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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?

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2  
Dear Matt, I think the tensor should be over $A$, not $R$. Regards, –  Matt E Jul 27 '12 at 12:33
    
Dear @MattE, yes of course, that is a typo, thank you! –  Matt N. Jul 27 '12 at 12:34
    
Think of A[X], A[Y] as free modules over A with the obvious bases of monomials in X, resp. Y. –  i. m. soloveichik Jul 27 '12 at 12:43
    
The same comments as to your previous question apply here: isomorphic as...algebra's? A-modules? groups? –  wildildildlife Jul 28 '12 at 13:33
    
@wildildildlife As $A$-modules. –  Matt N. Aug 5 '12 at 5:46

5 Answers 5

up vote 10 down vote accepted

I can see your problem. As Marlu suggested and in some of the comments above, the trick is to treat of the elements $x^iy^j$ as a "basis" for the polynomial ring $A[x,y]$. In fact, this is the trick because suppose you take some $\sum_{i,j} X^iY^j \in A[x,y]$. Suppose hypothetically that you already have a linear map $L : A[X,Y] \rightarrow N$. Then the action of $L$ on this polynomial being $\sum_{i,j}L(X^iY^j)$ so that the image of any polynomial in $A[X,Y]$ is in fact completely determined by the action of $L$ on the $X^iY^j$. Let us keep this idea in mind and consider the diagram below.

enter image description here

Because we want the diagram to commute, we should have just concentrating on $X^iY^j$ that

$$\begin{eqnarray*} b'(X^i,Y^j) &=& \ell \circ b(X^i,Y^j) \\ &=&\ell(X^iY^j) \end{eqnarray*} $$

Now from what I said in the first paragraph, you can extend $\ell$ additively. Let us check that $\ell$ is compatible with scalar multiplication. Take any $a \in A$. Then $$\begin{eqnarray*} \ell(aX^iY^j) &=& b'(aX^i,Y^j)\\ &=& ab'(X^i,Y^J) \\ &=& a\ell(X^iY^j) \end{eqnarray*}$$

I could take the $a$ out of $b'(\cdot, \cdot)$ because we are now considering $A[X]$ and $A[Y]$ as $A$ - modules and so $b'$ is $A$ - bilinear. We have now completed the check that $\ell$ is linear and uniqueness should be obvious. It follows you have shown that $A[X,Y]$ satisfies the universal property of the tensor product $A[X] \otimes_A A[Y]$ from which it follows that

$$A[X,Y] \cong A[X] \otimes_A A[Y].$$

$$\hspace{6in} \square$$

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I had to put this picture as my request on meta for support of the \begin{xymatrix} environment was deemed to be equivalent to the request for a pony.... –  user38268 Jul 27 '12 at 13:44
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Well you could also try drawing the picture in Latex and then taking a screenshot of the picture it produces. Although I agree that it would be nicer if one could just use such a package directly here. –  Adrián Barquero Jul 27 '12 at 15:05
    
@AdriánBarquero If you would like to you can draw the diagram in latex and edit my post above. –  user38268 Jul 27 '12 at 23:48
    
I added the diagram. I used $\ell$ instead of $l$, but I realized that after editing the post. I guess it won't be a problem. –  Adrián Barquero Jul 28 '12 at 3:43
    
@AdriánBarquero It's ok I have changed my $l$'s to $\ell$'s. –  user38268 Jul 28 '12 at 6:00

New version of the answer

I'll leave below the old version of the answer so that the comments remain understandable. Thank you to the commenters.

Let me prove a slightly more general statement:

(I) If $(X_i)_{i\in I}$ is a family of indeterminates, then we have a natural $A$-algebra isomorphism $$ A\left[(X_i)_{i\in I}\right]\simeq\bigotimes_{i\in I}A[X_i]. $$ Here and in the sequel, tensor products are taken over $A$.

Let $B$ and $C$ be the left and right hand side of the above display, and recall that the coproduct of a family $(A_i)_{i\in I}$ of $A$-algebras is their tensor product. Also note:

(II) For any $A$-algebra $D$ and any family $(d_i)_{i\in I}$ a family of elements of $D$, there is a unique $A$-algebra morphism from $B$ to $D$ mapping $X_i$ to $d_i$ for all $i$.

(III) For any $A$-algebra $D$ and any family $(d_i)_{i\in I}$ a family of elements of $D$, there is a unique $A$-algebra morphism from $C$ to $D$ mapping $$ x_i:=X_i\otimes\bigotimes_{j\neq i}1 $$ to $d_i$ for all $i$.

