Can k[x1,...,xn] be Isomorphically Mapped to k[x1]⊗k⋯⊗k[xn]?

[Euge]

Here is this week's POTW:

-----
Let $k$ be a field, and $x_1,\ldots, x_n$ indeterminates. Show that there is an isomorphism

$$k[x_1,\ldots, x_n] \approx k[x_1]\otimes_k \cdots \otimes_k k[x_n].$$
-----Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

No one answered this week's problem. You can read my solution below.

Spoiler

There is a $k$-multilinear map $\Phi : k[x_1]\times \cdots \times k[x_n] \to k[x_1,\ldots, x_n]$ given by $\Phi(f_1(x_1),\ldots, f_n(x_n)) = f_1(x_1)\cdots f_n(x_n)$. So $\Phi$ induces a $k$-linear map $F : k[x_1]\otimes_R \cdots \otimes_R k[x_n]$ such that $F(f_1(x_1)\otimes\cdots \otimes f_n(x_n)) = f_1(x_1)\cdots f_n(x_n)$. The map $G : k[x_1,\ldots, x_n] \to k[x_1]\otimes_k \cdots \otimes_k k[x_n]$ be defined by setting

$$G(\sum_{i_1,\ldots i_n} a_{i_1,\ldots, i_n}x_1^{i_1}\cdots x_n^{i_n}) = \sum_{i_1,\ldots, i_n} a_{i_1,\ldots, i_n} x_1^{i_1}\otimes \cdots \otimes x_n^{i_n}.$$

Then $GF$ is the identity on $k[x_1]\otimes_k \cdots \otimes_k k[x_n]$ and $FG$ is the identity on $k[x_1,\ldots, x_n]$. So $F$ is invertible, and hence $F$ is an isomorphism.

In general, if $R$ is a commutative ring, then there is an $R$-algebra isomorphism of $R[x_1,\ldots, x_n]$ onto $R[x_1]\otimes_R \cdots \otimes_R R[x_n]$.