Is the following ring isomorphism true? In recent few days, I have been trying to develop an intuitive idea for quotient rings and what I've learnt from my explorations is that if you want to quotient a ring by a principal ideal, you can use the trick of evaluating  polynomial in the ideal to zero. For example, you can show isomorphism between $F[x,y]/(y^2-x)$ and $F[y]$ by evaluation isomorphism $x \to y^2$ . From that intuition, I have got the following isomorphism and am confused if it is correct -
$F[x,y]/(y^2-x^2) \cong (F[x]/(x^2-1))[y]$
My explanation follows - if you use the evaluation isomorphism $x^2 \to y^2$,
you get $F[x,y]/(y^2-x^2) \cong R[y]$ with $R$ being ring of polynomials in $x$ with maximum degree $1$. But we can show $R \cong F[x]/(x^2-1)$ again by evaluation isomorphism $x^2 \to 1$.
Please tell me if this is true or otherwise where I'm going wrong.It would be helpful if you also explain how this intuition idea can be extended from principal ideals to finitely generated ideals for something like $Q[x,y]/(2,x,y)$. Thank you in advance.
 A: No, $x^2 \ne 1$. Actually I can't follow how you came to this conclusion at all. "Evaluation isomorphism $x^2 \to 1$"???? Why is it an isomorphism? Why $1$? Why is it a homomorphism even? Where does $x$ go?
Same with your "isomorphism" (??) $x^2 \to y^2$. You need to say at least where a generating set goes to have a homomorphism.
Geometrically, you want to think of $F[x,y]/(x^2 - y^2)$ as being the set of functions on the set $\{(a,b) : a^2 = b^2\}$. Which polynomial functions are the same on this set? Well if $f(x,y) = x^2$ and $g(x,y) = y^2$ then $f(a,b) = g(a,b)$ whenever $a^2 = b^2$. This correspondence between the set $\{(a,b) : a^2 = b^2\}$ and the ring $F[x,y]/(x^2 - y^2)$ is part of the subject of algebraic geometry.

Problem 0: show in detail that two polynomials represent the same function on $\{(a,b) : a^2 = b^2\}$ if and only if their difference is in the ideal $(x^2 - y^2)$. I.e., show that $f$ is identically $0$ on this set if and only if $f$ is a multiple of $x^2 - y^2$.

Some hints for Problem 0.


*

*Any polynomial in just a single variable of degree $n$, has at most $n$ roots. Therefore, a polynomial with infinitely many roots can only be the zero polynomial. (I didn't say it, but you need to assume that $F$ is infinite for Problem 0 to be true).

*Think about some of the later problems (especially 2 and 3). For instance, think about setting $x = y$ and $x = -y$.
The most general form of Problem 0 is the "Nullstellensatz" (German for zero-locus theorem). It is discussed and proved in algebraic geometry. The most general form of the Nullstellensatz requires some theory to prove, but small instances can be proved with more basic methods.

Question 1: Thinking about functions on $\{(a,b) : a^2 = b^2\}$, can you explain why $x^2 \ne 1$?

The equation $a^2 = b^2$ determines a pair of crossing lines: $a = b$ and $a = -b$. The first line corresponds to the ring $F[x,y]/(x - y)$ and the second corresponds to $F[x,y]/(x + y)$. This corresponds to the factorization $x^2 - y^2 = (x + y)(x - y)$.

Question 2: If $u = x + y$ and $v = x - y$, is there an isomorphism $F[x,y]/(x^2 - y^2) \cong F[u,v]/(uv)$? And try to prove it rigorously because your "intuition" is just churning out nonsense at this point.

Another thing you can do to think about these rings is: write down the monomials of each degree and cancel the redundant ones (e.g. we don't need $y^2$ because $y^2 = x^2$). This only works if you're quotient by homogeneous polynomials (i.e. polynomials in which every monomial has the same total degree [but this total degree does not need to be the same if your ideal is generated by more than one polynomial]).
For $x^2 = y^2$, we have
$$\{1\}, \{x,y\},\{x^2,xy\}, \{x^3,x^2y\}, \{x^4,x^3y\}, \dots, \{x^n,x^{n-1},y\},\dots$$

Problem 3: Prove that this list is correct: show that every monomial is either equivalent to $x^n$ or $x^{n-1}y$ and moreover, show that $x^n \ne x^{n-1}y$. (You can think about the set $\{(a,b) : a^2 = b^2\}$ again. Or you can try to argue why $x^n - x^{n-1}y$ is not a multiple of $x^2 - y^2$.)

