I'm trying to understand the answer here to the question of finding the induced maps of coordinate rings corresponding to explicit isomorphisms between $\mathcal Z(xy-z)$ and $\mathbb A^2$.

A morphism $\pi\colon \mathcal Z(xy-z) \to \mathbb A^2$ is given by $\pi(a_1,a_2,a_3) = (a_1,a_2)$, and its inverse is the morphism $\psi\colon \mathbb A^2 \to \mathcal Z(xy-z)$ given by $\psi(a_1,a_2) = (a_1, a_2, a_1a_2)$.

The induced maps of the coordinate rings are given by pre-composition. Since $$ (x\circ \psi)(a_1,a_2) = a_1, \quad (y\circ \psi)(a_1,a_2) = a_2, \quad \text{and} \quad (z\circ \psi)(a_1,a_2) = a_1a_2, $$ am I correct in understanding that $$ \tilde\psi(x) = x, \quad \tilde\psi(y) = y, \quad \text{and} \quad \tilde\psi(z) = xy? $$

Also, how would one go about determining whether or not $\mathcal Z(xy - z^2)$ is isomorphic to $\mathbb A^2$? I don't think the morphism $\pi\colon \mathcal Z(xy - z^2) \to \mathbb A^2$ that projects onto the first two coordinates is an isomorphism, since the third component of the inverse would involve $\sqrt{a_1a_2}$. However, this only shows that the map $\pi$ is not an isomorphism. How could one go about showing that there exists no isomorphism between the two algebraic sets?

  • $\begingroup$ $\mathbb{A}^2$ is non-singular while $xy-z^2=0$ is singular. So, they can not be isomorphic. $\endgroup$ – Mohan Mar 26 '16 at 4:08
  • $\begingroup$ @Mohan What is meant by a singular algebraic set? $\endgroup$ – justin Mar 26 '16 at 4:16
  • $\begingroup$ @justin A big theme in algebraic geometry is studying geometric objects by studying algebraic invariants associated to them. Two affine algebraic sets $X,Y$ are isomorphic if and only if their coordinate rings are isomorphic. What are the coordinate rings of $\mathbb{A}^2$ and $\mathcal{Z}(xy - z^2)$? Can you show that they are not isomorphic? $\endgroup$ – André 3000 Mar 26 '16 at 4:20
  • $\begingroup$ @SpamIAm If $k$ is infinite, then $\mathcal I(\mathbb A^2) = \{0\}$ so the coordinate ring of $\mathbb A^2$ is just $k[x,y]$. If $(a_1,a_2,a_3) \in \mathcal Z(xy - z^2)$, then $a_1a_2 = a_3^2$, but I'm not sure what the coordinate ring of this is. Do you suggest any references on this? $\endgroup$ – justin Mar 26 '16 at 4:41
  • $\begingroup$ Since $xy - z^2$ is irreducible, then $\mathcal{I}(\mathcal{Z}(xy-z^2)) = (xy-z^2)$. Thus its coordinate ring is $k[x,y,z]/(xy - z^2)$. Try to show that this is not isomorphic to $k[x,y]$. $\endgroup$ – André 3000 Mar 26 '16 at 6:06

As you were told in the comments the last question reduces to show that

$R_1=K[X,Y,Z]/(XY-Z^2)$ is not isomorphic to $R_2=K[X,Y]$.

One way to do this has been pointed out by Mohan: $R_1$ is not regular, while $R_2$ is.

Another one is to show that $R_1$ is not a UFD, and this is strongly suggested by the defining equation $xy=z^2$. (Show that $x$ is irreducible. Obviously it is not prime.)

  • $\begingroup$ As a side remark, these two rings are integrally closed. $\endgroup$ – user26857 Mar 26 '16 at 9:05

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