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I'm trying to do the following exercise from the book An Invitation to Algebraic Geometry:

Show that if $X \to Y$ is a surjective morphism of affine algebraic varieties, then the dimension of $X$ is at least as large as the dimension of $Y$.

Could anyone give me a hint? I'm looking for a proof that avoids commutative algebra since it is not mentionned in the book at this point.

Math536
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2 Answers2

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Let $f:X\to Y$ be your morphism of algebraic varieties over the field $k$.
a) By considering the restricted mapping $f^{-1}(Y_i) \to Y_i$ where $Y_i\subset Y$ is an irreducible component of maximal dimension , you reduce to the case where $Y$ is irreducible .

b) If $X_j$ are the finitely many irreducible components of $X$, you have $Y=\bigcup f(X_j)$, hence $Y=\bigcup \overline {f(X_j)}$, so that one of the $f(X_j)$ is dense in $Y$.
In other words by considering $f_{\big|X_j}: X_j\to Y $ you reduce to the case that $X$ is irreducible but you must weaken yout hypothesis to $f$ is dominant (that is with dense image) instead of $f$ is surjective. Don't worry : we'll still manage!

c) Since the morphism $f:X\to Y$ of irreducible varieties is dominant, it induces a morphism of the corresponding function fields $f^* :\operatorname{Rat}(Y) \to \operatorname{Rat}(X) $ .
This implies by pure algebra (field theory!) the inequality on transcendence degrees over $k$ : $$\operatorname{trdeg}_k (\operatorname{Rat}(Y))\leq \operatorname{trdeg}_k (\operatorname{Rat}(X)) \quad (*)$$
d) Finally, for an irreducible variety $Z$, we know that $\dim (Z)=\operatorname{trdeg}_k (\operatorname{Rat} (Z)$ (this is essentially a corollary of Noether's normalization theorem). If you take this into account, $(*)$ yields the required inequality $$ \dim(Y)\leq \dim(X) \quad (**) $$

Edit The above works for all algebraic varieties, affine or not. Actually I hadn't even noticed that the OP mentioned affine varieties when I answered the question!

Georges Elencwajg
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  • Thank you for your answer. I'm surprised that this question was in the first chapter of the book, but at least I won't lose my time trying to find an elementary solution. – Math536 Jan 02 '12 at 19:07
  • Dear @Math536, yes this is surprising. Of course, I can't prove that there is no elementary solution, but I would be very surprised if there were. Given the numerous extremely competent users on this site, I suggest that if no elementary solution is mentioned here within a few hours, we declare the question settled :-) – Georges Elencwajg Jan 03 '12 at 07:55
  • Does this result fail if we replace "affine variety" with "Noetherian space" and "morphism" with "continuous map"? – Daniel McLaury Jan 14 '12 at 20:06
  • Given that $Y$ is irreducible, why doesn't $Y=\bigcup f(X_j)$ imply that $Y = f(X_j)$ for some $X_j$? – Jon Sep 13 '21 at 12:00
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    @Jon: No, because you don't know that each $f(X_j)$ is closed. Think of the decomposition $\mathbb P^1=\mathbb A^1\cup {\infty}$ with $\mathbb P^1$ irreducible. – Georges Elencwajg Sep 13 '21 at 13:28
  • @GeorgesElencwajg I see, thanks! – Jon Sep 15 '21 at 11:32
  • I believe there is a typo in inequality (*) and should read as $\text{trdeg}_k(\text{Rat}(Y)) \le \text{trdeg}_k(\text{Rat}(X))$. – kindasorta Mar 26 '24 at 20:05
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    @kindasorta. You are right and I have corrected that typo. Thanks. – Georges Elencwajg Mar 27 '24 at 11:01
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Use the definition of dimension involving chains of subvarieties. Now, you can assume, without loss of generality, that both X and Y are irreducible (do you see why?). Induction on the dimension of Y should give it to you then.

Brooke Ullery
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    I can prove that if the statement is true for $Y$ of dimension $d$, then it is true for $Y$ of dimension $d+1$ and $X$ irreducible. The problem is that I don't see why you can assume $X$ is irreducible. – Math536 Jan 02 '12 at 00:14
  • If Y is irreducible, then one of the images of the irreducible components of X must contain an open (and thus dense) subset of Y, so you can throw away the rest of Y and the rest of X. – Brooke Ullery Jan 02 '12 at 08:50
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    I should like to emphasize that just using the definition of dimension will not even begin to scratch the problem. You have to make use of the fact that $X$ and $Y$ are algebraic varieties and invoke some non trivial results from commutative algebra/algebraic geometry . Like Chevalley's difficult theorem implicit in Brooke's comment, without which it is not at all clear why the image of an irreducible component of $X$ should contain an open subset of $Y$. – Georges Elencwajg Jan 02 '12 at 13:31