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As shown in this question here, the quotient of a ring by a sum of two ideals is isomorphic to the double quotient of the ring.

Is there any sort of similarly nice result that holds for either a product of ideals, or an intersection of ideas?

I'm ideally looking for something that would hold at least for polynomial rings and their quotients, even if they contain zero divisors.

I may have seen a theorem somewhere that holds for Noetherian rings and somehow yields a direct product of quotients, but I can no longer find it (and I'm not sure if I'm remembering correctly anyway).

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  • $\begingroup$ take a look at en.wikipedia.org/wiki/… $\endgroup$ – Jef L May 2 '17 at 7:37
  • $\begingroup$ You have the Chinese remainder theorem: if $I+J=R$, then $I\cap J=IJ$ and $$R/IJ\simeq R/I\times R/J.$$ $\endgroup$ – Bernard May 2 '17 at 8:03
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Well, it turns out at the question you linked that the result isn't really about a sum of ideals. It looks like the user perhaps did not apprehend fully that every ideal of $\frac{A}{\mathfrak a}$ is of the form $\frac{\mathfrak b'}{\mathfrak a}$, and that $\mathfrak b'$ already contains $\mathfrak a$, so that $\mathfrak a+\mathfrak b'=\mathfrak b'$.

But the result you are thinking of is probably the Chinese remainder theorem, which is often found in its "most concrete" case using integers. It can actually be expressed for arbitrary commutative rings this way (as seen in I.M. Isaacs Algebra: a graduate course):

enter image description here

A version even holds for noncommutative rings, but if I remember correctly 26.16 breaks down for noncommutative rings.

The more general construction is to take any family of ideals $I_i$ indexed by a set $S$ and to map $r\mapsto (\ldots, r+I_i,\ldots)\in \prod_{i\in S} R/I_i$, which is always a homomorphism with kernel $K=\bigcap_{i\in S} I_i$. The problem is that it is not necessarily onto the product, and the Chinese Remainder Theorem is a special case when you can connclude it is in fact onto. In this more general construction, $R/K$ is said to be a subdirect product of the $R/I_i$'s (because it has the extra nice property that it projects onto each of the $R/I_i$'s.)

Along the lines of isomorphism theorems with intersections: you are aware of the second isomorphism theorem, right? It says that for any two ideals $A,B$ in a ring $R$, $\frac{A+B}{A}\cong \frac{B}{A\cap B}$. (In fact it works for right and left ideals, or just plain modules too, as do the other two isomorphism theorems.)

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  • $\begingroup$ Thanks - I hadn't realized that quirk with the sum of ideals. Still, though, shouldn't an analogous result hold for double quotients in general, even if one ideal isn't contained in the other? It seems like the quotient by sum of ideals should be the pushout of the two quotient rings mod each ideal individually. $\endgroup$ – Mike Battaglia May 2 '17 at 17:47
  • $\begingroup$ @MikeBattaglia I don't think so... what about the localization $R=F[x,y]_{(x,y)}$? The ideal $(x)+(y)$ is the unique maximal ideal, and $R/(x,y)$ is indecomposable, so it could not bear a relationship to any direct sum related to $R/(x)$ and $R/(y)$. $\endgroup$ – rschwieb May 2 '17 at 22:08
  • $\begingroup$ Not a direct sum, a double quotient - I mean that I think $R[x,y]/(x,y) = R[x,y]/(x)/(y)$, where I'm abusing notation slightly because I'm on mobile. But hopefully the meaning is clear... $\endgroup$ – Mike Battaglia May 2 '17 at 22:12
  • $\begingroup$ @MikeBattaglia the only reasonable interpretation of the $(y)$ in what you wrote is as $((y)+(x))/(x)$ and then yes, the quotient would be $F[x,y]/((x)+(y))$ $\endgroup$ – rschwieb May 3 '17 at 2:59

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