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\begin{align*} \widetilde{\Bbb{A}^2}\subset\{(x,y)\times[y_1:y_2:y_3]\in\Bbb{A}^2\times\Bbb{P}^2:x^2y_2=xyy_1,xyy_3=y^2y_2,y^2y_1=x^2y_3\}=: Y. \end{align*} Consider the affine subset $U_i=\{(x,y)\in Y:y_1\neq0\}$ with affine coordinates $(x,y,y_2,y_3)$, where we set $y_1=1$. Therefore, the equations say that:

\begin{align*} x^2y_2&=xy\\y^2&=x^2y_3 \end{align*} and note that $xyy_3=y^2y_2$ can be obtained from these 2. Note that if $x=0$ then so is $x$. In the open set $U_2$, note that if $y=0$, then $x^2=xyy_1y_2^{-1}=0$ and when $y\neq0$ then $y^2y_2=xyy_3$ is nonzero and hence $x$ and $y_3$ are also not zero. In that case, we have $xy_2=yy_1$ and $xy_3=yy_2$ and $x^2y_3=y^2y_1$. All of this implies $y_1y_3=y_2^2$.

what to do now?

Gabrielle Rodriguez
  • 283

1 Answers1

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You have a faulty definition of the blowup here. It's not the case that if you're blowing up an ideal $(f_1,\cdots,f_m)\subset k[x_1,\cdots,x_n]$, you always get $V((f_iy_j-f_jy_i)_{1\leq i,j\leq m})\subset\Bbb A^n\times\Bbb P^{m-1}$. The correct definition of the blowup in an ideal $I$ is that it's $\operatorname{Proj} \bigoplus_{n\geq 0} I^n$, where $I^n$ is in degree $n$.

If your coordinate ring and your ideal are nice enough, this does reduce to your definition, but not always! You have found one such counterexample. Let's find what the equations for the blowup actually are: we can form a surjective graded homomorphism $$k[x,y][t,u,v]\to\bigoplus_{n\geq0} (x^2,xy,y^2)^n$$ by sending $t\mapsto x^2$, $u\mapsto xy$, $v\mapsto y^2$ where $t,u,v\in k[x,y][t,u,v]$ have degree 1, and we consider their images in $I$, the degree-one portion of the right hand side ring. Then the kernel of this homomorphism is $(u^2-tv,yu-xv,yt-xu)$, and indeed $V(u^2-tv,yu-xv,yt-xu)\subset \Bbb A^2\times\Bbb P^2$ works correctly: the affine open $D(t)$ has coordinate algebra $k[x,y,u,v]/(u^2-v,yu-xv,y-xu)$, which is isomorphic to $k[x,u,v]/(u^2-v,xu^2-xv) = k[x,u,v]/(u^2-v)$ after sending $y\mapsto xu$ and then finally $k[x,u]$ after sending $v\mapsto u^2$.

There is an even quicker solution, though. You can apply the fact that the graded map of graded rings $\bigoplus_{n\geq 0} I^{nd} \to \bigoplus_{m\geq 0} I^m$ by sending then $n$-graded piece $I^{nd}$ isomorphically on to the $(nd)$-graded piece $I^{nd}$ induces an isomorphism of homogeneous spectra, see for instance here.

KReiser
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