Let $Y$ be a regular surface (therefore a proper, $2$-dimensional scheme over base field $k$) and $y \in Y$ a rational closed point (so $k(y)=k$).

Denote by $I$ the ideal sheaf corresponding to $y$ interpreted as closed subscheme.

Performing the blowup at $y$ we obtain $b: X=Bl_y(Y):= Proj(\oplus_n I^n) \to Y$.

My question is how are the global sections $H^0(Y, O_Y), H^0(X, O_X)$ related to each other?

Indeed higher cohomologies are depending only on derived image sheaf $R^1f_* O_X$ via five term Leray Serre sequence

$$0 \to H^1(Y, f_*O_Y) \to H^1(X, O_X) \to H^0(Y, R^1f_* O_X) \to H^2(Y, f_*O_X) \to H^2(X, O_X) $$

The cruical point is what happens with global sections $H^0(-)$.

My motivatating example was the blowup $X:=Bl_{(0,0)}(\mathbb{A}^2_k)$ of affine surface $\mathbb{A}^2_k$ over basefield $k$ at $(0,0)$ (corresponds to max ideal $(x,y)$). Denote $R:=k[x,y), \mathfrak{a}=(x,y)$.

In order to avoid ambiguity we identify $\oplus_n I^n := R[\mathfrak{a} \cdot T]$ with bookkeeping undeterminant $T$.

Concrete calculations using Cech cohomology provide $H^0(Y, O_Y)=H^0(X, O_X)$.

So my question is under which conditions the blowing up of points preserve the property $H^0(Y, O_Y)=H^0(X, O_X)$?

What are sufficient conditions and why? Is it neccessary that the point $y$ has to be regular or/ and rational? In what dimension are the blowups with this property performed?

Are there recomendable sources which treat this question?

  • $\begingroup$ Suppose without loss of generality that $Y$ is connected and that $k$ is algebraically closed. Then in particular because $Y$ is connected and proper we find that $H^0(Y, \mathcal{O}_y=k$. Similarly one can show that $X$ is connected and proper and so that $H^0(X, \mathcal{O}_X=k$. $\endgroup$ Jul 2, 2019 at 13:20
  • $\begingroup$ Why wlog $k$ algebraically closed? $\endgroup$
    – user267839
    Jul 3, 2019 at 11:56
  • $\begingroup$ It suffices to check that the map is an isomorphism after tensoring up to an algebraically closed field (use flat base change + faithfully flatness of $k \to \overline{k}$. I just include this so there is no funny business with varieties being connected but not geometrically connected etc. $\endgroup$ Jul 3, 2019 at 17:29
  • $\begingroup$ @user45878: So essentially you want to use the stronger statement that for proper $k$-scheme $X$, coherent $F \in Coh(X)$ and field extension $k \subset L$ we have $dim_k H^i(X, F)=dim_L H^i(X \otimes Spec(L), pr^*F)$ where $pr: X \otimes Spec(L) \to X$, right? $\endgroup$
    – user267839
    Jul 4, 2019 at 9:47
  • $\begingroup$ I suggest using this, but that doesn't use properness of $X$, just the fact that it is finite type. $\endgroup$ Jul 4, 2019 at 12:38

1 Answer 1


The best general answer I know is that in the case when $Y$ is normal and you're blowing up an ideal sheaf with proper support in every irreducible component, the global sections are the same.

Observation: in a normal scheme, no point is on multiple irreducible components, as all the local rings of a normal scheme are integrally closed domains. So it suffices to treat the case of $Y$ irreducible.

Proof of statement: We'll show that the pushforwards of the structure sheaf is the structure sheaf. Let $\pi:X\to Y$ be the blowup map. This is proper, so $\pi_*\mathcal{O}_X$ is an $\mathcal{O}_Y$ algebra which is finite as a module. But the quotient fields of $X$ and $Y$ are the same at every point because the blowup map is birational (we chose our sheaf of ideals to have proper support, so the blowup map is an isomorphism on the dense open complement of the support of the sheaf of ideals). So at any point, $\mathcal{O}_{Y,y}\subset \pi_*(\mathcal{O}_X)_y$ is an integral extension of an integrally closed ring, and thus an equality. So we have that $\pi_*\mathcal{O}_X = \mathcal{O}_Y$.

To apply this to your stated example at hand, you're blowing up a proper subvariety of a normal variety and you can use exactly this statement.

If you're interested in seeing more arguments like the above, this is the sort of argument that shows up around the various versions of Zariski's main theorem, Zariski's connectedness theorem, and Stein factorization (although different sources will quote from one of something like ten different versions of ZMT - they're all essentially the same, but sometimes getting from one statement to the other can be a little bit of work).

In the case where $Y$ is not normal, interesting things can happen: consider using blowups to resolve the singularities of a curves. Another observation is that if you're working with proper varieties, blowing up won't change the global sections unless you alter the number of connected components. I would be surprised if there were more general things to say in the non-normal case - you'd probably have to do more specific casework and get your hands dirty there.

  • $\begingroup$ Hi, thank you for the answer. I have two questions: in the proof you say that $\pi_*O_X$ is finite as $O_Y$ module. The only critical point seems to be at $y$, the blowed up point , therefore I don't know how to see that $(\pi_*O_X)_y$ is finite $O_{Y,y}$ module. Unless $\pi$ is an immersion I don't know how explicitely calculate the stalk $(\pi_*O_X)_y$. Which argument you arer using here? $\endgroup$
    – user267839
    Jul 3, 2019 at 11:46
  • $\begingroup$ Secondly: Some days ago you posted another interesting approach via Hartogs theorem (in spirit Vakil's "fun in codimension one"...very amazing) where I saw some problems to "avoid" the generic point of a component of $\pi^{-1}(y)$ (since it has also codimension one). What concretely statement of Hartogs has you used? Unfortunately this one stacks.math.columbia.edu/tag/0BCS doesn't help at first glipse but thinking about various formulations of for example Nakayama or Hensel it might be possible that you used a version that handles the problem I have decrebed previously $\endgroup$
    – user267839
    Jul 3, 2019 at 11:48
  • $\begingroup$ do you remember? Or shall I explain the obstructions which I saw in the Hartogs approach to deduce $H^0(Y, O_Y)=H^0(X, O_X)$? Since I think that this is a very interesting point... $\endgroup$
    – user267839
    Jul 3, 2019 at 11:54
  • 1
    $\begingroup$ 1. The proper pushforward of a coherent sheaf is coherent, and the blowup map is proper. 2. I removed that answer and replaced it with this because the reference to algebraic Hartog's for "something nice happens on a normal scheme" was imprecise and not really what I wanted to say (of course, I realized this after typing it up and submitting it, oops). $\endgroup$
    – KReiser
    Jul 3, 2019 at 16:51

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