MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
Looks like I am about to do some reading. Thanks a lot for the detailed answer. – Jesko Hüttenhain Nov 4 '11 at 16:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.