4) Let \( F \) denote a field and consider the polynomial ring \( R = F[x, y] \). Consider the ideals \( I = R(x - y^2) \)
and \( J = R(x^2 - y^2) \) in \( R \). Prove that the quotient rings \( R / I \) and \( R / J \) are not isomorphic.
Proof. The key is that \( x - y^2 \) is not factorable in UFD \( F[x, y] \) and therefore there are no zero divisors
in \( R / I \), while \( x^2 - y^2 = (x - y)(x + y) \) is factorable, leading to zero divisors in \( R / J \).
For the second point, let \( \varphi: F[x, y] \rightarrow F[x, y] / J \) be the canonical homomorphism and note that \( \varphi(x - y) \neq 0,
\; \varphi(x + y) \neq 0 \), but \( \varphi((x - y)(x + y)) = \varphi(x^2 - y^2) = 0 \), showing that \( R / J \) has zero divisors.
To establish that \( x - y^2 \) is not factorable, set \( x - y^2 = (ax + by + c)(dx + ey + f) \) with coefficients from \( F \)
and show that a contradiction arises (a straightforward if somewhat painful calculation). Therefore \( x - y^2 \) is irreducible
and so prime in UFD \( F[x, y] \). It follows that \( R / I \) is an integral domain and therefore has no zero divisors, so
\( R / I \) and \( R / J \) cannot be isomorphic. QED.
Note: This is Ex. §9.1,#13 in Dummit & Foote.