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I am confused about a computation in exercise 24.4.L in Vakil's Foundations of Algebraic Geometry, where we are asked to explicitly compute a flat limit.

Let $X = \mathbb{A}^3 \times \mathbb{A}^1 \to Z = \mathbb{A}^1$ over a field $k$, where the coordinates on $\mathbb{A}^3$ are $x, y, z$, and the coordinates on $\mathbb{A}^1$ are $t$. Define $Y$ away from $t = 0$ as the union of the two lines $y = z = 0$ and $x = z - t = 0$. Find the flat limit at $t = 0$.

I think I am expected to get the union of two lines, plus some "nonreducedness" at the origin which points normal to the $xz$-plane due to the "motion" of the line $x = z - t = 0$ as we move $t \to 0$. However, my computation gives me the reduced union instead, and I'm not sure what I'm doing wrong.

Let $A = k[x, y, z, t]$ and $B = k[x, y, z, t, 1/t]/((y, z) \cap (x, z-t))$, so $B$ is the coordinate ring of $Y$. We first compute the scheme-theoretic closure of $Y$ in $X$, i.e. the kernel of the homomorphism $A \to B$. This is just the pullback of the ideal $(y, z) \cap (x, z-t) \subseteq A_t$ to $A$. Since $(y, z)$ and $(x, z-t)$ are both prime in $A_t$ and the primes of $A_t$ correspond bijectively to prime ideals in $A$ not containing $t$ via extension and contraction, the pullback of $(y, z) \cap (x, z - t)$ in $A$ is just $(y, z) \cap (x, z - t)$. The fiber is then computed by appending the equation $t = 0$, so the coordinate ring of $Y'\vert_0$ is $k[x, y, z, t]/((y, z) \cap (x, z - t), t) \cong k[x, y, z]/((y, z) \cap (x, z)) \cong k[x, y]/(xy)$, the reduced union of the two lines.

Surely the problem here arises from an erroneous computation of the pullback of $(y, z) \cap (x, z- t)$ to $A$, but I don't feel like my reasoning was flawed. What exactly is going wrong here?

e_s_
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1 Answers1

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$(y,z)\cap(x,z-t)=(xy,xz,y(t-z),z(z-t))=(xy,xz,yz-yt,z^2-tz)$, so $((y,z)\cap(x,z-t),t)=(xy,xz,yz,z^2,t)$, which is different from your conclusion of $(xy,z,t)$. The issue is that the ideal $(I\cap J)+K$ is not necessarily the same as $(I+K)\cap (J+K)$, which is a step that you used in your calculation with $I=(y,z)$, $J=(x,z-t)$, and $K=(t)$.

KReiser
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  • Thank you very much for your response. I'd mistakenly assumed that the extension of $(y, z) \cap (x, z - t)$ in $A/t$ preserved the intersection, but this is only true when $(y, z)$ and $(x, z- t)$ both contain $t$. How do we see that $(y, z) \cap (x, z - t) = (y, z)(x, z - t)$ in this case? I've always struggled to compute intersections. – e_s_ May 23 '23 at 00:39
  • @EmorySun well, one fully rigorous way is to use Grobner bases - this can be a bit tedious, but it's doable by computer and you can check your work. As for how I actually did it, I used a more geometric viewpoint: the two ideals here determine two 2-planes in 4-space intersecting transversely in one point, so they're two smooth subvarieties of a smooth variety intersecting properly and by something I learned long ago the intersection and the product are the same. – KReiser May 23 '23 at 03:06
  • Thank you very much for the reference and response—this was very helpful. – e_s_ May 23 '23 at 12:51
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    @EmorySun The map $k[x,y,z,t] \to k[x,y,z,t]$ taking $x,y,z$ to themselves and $t \mapsto z-t$ is an automorphism (an involution, even). Thus we can instead consider $(y,z) \cap (x,t)$, which is an intersection of monomial ideals. The intersection of monomial ideals is generated by the lcms of their generators: see here or here. The monomials in this example are all relatively prime, so we can simply take the pairwise products of the generators. – Viktor Vaughn May 23 '23 at 20:25
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    @ViktorVaughn that's a much better way to see it, thanks for the comment. – KReiser May 23 '23 at 20:32