TITLE: How to think of quotients of polynomial rings
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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.