# Fundamental ismorphism theorem

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!

• 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? – KReiser Nov 20 '18 at 6:27
• I'm thinking at $x \to t, z \to t^2$ but the kernel of this map is $\langle X^2-Z \rangle$. – mip Nov 20 '18 at 6:31
• Why not try sending $Z\mapsto X^2$ and sending $X\mapsto X$ as well as $Y\mapsto Y$? – KReiser Nov 20 '18 at 6:41
• But how is this related to the fundamental theorem of isomorphism? – mip Nov 20 '18 at 7:03
• 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. – KReiser Nov 20 '18 at 7:05

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 $$\blacksquare$$