0

Let $f:X\to Y$ be a morphism of finite type of (Noetherian) schemes . Is $X$ of finite dimension when $Y$ is? Could you please provide a reference, or a counter example?

It is rather obvious when $Y$ is the spectrum of a field, but I am not sure how to approach it in general.

user24453
  • 357
  • 1
  • 8
  • 1
    Can you do this when $X=\mathrm{Spec}B, Y=\mathrm{Spec} A$? Then, can you reduce to this case? – Mohan Jul 15 '17 at 00:14
  • 1
    Keeping in line with Mohan's recommendation, I imagine this is true given that any finitely generated algebra over a Noetherian ring of dimension $n$ should have dimension bounded above by $n+$ (number of variables) which is necessarily finite. Im saying that this is an upper bound since any finitely generated algebra should look like some quotient of $R[x_1, \dots,x_n]$ and quotient rings should have smaller dimension. – user7090 Jul 15 '17 at 18:43
  • 1
    Yes, this is basically the (well known I guess) statement $\dim R[X]=1+\dim R$ for a *noetherian* ring $R$. – MooS Jul 17 '17 at 06:35
  • Thank you, that answers my question (maybe I should've thought about it longer before posting). – user24453 Jul 17 '17 at 15:17

0 Answers0