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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.

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2 Answers

up vote 4 down vote accepted

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 $Y=\cup f(X_j)$, hence $Y=\cup \overline {f(X_j)}$, so that one of the $f(X_j)$ is dense in $Y$.
In other words by considering $f|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^* :Rat(Y) \to Rat(X) $ .
This implies by pure algebra (field theory!) the inequality on transcendence degrees over $k$ : $$trdeg_k (Rat(Y))\leq trdeg_k (Rat(Y)) \quad (*)$$
d) Finally, for an irreducible variety $Z$, we know that $dim (Z)=trdeg_k (Rat (Z)$ (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!

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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 2 '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 3 '12 at 7:55
    
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
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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.

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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 2 '12 at 0:14
    
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 2 '12 at 8:50
    
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 2 '12 at 13:31
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