MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $X$ is an algebraic variety and let $Z \subset X$ be a subvariety. Are there some useful criteria under which the blow-up $Bl_Z X$ becomes a nontrivial product $V \times W$ of the algebraic varieties $V$ and $W$? This means that these varieties should have positive dimensions.

Any example would be appreciated as well.

share|cite|improve this question

I am interpreting the question as also implying that $X$ itself is not a product. I do not have an answer in general, but I think that I have an example. It is a projective surface with a singular point: since you did not ask for the variety $X$ to be non-singular, this should qualify! Probably if $X$ is a surface with your property, then it must be singular, but I do not know how to show this.

Let $C$ be a non-hyperelliptic curve of genus at least three and let $Y := C \times C$. The diagonal $D$ in $Y$ is a curve that can be contracted: there is a morphism $\varphi \colon Y \to {\rm Jac}(C)$ given by $(x,y) \mapsto ([x]-[y])$. The morphism $\varphi$ contracts only the diagonal $D$, since the curve has genus at least three and it is non-hyperelliptic. The image of $\varphi$ is the surface $X$ we are looking for. It is a projective surface (with a unique singular point); its blow up is $Y$ and is therefore a product.

Presumably, the assumption that the curve is non-hyperlliptic is irrelevant: instead of letting $X$ be the image of $\varphi$, you can probably define $X$ to be the "intermediate" surface in the Stein factorization of $\varphi$.

share|cite|improve this answer
Actually, if a non-singular surface is a product of curves and is non-singular, then it is minimal, so it is not a blow up of a non-singular surface. This same argument might give some ideas of how to proceed in general... maybe! – damiano Oct 6 '10 at 11:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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