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

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  • $\begingroup$ The first ring enforces the relation that $X^2-Z=0$, yes? Can you define a map from the first ring to the second using this information? $\endgroup$ – KReiser Nov 20 '18 at 6:27
  • $\begingroup$ I'm thinking at $x \to t, z \to t^2$ but the kernel of this map is $\langle X^2-Z \rangle$. $\endgroup$ – mip Nov 20 '18 at 6:31
  • $\begingroup$ Why not try sending $Z\mapsto X^2$ and sending $X\mapsto X$ as well as $Y\mapsto Y$? $\endgroup$ – KReiser Nov 20 '18 at 6:41
  • $\begingroup$ But how is this related to the fundamental theorem of isomorphism? $\endgroup$ – mip Nov 20 '18 at 7:03
  • $\begingroup$ The first isomorphism theorem says that for $f:A\to B$ surjective, we have $B\cong A/\ker f$, yes? Do you see how you should fill in the blanks with $A,B,f$? If you complete the mad-lib appropriately, you should get an answer that satisfies you. $\endgroup$ – KReiser Nov 20 '18 at 7:05
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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 $\mathbb C[x,y,z]\to\mathbb C[x,y]$ that fixes $\mathbb C,x,y$ and maps $z\mapsto x^2$ and let $\eta:\mathbb C[x,y]\to\mathbb C[x,y]/(x^3-y^3)$ be the obvious quotient morphism. We claim that it is sufficient to take $\xi=\eta\circ k$.

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 $\forall \zeta\in \mathbb C,\ f'(\zeta)=f(\zeta)$, $f'(x)=f(x)$ and $f'(y)=f(y)$. Furthemore, we have that $\forall p\in \mathbb C[x,y,z],\ f'\circ k(z)-f(z)=f'(x)-f(x)=0$. 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 $$f=f'\circ k=(\tilde{f}\circ \eta)\circ k=\tilde{f}\circ (\eta\circ k)$$ And this $\tilde{f}$ can be readily seen as unique. So we're done $\blacksquare$

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