Question on an example about flatness in Hartshorne

I have been having trouble understanding some statements regarding flatness in Hartshorne - in particular relating to some of the examples in the text. Any help would be appreciated!

Here is the issue:

• In example III.9.8.4 Hartshorne discusses an example of a family of twisted cubics arising from the projection of $\mathbb{P}^3$ to $\mathbb{P}^2$ from a point. The result is that the flat limit of the twisted cubic is not only singular, but it has an embedded point at the singular point.

• In example III.9.10.1 he explains why if one takes the reduced induced structure on the fibers, then the family is not flat.

My question is: If one takes the flat family $Y\to\mathbb{A}^1$ from the first bullet, and uses the canonical map $Y_{red}\to Y$, then by composing one gets a family $Y_{red}\to \mathbb{A}^1$, which should be flat by proposition III.9.7 (which states that you have flatness over a smooth curve if every associated point of $Y$ maps to the generic point of the curve). Now, I expected $Y_{red}$ to be the family in the second bullet, but since it is not flat, this cannot be the case. What is going on?

Thanks!

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The problem lies in the sentence beginning "Now, I expected..." Taking the underlying reduced scheme is not the same thing as replacing the fibers by their nilreductions. There are lots of maps of reduced schemes with non-reduced fibers. Maybe try to write down a simple example. –  Ramsey Jun 20 '11 at 22:28
Thanks! Now everything makes sense! –  Enrique Jun 21 '11 at 19:24

Enrique, I think what's happening is that $Y$ does not have an embedded point, only its special fiber does. Taking $Y_{\mathrm red}$ does not change anything.

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Sandor is right. Indeed, in Hartshorne's example the total family $Y$ is defined by the ideal

$$I=(a^2(x+1)-z^2, ax(x+1)-yz, xz-ay, y^2-x^2(x+1)) \subset k[a, x,y,z],$$ whereas the central fibre (corresponding to $a=0$) is defined by

$$I_0 = ( z^2, yz, xz, y^2-x^2(x+1)) \subset k[x,y,z].$$

The following Macauley2 script shows that

• $I$ is a radical ideal, hence $Y$ is a reduced affine scheme, that is $Y=Y_{red}$.

• $I_0$ is not radical. In fact, the central fiber $Y_0$ has an embedded point at the node $(0,0,0)$ corresponding to the nilpotent element $z$.

i1 : k=ZZ/32003;

i2 : S=k[a,x,y,z];

i3 : I=ideal (a^2*(x+1)-z^2, a*x*(x+1)-y*z, x*z-a*y, y^2-x^2*(x+1));

o3 : Ideal of S

i4 : I==radical I

o4 = true

i5 : T=k[x,y,z];

i6 : I0=ideal (z^2, y*z, x*z, y^2-x^2*(x+1));

o6 : Ideal of T

i7 : I0 == radical I0

o7 = false

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You can see the same issue looking at the solutions of $y^2=t^2$ as $t$ varies. The total space is a reduced union of two lines, most fibers are two points, and one fiber is a double point. This is a case where the central fiber is again S1, which Hartshorne's example shows may fail. –  Allen Knutson Jun 21 '11 at 12:35
Thanks for computations! –  Enrique Jun 21 '11 at 19:17
And thank you for the example on the lines too! –  Enrique Jun 21 '11 at 19:23