Recalling result of tensor product of polynomial rings

Question

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:

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

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 03:25

score 9 · Accepted Answer · answered Jun 17 '12 at 03:23

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.

Recalling result of tensor product of polynomial rings

1 Answers1

Linked