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If $I(X)=(f_{1},\dots,f_{r})\subset k[x_{1},\dots x_{n}]$ and $I(Y)=(g_{1},\dots,g_{s})\subset k[y_{1},\dots ,y_{m}]$ then should $I(X\times Y)=(f_{1},\dots,f_{r},g_{1},\dots,g_{s})\subset k[x_{1},\dots x_{n},y_{1},\dots y_{m}]$? ($I(T)$ is the set of polynomials that vanish on $T$.)

One inclusion is trivial. $I(X\times Y)\supset (f_{1},\dots,f_{r},g_{1},\dots,g_{s})$ the other inclusion is troubling me. Any hints would be very nice.

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1 Answer 1

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Here is a hint.

Let $\widetilde{X} = Spec(k[x_1,...,x_n])$ and $\widetilde{Y} = Spec(k[y_1,...,y_m])$.

What are $I(X \times \widetilde{Y})$ and $I(\widetilde{X} \times Y)$?

Now use that $I(A \cap B) = I(A) + I(B)$ for algebraic sets A and B.

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  • $\begingroup$ So $Spec(k[x_{1},\cdots ,x_{n}])$ is the set of all prime ideals of $k[x_{1},\cdots,x_{n}]$? $\endgroup$ Feb 5, 2014 at 1:43
  • $\begingroup$ Yes, it is just $\mathbb{A}^n_k$. $\endgroup$
    – SomeEE
    Feb 5, 2014 at 2:02
  • $\begingroup$ I was unaware of the spectrum being so nice. It's not a PID so I didn't expect that to be true. $\endgroup$ Feb 5, 2014 at 4:12

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