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If $I,J$ are index sets, $R$ a commutative unital ring, $\mathfrak{a},\mathfrak{b}$ ideals of polynomial rings $R[x_i; i\!\in\!I]$, $R[y_j; j\!\in\!J]$, and $\langle\langle\ldots\rangle\rangle$ is the ideal generated by $\ldots$, is there an isomorphism of $R$-algebras

$$R[x_i; i\!\in\!I]/\mathfrak{a} \:\oplus\: R[y_j; j\!\in\!J]/\mathfrak{b} \;\cong\; R[x_i, y_j; i\!\in\!I, j\!\in\!J]/\langle\langle\mathfrak{a},\mathfrak{b},x_iy_j; i\!\in\!I,j\!\in\!J\rangle\rangle?$$

The map $(f(x)\!+\!\mathfrak{a},\,g(y)\!+\!\mathfrak{b})\longmapsto f(x)\!+\!g(y)\!+\!\langle\langle\ldots\rangle\rangle$ is not unital.

If $\cong$ does not hold, what other generators of the ideal $\langle\langle\ldots\rangle\rangle$ must I take?

Note: from what I understand, there is an isomorphism of $R$-algebras

$$R[x_i; i\!\in\!I]/\mathfrak{a} \:\otimes\: R[y_j; j\!\in\!J]/\mathfrak{b} \;\cong\; R[x_i, y_j; i\!\in\!I, j\!\in\!J]/\langle\langle\mathfrak{a},\mathfrak{b}\rangle\rangle.$$

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May I ask why a downvote? –  Leon Jan 7 '13 at 20:55

1 Answer 1

up vote 1 down vote accepted

Any time you have a decomposition of a ring as a direct sum of rings, you get an idempotent (an element $e$ with $e^2 = e$) given by the image of $1$ in each summand. So in order to get such an isomorphism, you'd need two nontrivial idempotents $e_1, e_2$ in the right-hand side (which moreover satisfy $e_1e_2 = 0$, $e_1 + e_2 = 1$), and in the ring you wrote down, there's no reason why you should expect this.

Adding more relations alone won't help: for instance, if $I$ and $J$ are both empty and $\mathfrak{a}$ and $\mathfrak{b}$ are both zero, you're asking for an isomorphism $R \oplus R \cong R$.

I think what you want is something like $$R[x_i,y_j,z]/\langle\langle z^2 - z, x_i(1-z), y_jz, x_iy_j, \mathfrak{a}, \mathfrak{b}\rangle\rangle$$ with the isomorphism sending $(1,0)$ to $z$ and $(0,1)$ to $1 - z$.

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Hmm, so you are suggesting a map $$(r+\sum_ir_ix_i+\sum_{i,i'}r_{i,i'}x_ix_{i'}+\ldots+\mathfrak{a},\:s+\sum_js_j‌​y_j+\sum_{j,j'}s_{j,j'}y_jy_{j'}+\ldots+\mathfrak{b}) \longmapsto z\big(r+\sum_ir_ix_i+\sum_{i,i'}r_{i,i'}x_ix_{i'}+\ldots\big)+(1\!-\!z)\big(s+ \sum_{j} s_jy_j +\sum_{j,j'}s_{j,j'}y_jy_{j'}+\ldots\big)+\langle\langle\ldots \rangle \rangle?$$ Why is this injective? Shouldn't we have $x_iz, y_i(1\!-\!z)$ in $\langle\langle\ldots \rangle \rangle$? –  Leon Dec 27 '12 at 16:00
    
The order I had it is right, since $x_i$ is annihilated by $(0,1)$ and $y_j$ by $(1,0)$. For your other question, it's probably easiest to work with the map going the other direction, and maybe taking $\mathfrak{a} = \mathfrak{b} = 0$ to start out. We then obviously have a map $R[x_i,y_j,z] \to R[x_i] \oplus R[y_j]$ sending $z$ to $(1,0)$ and the other variables to themselves, and you can check that the kernel is the ideal I gave, so the isomorphism exists in that case. (cont'd) –  Paul VanKoughnett Dec 28 '12 at 3:22
    
You can then check that for any ideals $\mathfrak{a}$ of $R[x_i]$, $\mathfrak{b}$ of $R[y_j]$, their preimages along this isomorphism are the obvious things. Thus the isomorphism exists in the more general case where you're quotienting by these ideals. –  Paul VanKoughnett Dec 28 '12 at 3:27

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