Uniqueness of flat limits 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 $U$.)
 A: 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.
