How to think of quotients of polynomial rings I'm studying for an algebra midterm and I'm really just having a hard time wrapping my head around quotients of polynomial rings, especially ones where the ideal being quotiented by is something non-principle (i.e an ideal of the form $(x^2 - 2, 3$) in an appropriate polynomial ring). 
For example this question Set of Ideals of a Polynomial Ring makes use of the fact that 
$$\mathbb{Z}[x]/(2,x^3 + 1) \cong \mathbb{Z}_2[x]/(x^3 + 1)$$
to arrive at a solution, but this isomorphism doesn't at all seem obvious to me (hopefully because I'm just not thinking about the quotient in the correct way). Another example, also a question from dummit and foote ($\S 9.1, 13$), is ''Prove that the rings $F[x,y]/(y^2 - x)$ and $F[x,y](y^2 - x^2)$ are not isomorphic for any field $F$ ''. Really I don't even see an obvious direction to proceed, but I think, on a more fundamental level, I really just have no intuitive notion as to what those fields even look like. 
So I was hoping for some helpful way(s) of thinking about these spaces. Any insight would be much appreciated.
 A: A quotient of rings is a structure where you add a new equation in the previous ring.
For example, $$\mathbb R[T]/(T^2 + 1)$$ is the ring of polynomials, with the new equation $$T^2 + 1 = 0$$so this is $\mathbb C$. So, making a quotient by an ideal generated by 2 elements gives you two new equations. That's all.


*

*For the first example, the ring is $\mathbb Z[x]$ with additional equations:
$$2 = 0 \ \ \& \ \ x^3 = 1$$
so this is $\mathbb Z_2[x]/(x^3 + 1)$ indeed.

*For the second, consider an isomorphism $f$ from $R_1$ to $R_2$;
$f(1) = 1$ so $f$ leaves $F$ invariant; it remains to find images of $x,y$ so take the relationship
$$x^2 = y^2 \ \ \ (R_1)$$
it implies that $(x+y)(x-y) = 0$, so it should be the case for the images of $x,y$ in $R_2$.


Let $f(x) = P(x,y) = P(y^2,y)$ and $f(y) = Q(x,y) = Q(y^2,y)$.
We have $$(P(y^2,y)+Q(y^2,y))(P(y^2,y)-Q(y^2,y))=0$$ but this is impossible, because it should be true in $\mathbb Z[y]$ which has no $0$ divisors.
A: The trick is that you need to think about doing algebra with rings. Not with elements of rings, but with the rings themselves.
The isomorphism you mention can be easily calculated:
$$ \mathbb{Z}[x] / (2, x^3 + 1) \cong \big(\mathbb{Z}[x] / (2)\big)  / (x^3 + 1) 
\cong \big(\mathbb{Z} /(2) \big)[x]  / (x^3 + 1)  $$
although you might have "seen" it by considering the most natural way to represent elements in the rings: you represent an element by writing down an integer polynomial in $x$, and in both cases, the equivalence relation that two integer polynomials represent the same element is the one generated by $2\equiv0$ and $x^3 + 1\equiv 0$.
Your example $F[x,y] / (y^2 - x)$ is a basic example of another particular sort of simplification: this is isomorphic to the ring $F[y]$ by the evaluation homomorphism $x \to y^2$. That is, the homomorphism $f(x,y) \mapsto f(y^2, y)$.
