# Surjective morphism of affine scheme

Let $f\colon X\rightarrow Y$ be a surjective morphism of affine schemes.

Is is true that the dimension of the fibers are equal to $\dim(X)-\dim(Y)$? Same question if $X$ and $Y$ are affine spaces?

Thanks

• How about a blow-up? More generally, what if the morphism is non-flat? What if $X$ and $Y$ are disconnected? – Ariyan Javanpeykar Jul 11 '17 at 21:07
• In my case, I have two affine spaces and a surjective map which given by quadratic polynomials. In this case I assume that the above claim is corrected? isn't it? – Z.A.Z.Z Jul 11 '17 at 21:11
• It's not true for affine schemes: E.g., $X$ and $Y$ disconnected; or as subsets of the plane, $X=V(xy)$, $Y=V(y)$, with projection. – Zach Teitler Jul 11 '17 at 21:16
• Here's some evidence that you should be looking at the adjective "flat": For a flat map, this is true. If $X$ is Cohen-Macaulay and $Y$ is regular, then the fibers being the correct dimension implies $f$ is flat. – KReiser Jul 12 '17 at 17:33

Here is an example (essentially blowing up). The example bertram wrote, which he has since deleted after my comment, can be modified. His example was $(x,y)\mapsto (x,xy)$ which unfortunately is not onto. But if one takes $(x,y)\to (xy, x(y^2+1))$, one can easily check it is a surjection from the plane to the plane and at least one fiber has dimension one.
Again, as bertram says, one has an inequlity, $\dim f^{-1}(y)\geq \dim X-\dim Y$ for all $y\in Y$ and for general $y\in Y$ one has equality.
One true statement in the direction you want is the following. (The hypotheses can surely be relaxed, but I am quoting the version you can find in Shafarevich Volume 1, Chapter I Section 6.) Let $X$ and $Y$ be varieties over an algebraically closed field and $f: X \rightarrow Y$ a morphism. Then every irreducible component of every fibre $f^{-1}(y)$ over a closed point $y \in Y$ has dimension at least $\operatorname{dim} X - \operatorname{dim} Y$.