# $R[x] \otimes_R R[y] \simeq R[x,y]$ as $R$-algebras.

Let $$R$$ be a commutative ring. I want to show that $$R[x] \otimes_R R[y]$$ is isomorphic to $$R[x,y]$$ as $$R$$-algebras.

First, I tried as follows:

$$R[x] \otimes _R R[y]$$ has a universal property (as a $$R$$-module), that for any bilinear map $$f: R[x] \times R[y] \to M$$ with $$M$$ an $$R$$-module, $$f$$ induces a unique $$R$$-module homomorphism $$\bar f : R[x] \otimes _R R[y] \to M$$. I showed that $$R[x,y]$$ also have this property (as an $$R$$-module), thereby showing $$R[x] \otimes_R R[y]$$ is isomorphic to $$R[x,y]$$ as $$R$$-modules.

$$R[x,y]$$ also has a universal property (as a ring), that any ring homomorphism $$R \to S$$ can be unqiuely extended to a ring homomorphism $$R[x,y] \to S$$. I showed that $$R[x] \otimes_R R[y]$$ also have this property (as a ring), and hence $$R[x] \otimes_R R[y]$$ is isomorphic to $$R[x,y]$$ as rings.

But I think these two does not imply that $$R[x] \otimes_R R[y]$$ is isomorphic to $$R[x,y]$$ as $$R$$-algebras. Namely, I don't expect that the following statement is true:

For two $$R$$-algebras $$A$$ and $$B$$, if $$A \simeq B$$ as $$R$$-modules and as rings, then $$A \simeq B$$ also as $$R$$-algebras.

How do I have to show that $$R[x] \otimes_R R[y] \simeq R[x,y]$$ as $$R$$-algebras? Should I just have to construct an explicit isomorphism?

$$R[x] \otimes_R R[y]$$ is an $$R$$-algebra

Both $$R[x]$$ and $$R[y]$$ are $$R$$-algebras. The map

$$R[x] \times R[y] \times R[x] \times R[y] \rightarrow R[x] \otimes R[y] \\ (r, s, r', s') \mapsto (rr') \otimes (ss')$$

is multilinear and, using associativity of the tensor product, induces a linear map

$$(R[x] \otimes R[y]) \otimes (R[x] \otimes R[y]) \rightarrow R[x] \otimes R[y]$$

By the universal mapping property, this map corresponds to a bilinear map

$$(R[x] \otimes R[y]) \times (R[x] \otimes R[y]) \rightarrow R[x] \otimes R[y] \\ (r \otimes s, r' \otimes s') \mapsto (r\otimes s)\cdot (r'\otimes s')$$

where $$(r\otimes s)\cdot (r'\otimes s') := (rr') \otimes (ss')$$. That gives us a multiplication on $$R[x] \otimes R[y]$$, turning it into an $$R$$-algebra.

$$R[x] \otimes_R R[y]$$ and $$R[x,y]$$ are isomorphic as $$R$$-algebras

To show this, construct the isomorphisms. The first we get from the universal mapping property of the tensor product:

$$F : R[x] \otimes R[y] \rightarrow R[x,y], r \otimes s \mapsto rs$$

For the other direction, define

$$G : R[x,y] \rightarrow R[x]\otimes R[y], x^i y^j \mapsto (x^i \otimes y^j)$$

on monomials and extend linearly.

Both $$x^iy^j$$ and $$x^i \otimes y^j$$ generate $$R[x,y]$$ and $$R[x]\otimes R[y]$$ as $$R$$-modules respectively, and on the generators we have

$$(F\circ G)(x^iy^j) = F(x^i \otimes y^j) = x^i y^j \\ (G \circ F)(x^i \otimes y^j) = G(x^i y^j) = x^i \otimes y^j$$

So $$F$$ and $$G$$ are inverse to each other, and F respects the ring structure:

\begin{align} F((r\otimes s)\cdot(r' \otimes s')) &= F((rr' \otimes (ss')) \\ &= rr'ss' \\ &= F(r\otimes s)\,F(r'\otimes s') \end{align}

Yes, you are right that these two isomorphism don't imply isomoprhism as $$R$$-algebras.

However there's an $$R$$-algebra variant of the universal property. So this is one way.

But there is also a simple explicit isomorphism

$$F:R[x]\otimes R[y]\to R[x,y]$$ $$F(W\otimes U):=W(x)\cdot U(y)$$