I was playing around with the definition of a blow up when I encountered something interesting.

Theorem IV-23 Eisenbud & Harris

Let $X$ be a scheme and $Y\subset X$ a closed subscheme. Let $\mathcal J \subset \mathcal O_Y$ be the ideal sheaf of $Y$ in $X$. If $\mathscr A$ is the sheaf of graded $\mathcal O_X$-algebras $$\mathscr A = \bigoplus_{n=0}^\infty \mathcal J^n$$ then the scheme $\operatorname {Proj}\mathscr A \to X$ is the blowup of $X$ along $Y$.

Now taking $X$ to be $\operatorname{Spec} k$ for some algebraically closed field $k$ we should have a variety with a single point. There is only one blow up that can be done, this gives the ideal sheaf as $k$ thus $\mathscr A \cong \mathcal k[t]$. This gives the blow up of $X$ along (essentially) $X$ to be $\operatorname{Proj}k[t]=\Bbb P_k^1$.

I have never heard of a description of protective space over an algebraically closed field $k$ as simply being the blow up of $\operatorname{Spec} k$.

So my questions are as follows:

Is $\Bbb P^1_k$ simply the blow up of $\operatorname{Spec}k$?

Does this generalise to $\Bbb P^1_S$ of some reduced scheme $S$ being the blow up of $S$ at a point? I suppose it isn't quite $\Bbb P^1$ every time...

Is this point of view helpful for understanding the geometry of $\Bbb P^1$?


1 Answer 1


[Thanks to TomGrubb for correcting some wrong statements that were previously in this answer!]

No: $\operatorname{Proj} k[t]$ is $\mathbb{P}^0_k\cong\operatorname{Spec} k$, not $\mathbb{P}^1_k$. To get $\mathbb{P}^1_k$, you would want a graded polynomial ring in two variables.

Note moreover that $k$ is the ideal sheaf of the empty subscheme of $X=\operatorname{Spec} k$, not of $X$ itself. The ideal sheaf of $X$ would be the zero ideal, and so the graded algebra you get would be the zero algebra whose Proj is empty.

More generally, blowing up a scheme along the empty subscheme never does anything, and blowing up a scheme along the whole scheme gives you the empty scheme. In the case of blowing up along the empty set, the graded algebra $\mathscr{A}$ will just be $\mathcal{O}_X[t]$, and when you take Proj that gives you back just $X$ itself. (In the case that $X=\operatorname{Spec} A$ is affine, $\operatorname{Proj} A[t]$ is covered by the single affine open set you get by inverting $t$, since $t$ generates the irrelevant ideal. Since the degree $0$ part of $A[t,t^{-1}]$ is just $A$, this gives you $\operatorname{Spec} A$.) In the case of blowing up along the whole scheme, $\mathscr{A}$ will just be $0$.

  • 2
    $\begingroup$ Hi Eric, sorry to bring up an old question like this, but blowing up along the entire scheme results in the empty scheme. The ideal sheaf of the entire scheme is the zero ideal and hence we would take $Proj(\mathcal{O}_X)$, not $Proj(\mathcal{O}_X[T])$. Blowing up along the empty set never does anything. $\endgroup$
    – TomGrubb
    Sep 27, 2019 at 3:09
  • $\begingroup$ Oof, you're right, thanks for catching that! I've fixed it now. $\endgroup$ Sep 27, 2019 at 3:44

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