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I am working on this problem and need a little guidance. I have 2 $k$-algebras $$\frac{k[x_1, \ldots, x_n]}{I} \quad \text{ and } \quad \frac{k[y_1, \ldots, y_m]}{J}$$ where $k$ is a field and $I$ is an ideal of $k[x_1, \ldots, x_n]$ and $J$ is an ideal of $k[y_1, \ldots, y_m]$

My goal here is to show that $$\frac{k[x_1, \ldots, x_n]}{I} \otimes_k \frac{k[y_1, \ldots, y_m]}{J} \cong_k \frac{k[x_1, \ldots, x_n, y_1, \ldots, y_m]}{(I,J)}.$$

My plan is to define a mapping: $$f: \frac{k[x_1, \ldots, x_n]}{I} \times \frac{k[y_1, \ldots, y_m]}{J} \rightarrow \frac{k[x_1, \ldots, x_n, y_1, \ldots, y_m]}{(I,J)}$$ If I can define a mapping that is bi-additive and k-balanced, then I can use the universal mapping property of tensor products to have a mapping $$\tilde{f} : \frac{k[x_1, \ldots, x_n]}{I} \otimes_k \frac{k[y_1, \ldots, y_m]}{J} \rightarrow \frac{k[x_1, \ldots, x_n, y_1, \ldots, y_m]}{(I,J)}.$$ I would still have to make the argument that $\tilde{f}$ is an isomorphism of $k$-algebras however.

Can anyone help me in defining $f$ explicitly.

Michael Hardy
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An Isomorphic Teen
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1 Answers1

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See if you can use this in the definition of $\tilde{f}$ to explicitly define a map?

Edit: This can also be seen as a very nice application of Yoneda's lemma. You can find a related post here

Arun Bharadwaj
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  • I read through both, and though slightly helpful, i still am shaky about what i should explicitly define $f$ to be. – An Isomorphic Teen Feb 13 '21 at 19:57
  • Would $f(x + I, y + J) = (xy + (I +J))$ work? – An Isomorphic Teen Feb 14 '21 at 00:22
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    (Sorry about the delayed reply) That was something along the lines of my first guess (except of course, we have more than two variables here). If $g$ and $h$ are in $k[x_1, \dots, x_n]$ and $k[y_1, \dots, y_m]$ respectively, $f(x_i + I, y_j +J) = x_iy_j + (I+J)$. It is bi-additive and $k-$balanced, so it checks off all the easy boxes. There is still the issue of showing that $\tilde{f}$ is an isomorphism though. – Arun Bharadwaj Feb 14 '21 at 02:44
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    (for some reason I can't edit my previous comment), so here is an add on: If $g$ and $h$ are in $k[x_1,…,x_n]$ and $k[y_1,…,y_m]$ respectively, $f(g+I,h+J)=gh+(I+J)$. It is bi-additive and k−balanced, so it checks off the easy boxes. There is the issue of showing that $\tilde{f}$ is an isomorphism though, I worked through the details, it is messy but doable if you show the isomorphism by constructing an inverse map and showing that composing them gives identity on both ends. Try $\prod_{i,j} x_iy_j \to \prod_i x_i \otimes \prod_j y_j$ – Arun Bharadwaj Feb 14 '21 at 02:56
  • Ill give that a go and see what pops out. Just by a quick inspection though, it should work. – An Isomorphic Teen Feb 14 '21 at 02:58
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    I did some more hunting, and here's a version of essentially the same exact question: https://math.stackexchange.com/questions/411428/isomorphism-including-a-tensor-product?rq=1 (The inverse map noted in the accepted answer there is more or less the one I suggested above). :D – Arun Bharadwaj Feb 14 '21 at 17:41