Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Let $X$ be a smooth scheme of finite type. I think ideal sheaf of codimension 1 subscheme of $X$ is locally free as it is locally defined by one equation. What about higher codimension cases?

share|cite|improve this question
For affine schemes, can you see what this would mean exactly about ideals in rings? – Mariano Suárez-Alvarez Oct 11 '12 at 5:53
Oh, it must be a principal ideal. So in general ideal sheaves corresponding to higher codimensional subschemes are not locally free, I guess. – M. K. Oct 11 '12 at 6:15
Do not guess. Find an actual example and add an answer to the question! :-) – Mariano Suárez-Alvarez Oct 11 '12 at 6:16
Thanks for pushing me to think about this with the hint. – M. K. Oct 11 '12 at 6:47
up vote 7 down vote accepted

Let $X=Spec\mathbb{C}[x,y]$. The ideal $(x,y)$ corresponding to the origin $X\cong \mathbb{C}^2$ is contained in the structure sheaf, so it is line bundle if it is locally free. However it needs two generators $x,y$ and cannot be a line bundle.

share|cite|improve this answer

No, ideal sheaves are not locally free in general.

The simplest example is given by $X=\mathbb A^2_k$ ($k$ a field) with the ideal sheaf $\mathcal I\subset \mathcal O$ of functions vanishing at the origin $P=(0,0)$, a codimension two subscheme of $X$.
If $\mathcal I$ were locally free it would be locally free of rank one (look at neighbouring stalks: on $U=X\setminus \lbrace P\rbrace$ we have $\mathcal I\mid U=\mathcal O\mid U$) .
However if $\mathcal I$ were free of rank one, we would have $\mathcal I_P\cong\mathcal O_{X,P}$ [an isomorphism of $\mathcal O_{X,P}$-modules].
This would mean that the ideal $(x,y)\subset k[x,y]_{(x,y)}$ is principal, and it is easy to check by hand that this is not true.
This contradiction proves that $\mathcal I$ is not locally free.

share|cite|improve this answer

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.