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.