Jtdylktuy

I don't understand how to apply the fundamental isomorphism theorem to polynomial quotient rings. For example is the ring $mathbb C[X,Y,Z]/langle X^2-Z,XZ-Y^3rangle$ isomorphic to $mathbb C[X,Y]/langle X^3-Y^3rangle$? Can you please elaborate a little? If yes, how is the theorem (the isomorphism theorem) applied here? Thank you!

FWIW a slightly different approach: It suffices to show by the UMP of $mathbb C[x,y,z]/(x^2-z,xz-y^3)$ that there exists a morphism $xi:mathbb C[x,y,z]to mathbb C[x,y]/(x^3-y^3)$ with the property that for each ring morphism $f:mathbb C[x,y,z]to R$ with $f(x^2-z)=f(xz-y^3)=0$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f=tilde{f}circ xi$.

In particular, let $k$ be the unique map $mathbb C[x,y,z]tomathbb C[x,y]$ that fixes $mathbb C,x,y$ and maps $zmapsto x^2$ and let $eta:mathbb C[x,y]tomathbb C[x,y]/(x^3-y^3)$ be the obvious quotient morphism. We claim that it is sufficient to take $xi=etacirc k$.

To that end, consider some arbitrary $f:mathbb C[x,y,z]to R$. By the UMP of polynomial rings, there exists a unique $f':mathbb C[x,y]to R$ such that $forall zetain mathbb C, f'(zeta)=f(zeta)$, $f'(x)=f(x)$ and $f'(y)=f(y)$. Furthemore, we have that $forall pin mathbb C[x,y,z], f'circ k(z)-f(z)=f'(x)-f(x)=0$. It follows (again from the UMP of polynomial rings) that $f=f'circ k$ and furthermore that $f'$ is the unique map with this property.

Now observe that $f'(x^3-y^3)=f'circ k(x^3-y^3)=f(x)f(x^2-z)+f(xz-y^3)=0$. By the UMP of $mathbb C[x,y]/(x^3,y^3)$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f'=tilde{f}circ eta$. It follows that $$f=f'circ k=(tilde{f}circ eta)circ k=tilde{f}circ (etacirc k)$$ And this $tilde{f}$ can be readily seen as unique. So we're done $blacksquare$