# Show that two quotient rings are not isomorphic

I looked at this thread but am not able to apply the answers there to my problem.

The problem is this:

Given an algebraically closed field $$k$$ of characteristic zero, consider the polynomial ring $$k[x,y]$$. Let $$I = (y^2 − x^3 − x^2)$$ and let $$J = (xy)$$ and define $$A = k[x, y]/I$$ and $$B = k[x, y]/J$$. Show that $$A$$ and $$B$$ are not isomorphic.

I don't even know where to begin, perhaps I can use some sort of parameterisation as in the other thread? If anyone can hint at how I can get started I will greatly appreciate it.

• Maybe $I$ is an integral domain as it contains no divisors of zero and $J$ is not an integral domain as the product of two nonzero polynomials say $p(x,y)=x$ and $q(x,y)=y$ is a zero polynomial $xy$ in $J$. – Sujit Bhattacharyya Feb 6 at 10:17
• Thank you! This seems like the way to go. – Wallgrenovic Feb 6 at 11:38

Over a field of characteristic $$0$$ the polynomial $$y^2-x^3-x^2$$ is irreducible, thus making the ideal $$I$$ a prime ideal and the quotient ring $$A$$ a domain.
On the other hand the ideal $$J$$ is not prime (just from the definition of prime ideal: $$x$$ and $$y$$ are not element of $$J$$ but their product $$xy$$ is) and so the quotient ring $$B$$ is not a domain.
Hence $$A$$ and $$B$$ cannot be isomorphic.
• @Wallgrenovic : Some attempts factorizing $y^2-x^3-x^2=P(x,y)Q(x,y)$ will soon convince you of the impossibility (note that $P$ and $Q$ need to be linear in $y$ and at most quadratic in $x$). Else, observe that the plane curve $y^2=x^3+x^2$ is irreducible: it is a plane cubic with a nodal singularity in the origin. – Andrea Mori Feb 6 at 11:44
• Why need the characteristic zero in order to conclude that $y^2-x^3-x^2$ is irreducible? This holds over every field (and over every integral domain). – user26857 Feb 6 at 19:30