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My prof loves exercises like: Determine which of these quotient rings are isomorphic:

$$ R_1 = \mathbb{Q}[X,Y]/(X^2) \\ R_2 = \mathbb{Q}[X,Y]/(XY) \\ R_3 = \mathbb{Q}[X,Y]/(X,Y) \\ R_4 = \mathbb{Q}[X,Y]/(XY+3X+2Y+6) \\ R_5 = \mathbb{Q}[X,Y,Z]/(X^2,Y-Z) \\ R_6 = \mathbb{Q}[X,Y,Z]/(XZ-5,Y^2,Z) $$

And I take way to long to solve them. By staring at them I kind of guessed the Isomorphisms: $R_2 \rightarrow R_4, X \mapsto X+2, Y \mapsto Y+3$ and $R_5 \rightarrow R_1, X \mapsto X, Y \mapsto Y, Z \mapsto Y$. Then $R_3 \cong \mathbb{Q}$ while the others aren’t.

But I don’t really have a systematic approach (if there even exists one). I‘d be very happy if someone could tell me how they‘d approach problem‘s like these. I think f.e. that it‘s a good to check for nilpotent elements or zero divisors: In $R_1$ we have $X$ is nilpotent but $R_2$ doesn’t contain any nilpotent elements thus $R_1 \ncong R_2$ but would nilpotent elements help me to find an isomorphism? F.e. $X$ in $R_5$ or $Y$ in $R_6$ are nilpotent in what way would that restrict possible isomorphisms between $R_1$ and each of the two?

Henry T.
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    I don’t have a full “answer,” but I would suggest trying to simplify each one separately before hunting for isomorphisms between them. Factoring the polynomial in $R_4$, for instance, would immediately give the isomorphism. Looking at $R_6$, we can see that $5$ is in the ideal, so the quotient is trivial. Checking for zero-divisors and nilpotent elements is a good step. Also thinking of the elements in the ideal as relations can help (e.g. quotienting by $(Y-Z)$ just means $Y=Z$ in the quotient). But I suppose there isn’t really a “systematic” way to do these problems that I know of… – Kenanski Bowspleefi Feb 04 '23 at 20:41
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    All rings can be modeled as quotient rings of some polynomial ring, so if there was a systematic answer here, it would be progress toward deciding if any two given rings are isomorphic, so I think you just have to do what you can and proceed ad hoc. Simplifying the rings (note when a generator is in the ideal, or 2 generators are equal in the quotient) will help reveal the underlying structure. Sounds like you got the trickiest one, so just keep going! – Justin Young Feb 04 '23 at 22:46
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    Re: your recently deleted question, you may also find helpful this answer which elaborates on ideas implicit in my first comment, esp. quotient reciprocity (which may also be useful in problems like those above). – Bill Dubuque Feb 06 '23 at 23:14
  • Thank you! I‘ll take a look at it! – Henry T. Feb 06 '23 at 23:49

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There is not "a" systematic way to decide this as has been pointed by Justin.

The following isomorphism theorem might be helpful. Let $A$ be any ring (commutative and unital) and $\mathfrak{a}\subset A$ an ideal and $\mathfrak{b}\subset A/\mathfrak{a}$ an ideal. Let $\pi:A\to A/\mathfrak{a}$ be the canonical projection. Then we have

$$(A/\mathfrak{a})/\mathfrak{b} \cong A/\pi^{-1}(\mathfrak{b}).$$

E.g. take $R_5 = \mathbb{Q}[X,Y,Z]/(X^2,Y-Z)$ let $\pi: \mathbb{Q}[X,Y,Z]\to \mathbb{Q}[X,Y,Z]/(Y-Z)$ be the projection map. Then $(X^2,Y-Z)= \pi^{-1}((\overline{X}^2))$ where $\overline{X}$ denotes the residue class of $X$. So we get $$ R_5 \cong (\mathbb{Q}[X,Y,Z]/(Y-Z))/(\overline{X}^2) \cong \mathbb{Q}[X,T]/(X^2).$$ (the last iso is induced by mapping both $Y,Z$ to $T$) Whereas in $R_6$ we have the relation $\overline{XZ}-5=0$ i.e. $\overline{X}\overline{Z}=5$. As $5$ is a unit in $R_6$ we get that $\overline{Z}$ is a unit but $\overline{Z}=0$ by definition of $R_6$ so $R_6 =0$. Thus, $R_5$ and $R_1$ are isomorphic to each other and not to $R_6$.

If you are familiar with localization of rings (modules) then what can also be helpful is that for a ring $A$, an element $0\neq f\in A$ we have $$A[X]/(Xf-1) \cong A_f$$ where $A_f$ is the localization of $A$ at $\{1,f,f^2,\ldots\}$. One sees this in the following way:

We get a map $A[X] \to A_f$ by mapping $X\to \frac{1}{f}$ this factors through $A[X] \to A[X]/(Xf-1)$ so we get a map $A[X]/(Xf-1)\to A_f$. On the other hand the map $A\to A[X] \to A[X]/(Xf-1)$ maps $f$ to the residue class $f \mod (Xf-1)$. But $f\mod (Xf-1)$ is a unit because we have the relation $Xf \equiv 1 \mod (Xf-1)$ so the above map factors through $A\to A_f$ (by universal property). One checks that these two maps are mutually inverse.

nilsw
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