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Let $X \subset \mathbb{A}^n$, $Y \subset \mathbb{A}^m$ be affine varieties and let us consider them embedded in disjoint subspaces of $\mathbb{A}^{n+m}$. Let $p \in k[x_1,\dots,x_n,y_1,\dots,y_m]$ be a polynomial that vanishes on $X \times Y$. How can we show directly that $p \in I_X+ I_Y$, where $I_X$ is the vanishing ideal of $X$ as a subvariety of $\mathbb{A}^{n+m}$? The way i see it is by using the fact that $A(X \times Y) = A(X) \otimes_k A(Y) \cong k[x_1,\dots,x_n,y_1,\dots,y_m] / (I_X + I_Y)$, but i am interested in a direct argument here. Hints are welcome.

Manos
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You should not think of $X$ as a subvariety of $\mathbb A^{n+m}$ because there is no canonical way to justify that. Here is instead how you may proceed:

Let $i(X)\subset k [x_1,\dots,x_n]$ be the ideal of $X\subset \mathbb A^n$ and $i(Y)\subset k [y_1,\dots,y_m]$ that of $Y\subset \mathbb A^m$.
The ideal $I_X \subset k [x_1,\dots,x_n;y_1,\dots, y_m]$ generated by $i(X)$ is the ideal of the subvariety $X\times\mathbb A^m\subset \mathbb A^{n+m}$ i.e. $$ I_X=i(X\times\mathbb A^m) .$$
Similarly, the ideal $I(Y) \subset k [x_1,\dots,x_n;y_1,\dots, y_m]$ generated by $i(Y)$ is the ideal of the subvariety $\mathbb A^n \times Y\subset \mathbb A^{n+m}$ i.e. $$ I_Y=i(\mathbb A^n \times Y) .$$ Everything now follows easily:
$$I(X)+I(Y)=i(X\times\mathbb A^m)+i(\mathbb A^n \times Y)=i((X\times\mathbb A^m)\cap (\mathbb A^n \times Y))=i(X\times Y)$$ just as you wished.

user26857
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Georges Elencwajg
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    Very instructive, many thanks. I also really think you should write a book. Say something like Harris but in your style. It would be a masterpiece. – Manos Jul 11 '15 at 23:11
  • Thanks for the kind words dear Manos, but our deleted exchange proves that my hint was sufficient for you to solve the problem on your own: bravo! – Georges Elencwajg Jul 11 '15 at 23:18
  • It just proves that you are a great teacher. But actually the notation in your answer is extremely clear. I feel this clarity is missing from the literature. – Manos Jul 11 '15 at 23:20
  • @GeorgesElencwajg Latex tip (hopefully useful for the future book I'm also hoping you will write; maybe such a book could be a good motivation for me to start a systematic study of AG): \cdots is regularly used for a succession of operations like in $a_1+\cdots+a_n$. (+1) – user26857 Jul 12 '15 at 07:56
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    @GeorgesElencwajg Why the second equality (in the last but one line) holds? – user26857 Jul 12 '15 at 08:00
  • @user26857: I didn't know about this convention and I use \cdots for aesthetic reasons, which are of course rather subjective. By the way I had at a time envisioned writing a book on algebraic geometry emphasizing concrete calculations but I won't do it because of the hassle of typing it. I'm very grateful to this site which allows me nevertheless to write about mathematics in pleasantly short little bursts and thus reach more people than would have cared about any book I could write. – Georges Elencwajg Jul 12 '15 at 08:43
  • @user26857: For subschemes $U,V\subset Spec A$ the equality $i(U\cap V)=i(U)\cap (V)$ is true by the definition of intersection of subschemes. In the elementary presentation without nilpotents you have to take radicals of ideals: it is not necessary to take radicals here, but you probably have to use algebra for proving that (and suppose that the base field is perfect) but I didn't check. – Georges Elencwajg Jul 12 '15 at 09:23
  • @GeorgesElencwajg Thanks for the explanation. (Instead I'm thinking seriously to write a book collecting the solutions to the interesting questions I've answered on M.SE.) – user26857 Jul 12 '15 at 10:47
  • @user26857 I'm already taking an option for buying that book ! – Georges Elencwajg Jul 12 '15 at 11:20
  • @user26857: Another customer here :) Here is an explanation i have in mind: $(X \times A^m) \cap (A^n \times Y) = Z(I_X+I_Y)$ and so $i((X \times A^m) \cap (A^n \times Y) ) = rad(I_X+I_Y)$. Now, the tensor product of the reduced $k$-algebras $A(X) \otimes_k A(Y)$ is a domain and so $I_X+I_Y$ is prime. (assuming $k$ of characteristic $0$). – Manos Jul 12 '15 at 15:48
  • I'm having some trouble with the first equality $ I_X = i(X \times \mathbb{A}^m)$. The inclusion $ \subset$ is kinda simple, but the other one doesn't seem to be obvious, could you help me with this ? – FMont Oct 29 '19 at 17:03