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:

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

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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

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

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