Let $X=\mathrm{Spec}(A)$ be an affine variety, $Z\subseteq X$ a closed, reduced subscheme. Let

$$\beta:Y=\mathrm{Bl}_Z(X)\to X$$

be the blow-up of $X$ in $Z$. In other words, $Y=\mathrm{Proj}(A[IT])$ for $I:=I(Z)$. Let $E:=\beta^{-1}(Z)$ be the exceptional divisor. For a point $Q\in E$, I am now wondering how the completion $\hat{\mathcal{O}}_{Y,Q}$ looks like. I would like to understand its relation to $\hat{\mathcal{O}}_{X,\beta(Q)}$, in particular.

Edit: Some remarks and my thoughts on the matter.

If the base field $k$ is algebraically closed and both $X$ and $Y$ are nonsingular, then $$\hat{\mathcal{O}}_{X,\beta(Q)}\cong k[[x_1,\ldots,x_d]]\cong\hat{\mathcal{O}}_{Y,Q}.$$

It becomes tricky when they are (possibly) nonsingular. I thought it would be nice to know something about the local rings. Since $X$ and $Y$ share the same function field $K$ and $\beta$ is dominant,

$$\mathcal{O}_{X,\beta(Q)}\hookrightarrow\mathcal{O}_{Y,Q}\hookrightarrow K.$$

We have adjoined the fractions $a/b$ for $a,b\in I_{\beta(Q)}$ and $bT\notin Q$.

Clearly, $\mathfrak{m}_{\beta(Q)}\cdot\mathcal{O}_{Y,Q}\ne\mathfrak{m}_Q$ which denies us access to the exactness property of the completion. We can write $\mathfrak{m}_{\beta(Q)}=(x_1,\ldots,x_d)$ with $x_i\in I_{\beta(Q)}$ iff $i\le r$. Then, I feel $\mathfrak{m}_Q$ should be equal to $$(z_1,\ldots,z_r,x_{r+1},\ldots,x_d)$$ with $z_i=x_i/b$ for some appropriate $b$, at least under certain conditions. However, I don't know if that is correct and if it even helps.


We can write $A = K[X_1, \ldots, X_n]/\mathfrak a$. Suppose that $I = (a_1, \ldots, a_m)$; then $A[It] \cong A[T_1, \ldots, T_m]/\mathfrak b$ for some ideal $\mathfrak b$. A description for $\mathfrak b$ is given in Section 1.1 of W. Vasconcelos, \textit{Integral Closure, Rees algebras, multiplicities and Algorithms}, (Springer).

Let us suppose, for simplicity, that $P := \beta(Q)$ is defined by $(X_1, \ldots, X_n)$ and that $Q$ is defined by $(X_1, \ldots, X_n, T_1, \ldots, T_m)$. Let us write $R$ and $S$ for the local rings at $P$ and $Q$, respectively. Since we are interested in completions, we may consider the affine open subset of $Y$ defined by $T_1 \neq 0$. This can be thought of as the spectrum of $A[\frac{a_2}{a_1}, \ldots, \frac{a_m}{a_1}] \cong A[Y_2, \ldots, Y_m]/\mathfrak c$ for some ideal $\mathfrak c$. Then $\mathfrak c$ can be obtained from $\mathfrak b$ by `dehomogenizing' with respect to $T_1$; see, \textit{e.g.}, Section 5.5 of \textit{Integral Closure of Ideals, Rings, and Modules} by I. Swanson and C. Huneke, (LMS Lecture Note Series 336). It contains the relations $a_1Y_i = a_i$ for all $1 \leq i \leq m$, but, in general, could have more. Hence $\widehat R \cong K[[X_1, \ldots, X_n]]/\mathfrak a$ and $\widehat S \cong\widehat R[[Y_1, \ldots, Y_m]]/\mathfrak c$. (By abuse of notation, we write $\mathfrak a$ for an ideal of $K[X_1, \ldots, X_n]$ and the ideal of $K[[X_1, \ldots, X_n]]$ generated by it.)

Therefore, in the non-singular situation, with $A = K[X_1, \ldots, X_n]$ and $I = (X_1, \ldots, X_n)$, we have $\widehat R \simeq K[[X_1, \ldots, X_n]]$ and $\widehat S \simeq K[[X_1, Y_2, \ldots, Y_n]]$. (We use the fact that since $X_1, \ldots, X_n$ is a regular sequence in $A$, the ideal $\mathfrak b$ is generated by their `Koszul syzygies', i.e., by $X_iT_j - X_jT_i, 1 \leq i < j \leq n$. Hence $\mathfrak c$ is generated by $X_1Y_j - X_j, 2 \leq j \leq n$.) They are (abstractly) isomorphic as formal power series rings, but the structure morphism is not an isomorphism.

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  • $\begingroup$ Looks like I am about to do some reading. Thanks a lot for the detailed answer. $\endgroup$ – Jesko Hüttenhain Nov 4 '11 at 16:26
  • $\begingroup$ @Manoj Kummini: Let X be a variety and W be a subvariety. We can define $\widehat{\mathcal{O}}_{X,W}$ which is the completion of $\mathcal{O}_X$ along W. Now consider the blow up of X along W. To compute the exceptional divisor can we work with the completion ring $\widehat{\mathcal{O}}_{X,W}$? Is the exceptional divior $\mathbb{P}(\widehat{\mathcal{O}}_{X,W}\oplus I/I^2 \oplus I^2/I^3\oplus......)$? $\endgroup$ – user100841 Jun 24 '17 at 21:16

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