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Let $k$ be a field (alg closed if you want). Now let $I_{i}$ be an ideal of $k[x_{i}]$ for every $i \in \{1,2,\ldots,n\}$. Is it always true that:

$$k[x_1,x_2,\ldots,x_n]/ \langle I_1,I_2,\ldots,I_n \rangle \cong k[x_1]/I_1 \otimes_k k[x_2]/I_2 \otimes_k \cdots \otimes_k k[x_n]/I_n$$

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Yes. The two sides represent the same functor. You can even replace the $k[x_i]$ with polynomial rings in any number of variables. –  Qiaochu Yuan Jun 17 '12 at 3:25
Is this still true over integral domain? –  Hamton Jul 14 '14 at 18:08

1 Answer 1

up vote 5 down vote accepted

Seems like it?

Compose the natural inclusion $k[x_i] \rightarrow k[x_1, ..., x_n]$ with the quotient map; the kernel is just $I_1$. So the universal property of tensor products induces a map from the RHS to the LHS.

The map in the other direction takes a monomial to the corresponding tensor product, and extends linearly. It is simple to check that these are inverses.

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