Let
- $R = \mathbb{C}[x,y]$
- $R^i \subset R$ be the abelian subgroup of $R$ generated by elements of $\mathbb{C}$ times monomials of degree at least $i$
- $I = (x^3+x^2-y^2)$
- $S = R/I$
- $S^i$ be the group generated by the image of $R^i$ in $S$
Consider the product on $S^i/S^{i+1} \times S^j/S^{j+1}$ given by $$ (a + S^{i+1})(b+S^{j+1}) = ab + S^{i+j+1} \in S^{i+j}/S^{i+j+1}. $$ Let
- $T = \oplus_{i= 0}^{\infty} S^i/S^{i+1}$. The operation above makes $T$ into a ring.
Prove that
- $S$ is an integral domain.
- $T$ is not an integral domain.
Edit (What I did): For part 1, I tried showing that $I$ is a prime ideal but didn't get far. For part 2, I tried generalizing something of the form $$ (a,0,0,...)(0,b,0,0,...) = (0,0,0,...) $$ where $a \neq 0 $ and $b \neq 0$.
$f=y^2-x^2(x+1)$ is an irreducible polynomial in $\mathbb C[x,y]$. This can be proved elementary by supposing that $f$ is a product of two polynomials necessarily of degree one in $y$ and getting a contradiction, or one can use the Eisenstein's criterion for the prime element $x+1$. To conclude, $S$ is indeed an integral domain.
$T$ instead is not an integral domain since $x^2=y^2$ in $T$, so $(x-y)(x+y)=0$, and $x\ne\pm y$.