Best strategies to check wether Quotient Polynomial Rings are isomorphic. 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?
 A: 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.
