Elementary Algebraic Geometry: Dummit & Foote Ch.15, Ex.24 Coordinate Ring

[Math Amateur]

I am trying to gain an understanding of the basics of elementary algebraic geometry and am reading Dummit and Foote Chapter 15: Commutative Rings and Algebraic Geometry ...

At present I am focused on Section 15.1 Noetherian Rings and Affine Algebraic Sets ... ...

I need help to get started on Exercise 24 of Section 15.1 ...Exercise 24 of Section 15.1 reads as follows:https://www.physicsforums.com/attachments/4763***NOTE***

I do not really fully understand the nature of a k-algebra ... so any help in making the notion of an algebra clearer will help ... as well as a significant start on the exercise ...Hope someone can help ...

Peter

 

[mathbalarka]

You should be able to do the first part of the exercise as I have already given hints in the other thread.

For the second part, I really don't see how you can do this without some machinery. $\mathcal{Z}(xy - z^2)$ looks like a cone in the affine $3$-space, if you try to draw it. You can try and you will convince yourself that the variety at $(0, 0, 0)$ is singular, while $\Bbb A^2$ is smooth everywhere. The formalization of this is that the Zariski tangent space of the variety at $(0, 0, 0)$ is a vector space of dimension $3$, whereas dimension of the variety is of dimension $2$. Thus, $\mathcal{Z}(xy - z^2)$ is not smooth, and hence cannot be isomorphic to $\Bbb A^2$.

EDIT : OK, figured out how to do this without fancy arguments. This one is tricky. Assume that $\mathcal{Z}(xy - z^2)$ is isomorphic to $\Bbb A^2$. Then the coordinate rings $k[x, y, z]/(xy - z^2)$ and $k[u, v]$ are isomorphic as affine $k$-algebras. This implies they must also be isomorphic as rings. But note that in $k[x, y, z]/(xy - z^2)$, $x, y, z$ are irreducible elements, but since $x \cdot y = z \cdot z$, it cannot be a P.I.D. But $k[u, v]$ is a P.I.D., contradiction.