TITLE: Question about proof of $A[X] \otimes_A A[Y] \cong A[X, Y] $
QUESTION [7 upvotes]: As far as I understand universal properties, one can prove $A[X] \otimes_A A[Y] \cong A[X, Y] $ where $A$ is a commutative unital ring in two ways:
(i) by showing that $A[X,Y]$ satisfies the universal property of $A[X] \otimes_A A[Y] $
(ii) by using the universal property of $A[X] \otimes_A A[Y] $ to obtain an isomorphism $\ell: A[X] \otimes_A A[Y] \to A[X,Y]$
Now surely these two must be interchangeable, meaning I can use either of the two to prove it. So I tried to do (i) as follows:
Define $b: A[X] \times A[Y] \to A[X,Y]$ as $(p(X), q(Y)) \mapsto p(X)q(Y)$. Then  $b$ is bilinear. Now let $N$ be any $R$-module and $b^\prime: A[X] \times A[Y] \to  N$ any bilinear map. 
I can't seem to define $\ell: A[X,Y] \to N$ suitably. The "usual" way to define it would've been $\ell: p(x,y) \mapsto b^\prime(1,p(x,y)) $ but that's not allowed in this case. 
Question: is it really not possible to prove the claim using (i) in this case?

REPLY [3 votes]: For a commutative unital ring $R$ we can show the following $R$-module isomorphism in two ways: $R[x,y] \cong R[x] \otimes R[y]$.

(i) Obtain an isomorphism by using the universal property of $(R[x] \otimes R[y], b)$ where $b: R[x] \times R[y] \to R[x] \otimes R[y]$ is the bilinear map $(p(x), q(y)) \mapsto p(x) \otimes q(y)$ (it looks like this by construction of the tensor product). For $R[x,y]$ we observe that $b^\prime: R[x]\times R[y] \to R[x,y]$, $(p(x),q(y)) \mapsto p(x)q(y)$ is bilinear hence there exists a unique linear map $l: R[x] \otimes R[y] \to R[x,y]$ such that $l \circ b = b^\prime$. We claim that $l$ is an isomorphism, that is, has a two sided inverse. To see this, define $f: R[x,y] \to R[x] \otimes R[y]$ as $\sum a_{ij} x^i y^j \mapsto \sum a_{ij} (x^i \otimes y^j)$. Then 
$$ l (f (\sum a_{ij} x^i y^j)) = \sum a_{ij} x^i y^j$$ and 
$$ f (l (\sum a_{ij} (x^i \otimes y^j))) = \sum a_{ij} (x^i \otimes y^j)$$

(ii) Show that $R[x,y]$ together with the map $b: R[x] \times R[y] \to R[x,y]$, $(p(x) , q(y)) \mapsto p(x) q(y)$ satisfies the universal property of $R[x] \otimes R[y]$. That is, for an $R$-module $M$ and a a bilinear map $b^\prime : R[x] \times R[y] \to M$ there is a unique linear map $l: R[x,y] \to M$ such that $l \circ b = b^\prime$. Define $l: \sum a_{ij} x^i y^j \mapsto \sum a_{ij}b^\prime (x^i, y^j)$. Then $l$ is linear, $l \circ b = b^\prime$ and $l$ is unique: let $l^\prime$ be such that $l^\prime \circ b = b^\prime$. Then $l^\prime \circ b (x^i, y^j) = b^\prime(x^i, y^j) = l \circ b (x^i, y^j)$ and hence $l(x^i y^j) = l^\prime(x^i y^j)$ and hence $l = l^\prime$.