely many disjoint irreducible locally closed affine subschemes?

An example: take the 4-dimensional affine space over a field and consider the union of two planes interesecting at a single point, a cover is {the first plane, the second plane minus a line passing through the point of intersection, the line passing through the point of intersection minus the point of intersection}.

  • $\begingroup$ For $X$ an algebraic set then $X = \bigcup_j X_j$ with $X_j\cong$ to an affine set and $X= \bigcup_j Y_j$ with $Y_j=X_j- (X_j \cap \bigcup_{i < j} X_i)$, if $Y_j$ is not irreducible then $\dim(Y_j) < \dim(X_j)$ so you can repeat the same decomposition with $Y_j$ instead of $X$, it will converge. $\endgroup$ – reuns Jul 23 at 13:22

Yes, and the assumption that $X$ is affine is unnecessary. Let $X$ be a Noetherian scheme, which we may assume is nonempty. By Noetherian induction, we may assume the result is known for every closed proper subscheme of $X$. Let $X_1,\dots,X_n$ be the irreducible components of $X$ and let $U=X\setminus (X_2\cup \dots\cup X_n)$. Then $U$ is open in $X$ and dense in $X_1$ and hence irreducible. Let $V$ be any nonempty affine open subset of $U$, which will still be irreducible. By the induction hypothesis, we can decompose $X\setminus V$ as a finite union of irreducible locally closed affine subschemes. Adding $V$ to this decomposition, we get a decomposition of $X$.

  • $\begingroup$ the base case of induction is when $X$ is a scheme whose underlying space is a point? $\endgroup$ – user690882 Jul 24 at 22:39
  • $\begingroup$ In general, Noetherian induction does not need a base case. In this particular case, the argument is different when $X$ is empty and so you can consider that as a base case. $\endgroup$ – Eric Wofsey Jul 24 at 22:44

Your Answer

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