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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?2013-01-07

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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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    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.2012-12-28