2
$\begingroup$

Let $R=k[x_1,..,x_n]/I$ and let $X=Spec(R)$ be it's associated affine scheme. Suppose that $X$ has only one isolated singularity, say at the origin $\mathfrak{m}=\langle x_1,...,x_n\rangle$. Now, let $R_{\mathfrak{m}}$ be the localization at $\mathfrak{m}$, which is a local ring with maximal ideal $\mathfrak{m}_{\mathfrak{m}}$ that is not regular based on the assumption that $X$ is singular at $\mathfrak{m}$. Take $Bl_{\mathfrak{m}_{\mathfrak{m}}}(Spec(R_{\mathfrak{m}}))=Proj(R_{\mathfrak{m}}[\mathfrak{m}_\mathfrak{m}t])$ to be the blowup of $Spec(R_{\mathfrak{m}})$ at the point $\mathfrak{m}_{\mathfrak{m}}$.

My question is: If the local blowup $Bl_{\mathfrak{m}_{\mathfrak{m}}}(Spec(R_{\mathfrak{m}_\mathfrak{m}}))$ is nonsingular, does that imply that the blowup $Bl_{\mathfrak{m}}(X)$ is also nonsingular? If not, is there an obvious counterexample? If so, can this be generalized to schemes with multiple isolated singularities or possibly to non-isolated singular loci?

$\endgroup$
3
$\begingroup$

Sure. This is true.

Indeed, the formation of the blowup is easily seen to commute with localization (your Rees algebra $R_{\mathfrak{m}}[\mathfrak{m}_{\mathfrak m} t]$s formation certainly commutes with localization of $R$. This generalizes to any sort of situation you'd like.

In particular, if $\mathfrak{m}$ is the only singular point, then the singularities of the blowup $Bl_{\mathfrak{m}}(X)$ certainly lie over $\mathfrak{m}$. Since all those points appear in the local blowup, the result you want holds. Indeed, the local blowup is obtained (say on charts) as a localization of the global blowup (and those charts are then reglued in the obvious way).

In terms of the greater generalization, if $Bl_{\mathfrak{m_m}}(R_{\mathfrak{m}_{\mathfrak{m}}})$ is regular, then the global blowup $f : Bl_{\mathfrak{m}}(X) \to X$ is regular in a neighborhood of $f^{-1}(\mathfrak{m})$. I don't think you can say more than that though.

$\endgroup$
4
  • $\begingroup$ This is sort of what I was thinking, but I wasn't totally convinced by my train of thought. To clarify, when we say that the blowup commutes with localization, we mean something like $Bl_{\mathfrak{m}_{\mathfrak{m}}}(R_{\mathfrak{m}})=Bl_{\mathfrak{m}}(X)_{f^{-1}(\mathfrak{m})}$ which is the localization of the blowup at the exceptional fiber? Or is it the localization at the fiber over $\mathfrak{m}$ in the strict transform? $\endgroup$
    – DavidWayne
    Jul 11 '13 at 21:50
  • $\begingroup$ Probably the best way to do it is as follows. Form the Cartesian product: $$Bl_m(X) \times_X \text{Spec} R_{\mathfrak m}.$$ It's not exactly localization in the usual sense. $\endgroup$ Jul 11 '13 at 22:11
  • $\begingroup$ I agree that this description of the localization makes the most sense geometrically, but I was having a hard time with the algebra in my argument. After reading more, I used what you said about localization commuting with the Rees algebra formation. Let $S=R-\mathfrak{m}$. Then $S$ is included into $R[\mathfrak{m}t]$ as $St^0$, and $R_{\mathfrak{m}}[\mathfrak{m}_{\mathfrak{m}}t] =R_S[\mathfrak{m}_St]=R[\mathfrak{m}t]_{St^0}$. Using this, it is more obvious to me exactly how blowups commute with localization, and I have an idea of how the details of the original question should work out. $\endgroup$
    – DavidWayne
    Jul 12 '13 at 3:59
  • $\begingroup$ Great! I guess the other way to do this is to deal with some sort of multicative system sheaf, form that product, and then take sheafy Spec. $\endgroup$ Jul 12 '13 at 6:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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