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Recalling result of tensor product of polynomial rings and Does the isomorphism $k[x]\otimes_k k[x]\cong k[x,y]$ hold? suggest that $k[x] \otimes_k k[y] = k[x,y]$.

I am sure that this is true if the tensor product is being done over $k[x,y]$. However, in $k[x] \otimes_k k[y]$ we have the non-equal tensors $x \otimes y$ and $ y \otimes x$. My confusion is that if I map $k[x] \times k[y] \to k[x,y]$, this factors through $k[x] \otimes_k k[y]$ with kernel containing $x \otimes y - y \otimes x$. Help?

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Elle Najt
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1 Answers1

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The tensor product over the field $k$ is the coproduct in the category of commutative $k$-algebras. This means that you are given maps $k[x] \to k[x] \otimes_k k[y]$ and $k[y] \to k[x] \otimes_k k[y]$, both given by $f(x) \mapsto f(x) \otimes 1$ and $g(y) \mapsto 1 \otimes g(y)$. Since you have inclusion maps $k[x] \to k[x,y]$ and $k[y] \to k[x,y]$, the universal property of the coproduct tells you that you get a map $$ k[x] \otimes_k k[y] \to k[x,y], \quad f(x) \otimes g(y) \mapsto f(x) g(y). $$ (The point of all this blabla is to show that this map is well-defined because you have to worry about this when defining maps where the domain is a tensor product. You could also do it "by hand", i.e. defining the corresponding $k$-balanced map and using the universal property of the tensor product.)

Can you show that this map is an isomorphism? (Hint : find the inverse map using the fact that $\{ f(x) g(y) \, | \, f(x) \in k[x], g(y) \in k[y] \}$ spans $k[x,y]$ as a $k$-vector space.)

Hope that helps,

Patrick Da Silva
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  • I think another way to see that it is an injection is to check that the induced map is a homomorphism of graded $k$ algebras, and since each of the degree $d$ homogeneous submodules is a finite dimensional vector space, surjectivity there implies injectivity.

    Is there some slick way to see that the tensor product over $k$ is the coproduct in the category of commutative $k$-algebras?

     – Elle Najt Oct 05 '14 at 20:22
  • @AreaMan : I don't know if you could call it slick, but it's definitely direct : given two commutative $k$-algebras $A$ and $B$, you have inclusion maps $A \to A \otimes_k B$ given by $a \mapsto a \otimes 1$ (similarly for $B$). Given maps $\varphi_A : A \to C$ and $\varphi_B : B \to C$ where $C$ is a $k$-algebra, use the map $\varphi_A \otimes \varphi_B : A \otimes_k B \to C$ defined by $(\varphi_A \otimes \varphi_B)(a \otimes b) = \varphi_A(a)\varphi_B(b)$, and make the checks. – Patrick Da Silva Oct 05 '14 at 20:47