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I just read about the following problem: Let $R:=\mathbb{Z}[X,Y]$ be the polynomial ring over $\mathbb{Z}$ in two variables. Determine whether $(X^2+Y,X+Y^2)$ is a prime ideal in $R$. To do this, one can show that $\mathbb{Z}[X,Y]/(X^2+Y,X+Y^2) \cong \mathbb{Z}[X]/(X^4+X)$ and since $\mathbb{Z}[X]/(X^4+X)$ is not an integral domain, it follows that the Ideal is not prime. My question is, how do I come up with the idea of those two rings being isomorphic? Is there a trick to see such things almost immediately or is it something one has to see in order to use it? I seem to struggle with tasks such as this one, which is why a "general trick" in order to determine isomorphic rings of that sort would really help me out.

eu271828
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    Well your ideal tells you that in the quotient, $X^2 = -Y$ and $Y^2 = -X$, so that $X^4 = Y^2 = -X$. So you have that relation. Moreover, $X$ generates the quotient (because it generates $X$ and $Y$). Then you just have to check that you don't get more relations – Maxime Ramzi Dec 14 '19 at 15:22
  • @Max Thank you for your comment! So I basically try to use the ideal to determine conditions and come up with an intuitive Ring, that might be isomorphic to the one I started with. There is no guarantee, however, is there? If not, then I still need to find a concrete Isomorphism from one ring to the other, right? – eu271828 Dec 14 '19 at 16:16
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    Yes, that's exactly right ! There is usually no guarantee, so the best is to actually describe the isomorphism in both directiond to check that it works – Maxime Ramzi Dec 14 '19 at 16:46
  • @Max Alright, thank you very much! Is the number of conditions I have to check the same as the number of generators of the ideal? What I mean by that is described by the following example: Let $(X^2+Y)$ be an ideal of $\mathbb{Z}[X,Y]$. Then $\mathbb{Z}[X,Y]/(X^2+Y) \cong \mathbb{Z}[X]$, right? Since I have $X^2=-Y$ and then $\mathbb{Z}[X,-X^2]=\mathbb{Z}[X].$ – eu271828 Dec 14 '19 at 18:43

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Generally, one can try constructing a ring homomorphism with kernel equal to the divisor and break it up using the isomorphism theorems. In this case, we can start with the homomorphism $\varphi: \mathbb{Z}[X,Y]\to \mathbb{Z}[X]$ $$\varphi(p(X, Y)) = p(X, -X^2)$$ And the kernel of this (surjective) transformation is precisely $(X^2+Y)$, and thus we get an isomorphism $\tilde{\varphi}$ $$\mathbb{Z}[X,Y]/(X^2+Y)\cong_\tilde{\varphi} \mathbb{Z}[X]$$ Now, note that we can divide the left side by $Y^2+X$ and the right side by $\tilde{\varphi}(Y^2+X) = X^4+X$ and we get $$\frac{\mathbb{Z}[X,Y]/(X^2+Y)}{(Y^2+X)}\cong \mathbb{Z}[X]/(X^4+X)$$ And finally, since $X^2 + Y$ and $Y^2 + X$ are relatively prime in $\mathbb{Z}[X,Y]$, we have $$\frac{\mathbb{Z}[X,Y]/(X^2+Y)}{(Y^2+X)}\cong \frac{\mathbb{Z}[X,Y]}{(X^2+Y, Y^2+X)}$$

Isaac Browne
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  • Thank you for your comment! However I have some quesitons. Why are you able to divide both sides by what you stated? What makes that mathematically legitimate? I have never seen someone do that tbh. Furthermore, for clarification, does the fraction you wrote mean that you divide the new ring (the one where the ideal is already factored out) by another ideal? – eu271828 Dec 14 '19 at 19:22
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    The point here is that we first establish a ring isomorphism, so the two rings are the same up to a renaming of elements via $\tilde{\varphi}$. Thus, we can divide on each side by the ideals of equivalent elements, as those ideals will also be equivalent. And yes, the multiple fractions mean we divide the quotient ring on top by the ideal on the bottom taken in that quotient ring – Isaac Browne Dec 14 '19 at 19:25
  • I see why it makes sense intuitively, however I have never seen and proven such a statement, therefore it's hard for me to apply it if that makes sense. – eu271828 Dec 14 '19 at 20:37
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    The proof is quite nice and shows the power of the isomorphism. All one must see is that the isomorphism is a renaming with exactly the same additive/multiplicative structure and that we define ideals and quotient rings and other objects using solely elements of the group and the additive/multiplicative structure of the ring. That is, if $R_1 \cong R_2$, and the element $x\in R_1$ is renamed as the element $y\in R_2$, then for any property defined in terms of ring theory (the elements and structure of the ring) $P$, we have $P(R_1, x)$, is the same as $P(R_2, y)$ up to this renaming. – Isaac Browne Dec 14 '19 at 21:07
  • Thank you for the effort trying to explain it to me! Do you have a link to the proof of it? I think it would help out quite a lot to see it. – eu271828 Dec 14 '19 at 21:27
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    This will probably help: https://math.stackexchange.com/questions/147632/isomorphisms-preserve-structure-operation-or-order – Isaac Browne Dec 14 '19 at 22:07