If I pull back a cycle of codimension $c$ along a morphism of schemes I can easily see that the codimension can stay the same or the codimension can drop to all the way to $0$. But intuitively it seems the codimension can never increase. Is there a precise statement of this form?

To state it more carefully suppose $X,Y$ are irreducible schemes and $f:X \to Y$ is a morphism and $Z\subset Y$ is an irreducible subscheme of codimension $c$. Is it always true that if $T\subset f^{-1}(Z)$ is an irreducible component then $codim(T) \leq codim(Z)$?

If this is not true then are there reasonable hypothesis under which it is true, $X,Y$ Noetherian, finite type, smooth?

If $f$ is flat the codimension should be preserved but the inequality above seems to be true more generally.


This is false as stated: taking for $f$ the embedding of a closed subscheme $X\subset Y$, it would mean that $\operatorname{codim}(X\cap Z,X)\leq \operatorname{codim}(Z,Y) $. There are well-known counter-examples: let $Y_0$ be a projective variety with two divisors $D_1,D_2$ which do not intersect, let $Y$ be the cone over $Y_0$, and let $Z,X$ be the subcones over $D_1$ and $D_2$.

To correct the statement you need to assume that $f$ is surjective. Then it is true if moreover $X$ and $Y$ are locally noetherian: this is Corollary 6.1.4 in EGA IV (Part 2).


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.