Blowup Computations

I did my first blowup computations this semester, and found it tremendously satisfying. Those were entirely blowups of affine plane curves at the origin, using what I’ll call the “classical” blowing up procedure (take the subset of $\mb A^2 \times \mb P^1$ with coordinates $(x, y, [z:w])$ cut out by $xw=zy$, and look at where the curve pulls back to under the projection map to $\mb A^2$). I might write up those examples eventually, but this page is mostly dedicated to blowing up a sheaf of ideals.

I begin here by working with the affine case, as for most surfaces that suffices (and the general case is just the gluing together of affines, as usual, so most of the “interesting” things happen in the affine case and the general case is in some ways a formal consequence.)

Before we begin our examples, we recall the affine blowup and an alternate characterization.

Definition (Affine Blowup):

If $X\cong \spec A$ is an affine $\spec k$-scheme of finite type and $Z\to X$ is a closed subscheme with ideal $I$, the blowup algebra is $A[tI]$ graded by degree in $t$ and the blowup $\tilde X$ of $X$ along $Z$ is $\proj A[tI]$.

Remark (Alternate Characterization):

By Eisenbud’s Commutative Algebra with a view towards Algebraic Geometry, §5.2, the blowup algebra is also the homeomorphic image of $k[x_1, ..., x_n, y_1, ..., y_m]$ under the map $x_i\mapsto g_i$, $y_i\mapsto th_i$ where $g_i$ are the generators of $A$ and $h_i$ the generators of $I$.

(For a few more examples, see my notes on intersection theory)

The cone

Let $X = \spec k[x, y, z]/(x^2-y^2-z^2)$ be the affine cone, and first take $Z = V(x, y)$ to be the point at the origin. This has ideal given by $(x, y, z)$, so the blowup algebra is $(k[x, y, z]/(x^2-y^2-z^2))[tx, ty, tz]$ and the blowup is $\proj( (k[x, y, z]/(x^2-y^2-z^2))[tx, ty, tz])$. By the alternate characterization, we can view this as

where the grading is given by the uppercase variables. Cover $\tilde X$ with three patches $U(X)$ (all homogeneous prime ideals which do not contain $X$), $U(Y)$, and $U(Z)$ (symmetrically). Dehomogenizing gives maps

For the first, we perform substitutions and simplifications to obtain

Compute the Jacobian

The image of this matrix can have dimension no more than 2; we wish to determine when it is exactly two, as then the corank of the matrix is the dimension of the variety ($\tilde X$ is a surface), and we have a regular variety. But the first row has a zero and the second does not, so the are linearly independent, and we are done.

The cone at a non-reduced point

Let $X = \spec k[x, y, z]/(x^2-y^2-z^2)$ be the affine cone, and first take $Z = V(x, y)$ to be the point at the origin. This has ideal given by $(x, y)$, so the blowup algebra is $(k[x, y, z]/(x^2-y^2-z^2))[tx, ty]$ and the blowup is $\proj( (k[x, y, z]/(x^2-y^2-z^2))[tx, ty])$. By the alternate characterization, we can view this as

where the grading is given by the uppercase variables.

Cover $\tilde X = \proj (k[x, y, z, X, Y]/(x^2-z^2-y^2, xY-yX))$ with two patches $U(X)$ (all homogeneous prime ideals which do not contain $X$) and $U(Y)$ (symmetrically). Dehomogenizing gives maps $U(X)\cong \spec k[x, y, z, w] / (x^2-y^2-z^2, xw-y)$ and $U(U)\cong \spec k[x, y, z, w] / (x^2-y^2-z^2, x-yw)$. By substitution we obtain that these are $U(X)\cong \spec k[x, z, w] / (x^2-(xw)^2-z^2)$ and $U(U)\cong \spec k[y, z, w] / ((yw)^2-y^2-z^2)$, respectively; these are both affine schemes cut out by a single equation, but the equation’s partials all vanish simultaneously at $V(x, y, z)$ still (when $x=y=z=0$), and so the blowup is still singular.

Remark:

The point of this example is that $\spec(k[x,y,z]/(x^2-y^2-z^2))/(x,y)$ is $\spec k[z]/(z^2)$, a non-reduced scheme. The underlying set is still just a point, because the only prime ideal of $k[z]/(z^2)$ is the ideal $(z)$ (in particular, the zero ideal is not prime), but this is NOT what we mean when we say the “origin”. It’s also interesting to note that blowing up at this scheme gives a singular scheme. I didn’t realize this at first, and just blew up at $(x, y)$, thinking it would give the same thing. I spent a tremendously long time wondering why the scheme I was getting out was not regular.