Irreducible components of the seminormalization

Question

Let $X$ be a scheme with finitely many irreducible components $V_1,\dots,V_r\subset X$. Its normalization $X^\nu\to X$ is the morphism obtained as follows: $$X^\nu=\coprod_iV_i^\nu\to X_{red}\to X.$$ Here $V_i^\nu$ is the normalization of $V_i$. They are the irreducible components of $X^\nu$.

However, there is an intermediate scheme, the semi-normalization of $X$, which fits in $$X^\nu\to X^{sn}\to X.$$ There are of course factorizations $$V_i^\nu\to V_i^{sn}\to V_i.$$

Question. Is it true that $V_i^{sn}$ are the irreducible components of $X^{sn}$?

Thanks!

Answer 1

I think your question is equivalent to this one:

Is it true that the irreducible components of a seminormal scheme are themselves seminormal?

A curve is seminormal if it is locally analytically isomorphic to the coordinate axis in an affine space. Any subset of that has the same property (for a smaller dimensional space) so a partial answer is

Yes, if $\dim X=1$.

However, this fails in higher dimensions, because there exist reducible seminormal schemes whose irreducible components are not seminormal. The examples are not entirely simple. See (7.9.3) and (10.12) of the reference below. So the answer is

No, in general.

On the other hand, there is an almost obvious Chinese remainder theorem type exact sequence connecting the seminormalization of the scheme, the seminormalization of its irreducible components and the semi-normalization of the pairwise intersections of the irreducible components. (I suppose one can probably continue that sequence with more intersections, so it's perhaps more like a Cech resolution type exact sequence).

A good resource for the statements I mentioned above is section 10.2 of this book by Kollár.