Which "known" ring is $K = \frac{\mathbb{Z}_5[X,Y]}{(Y-X^2, XY + Y + 2)}$ isomorphic to? Which "known" ring is $K = \frac{\mathbb{Z}_5[X,Y]}{(Y-X^2, XY + Y + 2)}$ isomorphic to?

I am not 100% sure how to solve this. So far this is what I have:
I know that $(Y-X^2, XY + Y + 2)$ is the ideal, which means that $Y-X^2$ gets 'projected' to 0, in other words $Y=X^2$.
This then gives me that $XY + Y + 2 = X*X^2 + X^2 +2 = X^3 + X^2 + 2 = 0$.
But now I don't know what I have to do.
I have checked and 1,2,3,4 are not roots of the equation in $\mathbb{Z}_5$. So I'm a bit stuck on how to solve this question. Is what I have done so far correct, and how do I continue?
 A: Very good.
You can prove that it's isomorphic to $\Bbb Z_5[X]/(X^3+X^2+2)$ and you already showed that polynomial has no roots in $\Bbb Z_5$.
Hence, it is irreducible (being of degree 3), so the quotient is the field of $5^3$ elements.
A: Note that by your remark $(Y-X^2,XY+Y+2) = (Y-X^2,X^3+X^2+2)$.
The assignment $X \to X, Y\to X^2$ defines an arrow $\tau \colon \Bbb Z_5[X,Y]/(Y-X^2) \to \Bbb Z_5[X]$. Surjectivity is clear, as for injectivity: pick $p(X,Y)$ in the kernel of $\tau$ and
via the identification $\Bbb Z_5[X,Y] = \Bbb Z_5[X][Y]$ divide $p$ by $Y-X^2$. Note that even though polynomial rings in several variables need not have euclidean algorithms, we can always divide by monomials (as in this case). Hence we have
$$
p(X,Y) = (Y-X^2)q(X,Y) + r(X).
$$
But then applying $\tau$ we get $r(X) = 0$ as desired.
Finally, by the third isomorphism theorem we have
$$
\Bbb Z_5[X,Y]/(Y-X^2, X^3+X+2) \simeq \Bbb Z_5[X]/(X^3+X^2+2).
$$
Note that the polynomial $X^3+X^2+2$ has no roots modulo $5$ and is thus irreducible in $\Bbb Z_5$. Therefore, the quotient is a field. To compute the order, you can note for example that it is a $\Bbb Z_5$ vector space of dimension $3$ over $\Bbb Z_5$.
Hence the quotient can be characterized as the field with $5^3$ elements.
