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Let $X = Z(xy - zw)$ in $\mathbb{A}^4$ and $Y = Z(x, z)$. If $\pi: B \rightarrow \mathbb{A}^4$ is the blow up of $Y$, then how can I find the strict transform of $X$ and the exceptional divisor?

I honestly have no idea how to find the strict transform in this case. I'm only fairly comfortable with blowing up single points. This is what I have so far, gathered from random lecture notes. It only deals with the exceptional divisor and is rather messy:

Let $E = \pi^{-1}(Y)$ and $J = I(Y) = (x, z)$. By Theorem 14.7, $B(J) = U_1 \cup U_2$ where $U_1, U_2$ are affine and $\mathcal{O}_{B(J)}(U_1) = k[\mathbb{A}^{4}][z/x] = k[x, y, z/x]$, $\mathcal{O}_{B(J)}(U_2) = k[\mathbb{A}^{4}][x/z] = k[y, z, x/z].$ We have $x = 0$ is a local equation of $E$ in $U_1$, $z = 0$ is a local equation of $E$ in $U_2$, and $k[E \cap U_1] = k[x, y, z/x]/(x)k[x, y, z/x] = k[y, z/x]$, $k[E \cap U_2] = k[y, z, x/z]/(x)k[y, z, x/z] = k[y, x/z].$ (I do not recognize what this means about $E$, but $E \not = \mathbb{A}^{2}$. Random guess: $E$ is the subset of $\mathbb{P}^{2}$ with coordinates $x, y, z$ described by $(x : xy : z)$ for some $(x, y, z) \in U_1$ and $(x : yz : z)$ for some $(x, y, z) \in U_2$. Using $U_1 \cap E = Z(J\mathcal{O}_{B(J)}(U_1)) = Z(x)$ and $U_2 \cap E = Z(J\mathcal{O}_{B(J)}(U_2)) = Z(z)$, we get $E = \{(x : 0 : z) \mid x \in \pi_1(U_1), z \in \pi_3(U_2)\}$ . . . )

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    $\begingroup$ Do you blow up $X$ along $Y$? $\endgroup$
    – JacobI
    Feb 24, 2012 at 10:27

1 Answer 1

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(Sorry for not using your notations, I prefer use the standard ones : X the base variety, A and B are for rings, Z is a center, etc.)

You have a affine variety $X=\operatorname{Spec}A$, and a closed set $Z$ of $X$ given by an ideal $I$ of $A$. You consider the blow-up $\epsilon : Y \to X$ of $X$ with centre $Z$.

Choose a generating set of $I$, say $(f_1,\dotsc,f_n)$. Then $Y$ is covered by affine charts $U_i = \operatorname{B_i}$, with $B_i = B[I/f_i]$, i.e. $B[f_1/f,\dotsc,f_n/f]$. You can think of thing ring as a subring of its total ring of fractions, if $f$ is not a zero divisor.

Now consider $W$ a hypersurface of $X$, defined by an equation $h\in A$. The pull-back of $W$ in the chart $U_i$ of $Y$ is given by the ideal of $B_i$ generated by $h$. In the chart $U_i$, the exceptionnal divisor is given by $f_i$. (It is a hypersurface, in a sense, this is the universal property of the blow-up.)

In $B_i$, factor $h$ as $f_i^n h_0$, with $n$ a non-negative integer and maximal. Then $h_0$ is the equation of the strict transform of $W$.

There is one case where the computation is particulary easy : $X$ is the affine space $\mathbb A^n = \operatorname{Spec} k[x_1,\dotsc,x_n]$ and $Z$ is given by the ideal $(x_1,\dotsc,x_p)$. Each chart of the blow-up is itself isomorphic to the affine space $\mathbb A^n$. Indeed, the ring $$ k[x_1,\dotsc,x_n][x_1/x_k,\dotsc,x_p/x_k] $$ is isomorphic to the ring $k[y_1,\dotsc,y_n]$ by with $y_i\mapsto x_i/x_k$ if $i\leq p$ and $i\neq k$, and with $y_i\mapsto x_i$ if not.

Consider $h$, the equation of $W$, and $k$ its order along $Z$ — i.e. the least integer $p$ such that $h$ has a non zero monom with total degree $p$ with respect to the variables $x_1,\dotsc,x_n$.

In $B_p$ (and it is similar in the other $B_i$'s) , write $h$ as $h$ as $$ h\left(x_p \frac{x_1}{x_p}\dotsc,x_p\frac{x_{p-1}}{x_p},x_p,\dotsc,x_n\right).$$ One easily that $x_p$ factors out with an exponent $k$. What remains is the equation of the strict transform.

In your example $h$ is $x y-z w$, and we are blowing up $(x,z)$. In the chart $U_x$, we have $$ x y-z w = x y - x\frac zx w = x(y-\frac zx w)$$

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