Hartshorne III Proposition 9.5 states:

Let $f:X\to Y$ be a flat morphism of schemes of finite type over a field $k$. For any point $x\in X$ let $y=f(x)$. Then $$\dim_x(X_y)=\dim_x X-\dim_y Y$$ where for any scheme $X$, by $\dim_x X$ we mean the dimension of the local ring at $\mathcal{O}_{x,X}$.

The proof involves several statements about base change which I am unable to understand. First, making the base change $Y'\to Y$ where $Y'=\text{Spec } \mathcal{O}_{y,Y}$ we get $X'=X\times_Y Y'$ and $f':X'\to Y'$. We lift $x$ to $X'$ and Hartshorne states "the three numbers in question are the same." I do not understand at all why they would be the same, and unfortunately I don't understand base change very well either intuitively or from a technical perspective, so I don't really know how I would go about seeing this.

Later on, a base extension is made to $Y_{\text{red}}$ and again "nothing changes" according to Hartshorne. Also it is mentioned that the fibre $X_y$ does not change under base extension.

Can someone help me understand these statements and why they should be so clear (but aren't)?


Base "extension" is not a great description of this example, I agree. What Hartshorne is doing geometrically is restricting, not extending.

For example, if $U\subset Y$ is an open set containing $y$, we can "base extend along $U \to Y$ and get $X_U = X \times_Y U$", but what we really mean is we are restricting the morphism to $f^{-1}(U)$ (this is what $X_U$ is). Of course, this does not change the fiber $X_y$ or the numbers in question, since $U$ is an open neighborhood of $y$ (so we still have a safe 'window' around $y$ to look at dimensions in, and in any case the fiber $X_y$ is ultimately obtained by "restricting to just $y$" -- it doesn't matter if we decide to first pass to a neighborhood of $y$.)

What Hartshorne has done is to first restrict the morphism to Spec of the local ring of $y$. This is like passing to an arbitrarily small neighborhood of $y$, just like in the example above.

Then, he passes to the reduction, since reducing a ring does not change its dimension.

  • $\begingroup$ Thanks, this helped me gain some intuition. I guess when confused, pretend that the fibre product is the same as it is in the category of sets to get some idea of what is supposed to be going on. Unfortunately, I still don't really have any formal understanding here. $\endgroup$ – Seth Mar 15 '15 at 0:29
  • $\begingroup$ Ah, I think I understand it formally now, I don't think it's actually as hard as I thought. $\endgroup$ – Seth Mar 20 '15 at 15:00
  • $\begingroup$ Glad to hear it! $\endgroup$ – Jake Levinson Mar 21 '15 at 14:40

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.