Let $A$ be a DVR, and consider a Noetherian scheme $X$ over $B = spec A$. Let $\eta$ denote the generic point of $B$, and $X_{\eta}$ the generic fiber. Suppose that $i : Y \to X_{\eta}$ is some closed subscheme. Then we consider the scheme theoretic closure of $Y$, $i' : Y' \to X$.

Claim: $Y' \to X \to Spec(A)$ is a flat morphism, and $Y$' is the unique closed subscheme having this property and restricting to $Y$ on $X_{\eta}$.

I can prove that $Y'$ is flat, but I am confused about uniqueness, which seems to come down to the following statement:

Let $R$ be a Noetherian $A$-algebra over a DVR $A$, and $I, J$ ideals of $R$. Suppose that $R/I$ and $R/J$ are flat over $A$, and that $R/I \otimes_A A_{\eta}$ and $R/J \otimes_A A_{\eta}$ are isomorphic as $R \otimes_A A_{\eta}$ algebras. Then $I = J$.

Is such a statement true? I am stuck - I would appreciate a hint!

(I can prove that $Y'$ is flat:

Firstly, $i_* O_Y$ is a quasicoherent sheaf, because a locally closed embedding into a Noetherian scheme is quasi-compact and quasi-separated. Then $Y'$ is defined as the vanishing of the quasi-coherent ideal sheaf which is the kernel of the map $O_X \to i_*(O_Y)$.

We reduce to the case where $X$ is affine, say $X = Spec B$. Then $Y = (B \otimes_A A_{\eta})/I$, and this is flat over the field $A_{\eta}$. Since $Spec(A_{\eta})$ is an open subscheme of $Spec A$, it follows that $Y$ is flat over $Spec(A)$. Hence if we denote by $t$ the uniformizer for $A$, then $(B \otimes_A A_{\eta})/I$ has no $t$-torsion.

We are given $B \to B \otimes_A A_{\eta} /I$, with kernel $I'$. Then the closed subscheme $Y'$ is $Spec(B / I')$. But as $B / I'$ is a subring of $ B \otimes_A A_{\eta} /I$, and the latter has no $t$-torsion, it follows that $B / I'$ has no $t$-torsion. This implies that $Y'$ is flat over $A$.

I can also prove that that $Y'$ restricts to $Y$ on $A_{\eta}$ - I think that this is a general property of the scheme theoretic closure. If $Y \to U \to X$ is a locally closed embedding, then the scheme theoretic closure fits into a fiber diagram in the natural way. It should follow because of the quasicoherence of the ideal sheaf $I'$, and the fact that the map $O_X \to i_* O_Y$ restricted to $U$ fits into the short exact sequence defining the closed subscheme $Y$ of $U$.)

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    $\begingroup$ It does not say $R/I\otimes A_{\eta}$ and $R/J\otimes A_{eta}$ are isomorphic, but actually equal. $\endgroup$ – Mohan Apr 2 '16 at 15:26
  • $\begingroup$ @Mohan Being isomorphic over $R_{\eta}$ (not just isomorphic as rings or whatever) is an equivalent statement. I don't really understand how intepreting as equality helps though, I would appreciate it if you could say more. $\endgroup$ – Lorenzo Apr 2 '16 at 18:33
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    $\begingroup$ We have the natural map $R\to R/I\otimes A_{\eta}$ and $I$ is the kernel. Since the latter is same for $I, J$, clearly being kernels of the same map, they must be same. $\endgroup$ – Mohan Apr 2 '16 at 21:48
  • $\begingroup$ @Mohan Why are they the kernel of that map? The only thing that can be said is that $I_{\eta} = J_{\eta}$... the result is not true without the flatness assumption - take (tx) and (x) in $K[x,t]_{(t)}$ over $K[t]_{(t)}$. They agree over the generic point, but they are not the same ideal. The quotient by the former is not flat ( x is t torsion). $\endgroup$ – Lorenzo Apr 2 '16 at 23:32
  • $\begingroup$ I thought you are assuming $R/I,R/J$ are flat over $A$. That is immediate if you take closures. If you take a n arbitrary scheme, not necessarily flat, of course it is false. $\endgroup$ – Mohan Apr 3 '16 at 2:12

This is essentially what Mohan is saying in the comments. I hope this clears up the confusion.

Lemma. Let $A$ be a DVR with fraction field $K$, and let $R$ be an $A$-algebra. If $I, J \subseteq R$ are ideals such that $R/I$ and $R/J$ are flat, and $R/I \otimes_A K \cong R/J \otimes_A K$ (as $R \otimes_A K$-algebras), then $I = J$.

Proof. Since $R/I$ is flat, the map $R/I \to R/I \otimes_A K$ is injective. Thus, the kernel of $R \to R/I \otimes_A K$ is just $I$. Similarly, the kernel of $R \to R/J \otimes_A K$ is $J$. Since $R/I \otimes_A K \cong R/J \otimes_A K$ as $R$-algebras, the diagram $$\begin{array}{ccccc} & & R & & \\ &\swarrow & &\searrow& \\ R/I \otimes_A K & & \stackrel\sim\longrightarrow & & R/J \otimes_A K \end{array}$$ commutes, so $I = J$. $\square$

Remark. The isomorphism as $R \otimes_A K$-algebras, rather than just $K$-algebras, is what Mohan calls them being 'equal'. This is essential in the proof: we use the maps $R \to R/I \otimes_A K$; you couldn't compare the two if you only remembered the $K$-structure.

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    $\begingroup$ Thanks. It was the first line of your argument that had not occurred to me, even though that is an obvious consequence of flatness and the inclusion of A into its fraction field. I made the point that they were isomorphic as $R \otimes K$ algebras in the original statement of my question, I don't really understand what it was about the presentation of my question that led people to believe that I was especially confused on that point... anyway, thanks for taking the time to clear this up! $\endgroup$ – Lorenzo Apr 3 '16 at 14:00

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