# Tensor product of 2 coordinate rings

For the term variety, I mean the irreducible algebraic set.

My question is, if $V$ and $W$ are 2 varieties over a field $\Bbbk$, then does $\Bbb{k}[V]\otimes \Bbb{k}[W]$ has special structure?

I try to prove that, it is another coordinate ring of another variety, which depends on $V$ and $W$. Then from the irreducible of $V$ and $W$ I reduced the problem to understand the tensor product of $\mathfrak{R}/\mathfrak{p} \otimes \mathfrak{R}/\mathfrak{q}$, which is known that isomorphic to $\mathfrak{R}/\mathfrak{(p+q)}$, where $\mathfrak{R}$ is a Noetherian ring(or a finitly generated algebra), and $\mathfrak{p},\mathfrak{q}$ are prime ideals of $\mathfrak{R}$.

So, what can we imagine the element of $\mathfrak{R}/\mathfrak{p} \otimes \mathfrak{R}/\mathfrak{q}$? I just can understand the tensor product for 2 things(roughly speaking): making bilinear map into linear map, and playing the role of a funtor acting on exact sequence. I could not imagine concretely the element of tensor product(for example, $u\otimes v$ where $u,v$ are two vectors).

Therefore, I could not define the tensor product of 2 coordinate rings( there may be some different ways, please answer here if you got them).

The above sentences describe my situation. Please help me point it out the intuitive picture of tensor product and help me to solve(prove/disprove) my initial question on tensor product of 2 coordinate rings.

Thanks.

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What do you mean by "make sense"? You tensor over $\mathbb k$ and that's defined. –  Dylan Moreland Feb 29 '12 at 19:48
Dear Dylan, I has just edited my post. For the words "make sense" does the tensor product above has some nice structure, properties...Thanks for reminding me. –  Arsenaler Feb 29 '12 at 19:54
Yes, it is the coordinate ring of another variety. Would you like to speculate about which variety it is the coordinate ring of? –  Zhen Lin Feb 29 '12 at 20:04

In general, the tensor product of two vector spaces over a field is spanned by products of the form $v \otimes w$. In your case, this amounts to products $f \otimes g$, where $f$ is a function on $V$ and $g$ is a function on $W$. Now $f \otimes g$ defines a function on $V \times W$ by the rule $$(f \otimes g) (v,w)=f(v)g(w),$$ and so we get a map from $k[V] \otimes_k k[W]$ to $k[V \times W]$. This map turns out to be an isomorphism: the tensor product of the coordinate rings of affine varieties is the coordinate ring of their product.

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If $k$ is an algebraically closed field (of any characteristic), the classical view of algebraic geometry is that there is a perfect correspondence ("an antiequivalence of categories") between affine algebraic varieties over $k$ and reduced finitely generated $k$-algebras.

In this correspondence the affine irreducible subvariety $V\subset \mathbb A^n_k$ given by the equations $f_1=f_2=...=f_r=0 \quad (f_i\in k[T_1,...,T_n])$ corresponds to the domain $k[V]=[T_1,...,T_n]/\langle f_1,f_2,...,f_r \rangle$.
Similarly an irreducible subvariety $W \subset \mathbb A^m_k$ will correspond to a domain $k[W]=[T_1,...,T_m]/\langle g_1,g_2,...,g_s \rangle$.

The product $V\times W\subset \mathbb A^{n+m}_k$ of these varieties, given by the equations $f_1=f_2=...=f_r= g_1=g_2=...=g_s =0$, corresponds to the $k$-algebra
$$k[V\times W]=k[V]\otimes_k k[W]=k[T_1,...,T_n,T_{n+1},...,T_{n+m}]/\langle f_1,f_2,...,f_r, g_1,g_2,...,g_s \rangle$$

Is all this purely formal and trivial? No! There is a real (albeit not very difficult) theorem in algebra behind this: the tensor product of two finitely generated domains is a domain, so that $k[V]\otimes_k k[W]$ is also a domain.
This is false if the base field is not algebraically closed, a simple counterexample being the $\mathbb R$- algebra $\mathbb C\otimes_\mathbb R \mathbb C\cong \mathbb C\times \mathbb C$, which is clearly not a domain since it is a product of rings.

Edit
At the OP's request, I'll try to make more explicit why the $k$-algebra corresponding to $V\times W$ is $k[V] \otimes_k k[W]$.
A point $(v,w)\in \mathbb A^{n+m}_k$ is in $V\times W$ iff $v\in V$ and $w\in W$, in other words iff $f_1(v)=...=f_r(v)=0$ and $g_1(w)=...=g_s(w)=0$.
This means that $V\times W$ is defined by the vanishing of the set of $r+s$ polynomials
$$f_1,...,f_r,g_1,...,g_s\in k[T_1,...,T_n,T_{n+1},...,T_{n+m}]$$
and thus that
$$k[V\times W]= (k[T_1,...,T_n,T_{n+1},...,T_{n+m}]/\langle f_1,f_2,...,f_r, g_1,g_2,...,g_s \rangle )_{red}$$

[It should be noticed that the polynomial $f_1(T_1,...,T_n)$, for example, is now considered as a polynomial in $k[T_1,...,T_n,T_{n+1},...,T_{n+m}]$, although it actually doesn't contain the variables $T_{n+1},...,T_{n+m}$)]

It is then a purely algebraic fact that $k[T_1,...,T_n,T_{n+1},...,T_{n+m}]/\langle f_1,f_2,...,f_r, g_1,g_2,...,g_s \rangle$ is already a reduced algebra (actually even a domain) and isomorphic to $k[V]\otimes_k k[W]$ , so that we have $k[V\times W]= k[V]\otimes_k k[W]$.

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Thank you very much for your answer. How ever, can you explaine more at $k[VxW]=k[v]\otimes_{k} k[W]$. I do not think that's trivial. –  Arsenaler Mar 2 '12 at 9:20
I have added an edit with more explanations. –  Georges Elencwajg Mar 9 '12 at 14:06

I wanted to say more about the approach suggested in Steve's answer. The first thing to show is that this is well defined, which is not too hard. It's surjective, since all of the coordinate functions on $V \times W$ are in the image.

Injectivity is less obvious, but remember that you have a field, over which all modules are free, on hand. If $\{f_i\}$ is a (gigantic) basis for $k[V]$ over $k$, and $\{g_j\}$ a basis for $k[W]$, then $\{f_i \otimes g_j\}$ is a $k$-basis for the tensor product. If you have some expression $\sum a_{ij}f_i\otimes g_j$, $a_{ij} \in k$, which maps to zero in $k[V \times W]$, then for any fixed $Q \in W$ the polynomial $\sum a_{ij}g_j(Q)f_i$ vanishes on $V$. What can you conclude?

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Thank you Dylan Moreland. –  Arsenaler Mar 20 '12 at 9:59