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First I should say that I am aware of the existence of this question here and this question here. My question is a little different from these two because I am asking about a certain detail in the proof and besides $V$ and $W$ are now just algebraic sets and not affine varieties.

Now $V \times W$ is naturally an algebraic set as follows. If $V$ is the zero locus of some $f_1,\ldots f_k\in k[\Bbb{A}^n]$ and $W$ of $g_1,\ldots,g_l$ in $k[\Bbb{A}^m]$ I believe that $V \times W$ is now the zero locus of $f_1,\ldots,f_k,g_1,\ldots,g_l$ now considered as polynomials in $k[\Bbb{A}^n \times \Bbb{A}^m]$. For notational purposes I will now define

$$\begin{eqnarray*} R &\stackrel{\text{def}}{\equiv}& k[x_1,\ldots,x_n]\\ S &\stackrel{\text{def}}{\equiv}& k[x_{n+1},\ldots,x_{m+n}]\\ T&\stackrel{\text{def}}{\equiv}& k[x_1,\ldots,x_{m+n}].\end{eqnarray*}$$

Let $\mathcal{I}(V)$ denote the ideal of functions that vanish on $V$ in $R$, $\mathcal{I}(W)$ an ideal of $S$ similarly defined. Then I have shown that as $k$ - algebras, we have $$\frac{T}{\mathcal{I}(V)T + \mathcal{I}(W)T} \cong \frac{R}{\mathcal{I}(V)} \otimes_k \frac{S}{\mathcal{I}(W)} $$ subject to the validity of:

My question is: Is the isomorphism $$\frac{T}{\mathcal{I}(V)T} \cong T \otimes_k \frac{R}{\mathcal{I}(V)}$$ of $k$ - algebras valid? I know of several related results concerning isomorphisms of polynomial algebras but they don't seem to apply to the result that I want above.

Here's a proof that we have an isomorphism of $k$ - algebras as claimed if what I ask in my question is true. We have

$$\begin{eqnarray*} \frac{T}{\mathcal{I}(V)T + \mathcal{I}(W)T} &\cong& \frac{T}{\mathcal{I}(V)T} \otimes_T \frac{T}{\mathcal{I}(W)T} \\ &\cong& \left(T \otimes_k \frac{R}{\mathcal{I}(V)}\right) \otimes_T\frac{T}{\mathcal{I}(W)T} \\ &\cong& \frac{R}{\mathcal{I}(V)} \otimes_k \frac{T}{\mathcal{I}(W)T} \\ &\cong& \frac{R}{\mathcal{I}(V)} \otimes_k \left( T \otimes_k \frac{S}{\mathcal{I}(W)} \right) \\ &\cong& \frac{R}{\mathcal{I}(V)} \otimes_k \frac{S}{\mathcal{I}(W)} \end{eqnarray*}$$

as $k$ - algebras.

Community
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2 Answers2

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This is not quite correct. The proposed isomorphism holds if the tensor is taken over $R$, but over $k$ we have a counterexample by choosing $R=k[x], S=k[y], I(V)=(x).$

Andrew
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Take $R=k[X]$, $I=Xk[X]$ and $T=k[X,Y]$. Then $$\frac{T}{I(V)T}=k[Y],$$ $$T \otimes_k \frac{R}{I} =k[X,Y],$$ hence your claim is not true.

For your problem, note that $T=R \otimes S$, and show that the maps $$\frac{R \otimes S}{I \otimes S + R \otimes J} \rightarrow \frac{R}{I} \otimes \frac{S}{J}, x \otimes y \mod (I \otimes S + R \otimes J) \mapsto (x \mod I) \otimes (y \mod J), $$ $$\frac{R}{I} \otimes \frac{S}{J} \rightarrow \frac{R \otimes S}{I \otimes S + R \otimes J} , (x \mod I) \otimes (y \mod J) \mapsto x \otimes y \mod (I \otimes S + R \otimes J)$$ are well defined and inverse from each other.

user10676
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  • Thanks for your answer. Why is the extension of the ideal I of R in $ R\otimes S $ equal to $ I\otimes S$? –  Feb 06 '13 at 17:07
  • In my notation, $I \otimes S$ is the subspace of $R \otimes S$ defined by the image of the natural map $I \otimes S \rightarrow R \otimes S$. You can easily check that this image is also $I.(R \otimes S)$. If you have trouble, then replace $I \otimes S$ by $I.T$. – user10676 Feb 06 '13 at 17:52
  • I have used universal properties to get the maps are you claim above. Once we know both of them are well-defined, it is easily seen that they are inverses of each other. –  Feb 07 '13 at 03:40
  • When we write $R \otimes S$ we mean $R \otimes_k S$. Now for $I \otimes_k S = I\cdot (R\otimes_k S)$ I need to be able to say for any $a \in I$, $r \in R$ and $s \in S$ that $a(r\otimes s) = (ar) \otimes s$. However I am a little confused because in order to say this do I need to change the base ring that I am tensoring over? At the moment the base ring is $k$. –  Feb 07 '13 at 04:05
  • Ah it's ok don't worry that is well-defined, it's just our usual extension of scalars :D –  Feb 07 '13 at 04:07