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Let $X,Y \subset \mathbb{A}^n_k$ be be subvarieties of pure dimension r,s respectively, K a field. How could I show that $X \times_k Y$ is of pure dimension $r+s$? I am self-studying so anything is welcome. I have tried Noether normalization but I couldn't make it work.

RogozjinRob
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    See http://math.stackexchange.com/questions/191667/dimension-of-a-tensor-product-of-affine-rings – Cantlog Dec 13 '13 at 22:48
  • @Cantlog: That doesn't show it is of pure dimension r+s, or am I wrong? – RogozjinRob Dec 13 '13 at 22:49
  • Hint: If $X$ and $Y$ are irreducible over $k=\overline{k}$, then $X\times_k Y$ is irreducible and of dimension $\dim X+\dim Y$. – rfauffar Dec 14 '13 at 04:02
  • @RobertAuffarth: Sure, but why does this Apply in full to this situation? The irreducibles of the fiber product over K where K is not algebraically closed, that is. – RogozjinRob Dec 14 '13 at 12:26
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    @RogozjinRob take a look at the answer here: http://math.stackexchange.com/questions/152056/is-fibre-product-of-varieties-irreducible-integral – rfauffar Dec 14 '13 at 13:41
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    @RobertAuffarth: I agree that the product, over an alg. closed field is irreducible, but if we are in the following situation: $X,Y$ not neccesarily irreducible and we form the fiber product over k (not alg. Closed), I can't see that the result you are quoting will give us that $X \times_k Y$ is pure. There is probably some reduction, but I can't seem to find it. An answer would be very appreciated. – RogozjinRob Dec 14 '13 at 14:28
  • @RogozjinRob: you are right. I thought your $k$ is algebraically closed as for most of the questions here. So for general fields $k$, one has to know that if $X$ is pure of dimension $r$, then $X_{\bar{k}}$ is also pure of dimension $r$ (and conversely too). – Cantlog Dec 15 '13 at 00:34
  • Dear RogozjinRob, Noether normalization is certainly a good approach, and the one I would suggest; why couldn't you make it work? Regards, – Matt E Dec 16 '13 at 03:14

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I agree with the OP that there is a bit of an issue here: if you use the argument with Noether normalization, then you have to show there aren't irreducible components of $X \times Y$ of lower dimension.

If $k$ is algebraically closed you can first prove the product is irreducible, see Lemma Tag 05P3). In the case that $k$ is not algebraically closed, you can work around the issue, by extending $k$ to the algebraic closure and working with one irreducible component at a time (for each $X$ and $Y$). This will work and it is the right thing to do for most students of this material.

However, if you want, you can try to prove something about minimal primes of $A \otimes_k B$ for domains $A$ and $B$ over $k$. For example, since $k \to B$ is flat, we see that $A \to A \otimes_k B$ is flat. Hence, by going down for flat ring maps (Lemma Tag 00HS), any minimal prime of $A \otimes_k B$ lies over $(0) \subset A$. Similarly, for $B \to A \otimes_k B$. Thus now we reduce to thinking about minimal primes of $K \otimes_k L$ where $k \subset K$ and $k \subset L$ are field extensions (they are the fraction fields of $A$ and $B$). In this case, if $K' \subset K$ and $L' \subset L$ are subextensions, then $K' \otimes_k L' \to K \otimes_k L$ is flat and we conclude that minimal primes of $K \otimes_k L$ lie over minimal primes of $K' \otimes_k L'$.

Finally, it is easy to see that $k(x_1, \ldots, x_n) \otimes_k k(y_1, \ldots, y_m)$ has a unique minimal prime with residue field $k(x_1, \ldots, x_n, y_1, \ldots, y_m)$.

answer_bot
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