Apparent discrepency of statements in Hartshorne exercise III.9.3c

algebraic-geometry

In exercise III.9.3c, we are asked to prove that for the affine scheme $X=\operatorname{Spec}k[x,y,z,w]/(z^2,zw,w^2,xz-yw)$ over $Y=\operatorname{Spec}k[x,y]$,

  1. $X_{\text{red}}=Y$;
  2. $X$ has no embedded points;
  3. The natural morphism $f:X\to Y$ induced by the inclusion homomorphism is not flat.

I had a hard time in proving statement 2 and 3.

My approach is as follows: Find $P\in X$ such that $\mathcal{O}_{X,P}$ is not a free $\mathcal{O}_{Y,f(P)}$-module. But I had a hard time in showing any of the statement, I also haven't try the case when $P$ is not a closed point.

From the ideal of definition, points in $X$ written in affine coordinates has the form $(a,b,0,0)$ for some $a,b\in k$.

At the point $(1,0,0,0)$ (similarly for $(0,1,0,0)$), using the relation $xz-yw=0$ and the invertibility of $x$, $\mathcal{O}_{X,P}=k[x,y,z]_{(x-1,y,z)}/(z^2,az^2,a^2z^2)=k[x,y,z]_{(x-1,y,z)}/(z^2)$, where $a=y/x$. This apparently shows that $(1,0,0,0)$ is an embedded point of $X$ (if I get the definition correctly, I couldn't find a real good definition). However, it is still a free $\mathcal{O}_{Y,f(P)}=k[x,y]_{(x-1,y)}$-module.

I couldn't show whether $(0,0,0,0)$ is a flat point, but apparently it is also an embedded point.

It will be great if someone can give me a hint on this problem.

Best Answer

First, the definition of an embedded point. For an affine scheme $\operatorname{Spec} A$ and a quasi-coherent sheaf $\widetilde{M}$, an associated point is an associated prime of $M$, or a prime ideal $\mathfrak{p}\subset A$ such that $M$ has an element $m$ with $\operatorname{Ann}(m)=\mathfrak{p}$. An embedded point is a non-minimal associated prime of the structure sheaf; in this case, it would be a non-minimal associated prime of $k[X]$.

To show that $X$ has no embedded points, find the minimal primes of $k[X]$ (there's a unique minimal prime) and classify annihilators of elements of $k[X]$ (you'll end up with two options, one of which is the unique minimal prime and one which is not a prime ideal).

To show that the obvious map $f:X\to Y$ is not flat, use the same method as (b): since $f$ is finite, the length of the fiber over any two points should be the same by theorem III.9.9. Find two points with fibers of different lengths.

If this isn't quite enough hinting for you, please leave a comment with the progress you make and where you're still stuck.

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