Proof of (I): By (II) and (III) the $A$-algebra morphism from $B$ to $C$ mapping $X_i$ to $x_i$ for all $i$ and the $A$-algebra morphism from $C$ to $B$ mapping $x_i$ to $X_i$ for all $i$ are inverse isomorphisms.

Additional remarks.

$\bullet$ Inductive limits of $A$-algebras exist. In the terminology of

Categories and Sheaves by Kashiwara and Schapira (Springer 2006), Google preview, Amazon preview,

we have: Let $\mathcal U$ be a universe, let $\mathcal A$ be the category of $A$-algebras whose underlying set belongs to $\mathcal U$, let $I$ be a small category, and let $\alpha:I\to\mathcal A$ be a functor. Then the inductive limit of $\alpha$ exists in $\mathcal A$.

$\bullet$ If $(M_i)_{i\in I}$ is a family of $A$-modules, the the tensor product $\bigotimes_{i\in I}M_i$ is well defined, and satisfies the usual universal property.

Old version of the answer

$A[X,Y]$ and $A[X]\otimes_AA[Y]$ can viewed be as rings, as $A$-algebras, as $A$-modules, as $A[X]$-algebras, as $A[X]$-modules, as $A[Y]$-algebras, as $A[Y]$-modules, and in many others ways. I'll view them as $A$-algebras. Then the natural isomorphism $A[X,Y]\simeq A[X]\otimes_AA[Y]$ is an immediate consequence of the following two more general facts.

If $B$ and $C$ are $A$-algebras, let me denote by $\mathcal A(B,C)$ the set of $A$-algebra morphisms from $B$ to $C$.

Fact $1$. If $B,C,D$ are $A$-algebras, then $B\otimes_AC$ is the coproduct of $B$ and $C$, that is, we have a canonical bijection $$ \mathcal A(B\otimes_AC,D)\simeq\mathcal A(B,D)\times\mathcal A(C,D). $$ Fact $2$. If $(X_i)_{i\in I}$ is a family of indeterminates, then $A\left[(X_i)_{i\in I}\right]$ is free over $I$, that is we have a canonical bijection $$ \mathcal A\left(A\left[(X_i)_{i\in I}\right],D\right)\simeq D^I, $$ where $D^I$ is the set of maps from $I$ to $D$.

EDIT A. This is to answer Marc's comment. Let $f:B\to C$ be an $A$-algebra morphism. For any $A$-algebra $D$ let $$ f_D:\mathcal A(C,D)\to\mathcal A(B,D) $$ be the induced map. Assume that $(1)$ $f_B$ is surjective and $(2)$ $f_C$ injective. Then, by $(1)$, there is a $g:C\to B$ such that $g\circ f=\text{id}_B$, and $(2)$ implies $f\circ g=\text{id}_C$. (This is a general trick.)

EDIT B. To complete the argument, one needs a morphism between $A[X,Y]$ and $A[X]\otimes_AA[Y]$. This is obtained by using the (omitted but obvious) description of the above canonical bijections.

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Wouldn't this argument just show that $\mathcal A(A[X]\otimes_AA[Y],D)\simeq\mathcal A(A[X,Y],D)$ for all $A$-algebras $D$? That is certainly close to the goal, but not quite there. –  Marc van Leeuwen Jul 28 '12 at 8:57
    
Dear @Marc: Thanks for your comment. I edited the answer. –  Pierre-Yves Gaillard Jul 28 '12 at 9:35
    
The main point I tried to make is the that a statement like "$A[X,Y]$ and $A[X]\otimes_AA[Y]$ are isomorphic" really depends on the underlying category (for which there are many choices). (At the time of writing this category is not spelled out in the Question.) The second thing I tried to stress is that, whatever the choice of the category, the natural way to prove the isomorphism is to look at the functors represented by the two objects. –  Pierre-Yves Gaillard Jul 28 '12 at 11:01
    
@MarcvanLeeuwen: right; we could appeal to the Yoneda lemma (after showing the isomorphism to be natural) to finish the argument. –  wildildildlife Jul 28 '12 at 13:39
    
@Pierre-YvesGaillard This is indeed a high-level approach to the problem, very eye opening..... –  user38268 Jul 29 '12 at 13:31

Define $l(X^i Y^j) := b'(X^i, Y^j)$ and continue this to an $A$-linear map $A[X,Y] \to N$ by $$l\left(\sum_{i,j} a_{ij} X^i Y^j\right) = \sum_{i,j} a_{ij} b'(X^i,Y^j).$$

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Why Not show that the tensor product has the universal property of the polynomial algebra? For me that is the more intuitive way of proving this fact.

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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 $l = l^\prime$.

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