Let $X\subset\mathbb{P}^n$ be a variety singular along a smooth subvariety $Z\subset X$ of positive dimension. Let us assume that $X$ has ordinary singularities along $Z$. Now, let $\pi:Y\rightarrow \mathbb{P}^n$ be the blow-up of a point $p\in Z$, and let $\widetilde{X}\subset Y$ be the strict transform of $X$ with exceptional divisor $E$. Then $\widetilde{X}$ is singular along the strict transform $\widetilde{Z}$ of $Z$.

Here, ordinary means that when we blow-up just $Z$ the strict transform of $X$ is smooth and intersects transversally the exceptional divisor.

Could it happen that $\widetilde{X}$ is singular somewhere else in the intersection $\widetilde{X}\cap E$ ?

  • $\begingroup$ Can you please add the definition of an ordinary singularity? $\endgroup$
    – pinaki
    Feb 28, 2015 at 13:21
  • $\begingroup$ Sure. I added it. $\endgroup$
    – user58018
    Mar 7, 2015 at 17:05
  • $\begingroup$ You completely changed the question so that the previously accepted answer didn't make sense, I rolled it back. If you have a new question, you should post a different question. $\endgroup$ Mar 20, 2015 at 23:31

1 Answer 1


Note that without any assumption on the singularities the answer to your question is yes. For instance, consider the hypersurface $Y = \{x_0^2+x_1^3+x_2^4\}\subset\mathbb{P}^3$. Then $Sing(Y) = [0:0:0:1]$. The projective tangent cone of $Y$ in the singular point is not reduced. Therefore, the singularity is not ordinary. Let $\pi:X\rightarrow\mathbb{P}^3$ be the blow-up of $[0:0:0:1]$. It is easy to check that the singular locus of the strict transform $\widetilde{Y}\subset X$ contains a curve.

On the other hand, if the singularities are ordinary this can not happen.

More precisely:

Let $W\subset Z\subset X$ be smooth projective varieties, and let $Y\subset X$ be a projective variety such that $Sing(Y) = Z$ and $Y$ has ordinary singularities along $Z$. Let $\pi_W:X_W\rightarrow X$ be the blow-up of $W$, and $Z_W$, $Y_W$ the strict transforms of $Z$ and $Y$ respectively. Then $Sing(Y_W) = Z_W$ and $Y_W$ has ordinary singularities along $Z_W$.

Here is the proof:

The inverse image via $\pi_W$ of $Z$ is given by $Z_W\cup E_W$, where $E_W$ is the exceptional divisor of $\pi_W$. Now, we may consider the blow-up $\pi_{Z_w}:X_{Z_W}\rightarrow X_W$ of $Z_W\cup E_W$ in $X_W$, and the blow-up $\pi_Z:X_Z\rightarrow X$ of $Z$ in $X$. Note that since $E_W$ is a Cartier divisor in $X_W$ its blow-up does not produce any effect. By Corollary 7.15 in Harshorne there exists a morphism $f:X_{Z_W}\rightarrow X_Z$ such that $\pi_Z\circ f = \pi_W\circ \pi_{Z_{W}}$. Therefore, the morphism $f$ must map the exceptional divisor $E_{Z_W}$ of $\pi_{Z_W}$ onto the exceptional divisor $E_{Z}$ of $\pi_Z$. Furthermore, $f$ contracts the strict transform $\tilde{E}_W$ of $E_W$ in $X_{Z_W}$ onto $\pi_{Z}^{-1}(W)\subset E_Z$.

Now, let $g:\widetilde{X}_Z\rightarrow X_Z$ be the blow-up of $\pi_{Z}^{-1}(W)$. By the universal property of the blow-up, see Proposition 7.14 Hartshorne, there exits a unique morphism $\phi:X_{Z_W}\rightarrow \widetilde{X}_Z$ such that $g\circ\phi = f$.

Note that since we just blew-up smooth subvarieties both $X_{Z_W}$ and $\widetilde{X}_Z$ are smooth with Picard number $\rho(X_{Z_W}) = \rho(\widetilde{X}_Z) = \rho(X)+2$. In particular, the birational morphism $\phi:X_{Z_W}\rightarrow \widetilde{X}_Z$ can not be a divisorial contraction. Notice that $\phi$ can not be a small contraction either because otherwise $\widetilde{X}_Z$ would be singular. We conclude that $\phi$ is a bijective morphism, and since both $X_{Z_W}$ and $\widetilde{X}_Z$ are smooth $\phi$ is an isomorphism.

Now, let $Y_W$ be the strict transform of $Y$ in $X_W$, and assume that $Z_W\subsetneqq Sing(Y_W)$. Therefore, the strict transform $Y_{Z_W}$ of $Y_W$ in $X_{Z_W}$ is singular, and since $g$ is just the blow-up of a smooth variety the image $(g\circ \phi)(Y_{Z_W}) = f(Y_{Z_W})\subset X_Z$ is singular as well. Note that $f(Y_{Z_W})\subset X_Z$ is nothing but the strict transform of $Y$ via $\pi_Z$, and since $Sing(Y) = Z$ and $Y$ has ordinary singularities along $Z$ the blow-up $\pi_Z$ resolves the singularities of $Y$. A contradiction. We conclude that $Sing(Y_W) = Z_W$. Finally, if the intersection $Y_{Z_W}\cap E_{Z_W}$ is not transversal then the intersection $Y_Z\cap E_Z$ is not transversal as well. But this can not happen because the singularities of $Y$ are ordinary. Therefore, $Y_W$ has ordinary singularities along $Z_W$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.