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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.

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  • $\begingroup$ 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 $J$. $\endgroup$ – Sujit Bhattacharyya Feb 6 at 10:17
  • $\begingroup$ Thank you! This seems like the way to go. $\endgroup$ – Wallgrenovic Feb 6 at 11:38
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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.

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  • $\begingroup$ Thank you! How can one see that that polynomial is irreducible? Eisensteins criterion? $\endgroup$ – Wallgrenovic Feb 6 at 11:24
  • $\begingroup$ @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. $\endgroup$ – Andrea Mori Feb 6 at 11:44
  • $\begingroup$ Thanks again! Factorizing led me to contradicitons like you suggested. $\endgroup$ – Wallgrenovic Feb 6 at 12:09
  • $\begingroup$ 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). $\endgroup$ – user26857 Feb 6 at 19:30

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