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I am a beginner to sheaf theory; so please bear with me.

Consider the sheaf of differentiable functions on $\mathbb R^2$, and the subsheaf consisting of functions vanishing at $0$.

Question 1:

How would you show that $\mathcal F$ is not a locally free sheaf over $\mathcal A$?

Question 2:

If instead of $\mathbb R^2$ we were considering $\mathbb R$, then is the ideal sheaf at $0$ locally free?

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1 Answer

Let $\mathcal{F}$ denote the subsheaf in question. If it were locally free then there would be some open set containing $0$ on which $\mathcal{F}\vert_U \cong \mathcal{O}^{J}_{\mathbb{R}^2}\vert_U$, where $J$ is some set. But on every open set containing zero, the latter sheaf has either all nonzero stalks or all zero stalks (if $J = \varnothing$) whereas the former has stalks of both types. Isomorphisms of sheaves induce isomorphisms of stalks, so we have a contradiction.

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The stalk at $0$ of the ideal sheaf is not zero, if that is what you are suggesting. The stalks consists of equivalence classes of smooth functions vanishing at $0$, with the relation that they agree on some neighborhood of $0$. It is easy to find two such functions that are not equivalent under this relation. For example, $f(x, y)=x $ and $g(x,y) =x^2$ will do. –  Shinigami Feb 10 '13 at 6:03
    
Ah, I think your example can be fixed in the following way. Consider instead $\mathcal F_x/\mathcal M_x \mathcal F_x$ where $\mathcal M_x$ is the maximal ideal in the stalk of the structure sheaf at $x$, and $\mathcal F_x$ is the stalk of $\mathcal F$ at $x$. Then, this has dimension $2$ at $0$ and $1$ everywhere else. This also suggests that if we were to consider $\mathbb R$ instead of $\mathbb R^2$, the ideal sheaf might indeed be locally free. –  Shinigami Feb 10 '13 at 6:19
    
Oh dear- I was being silly! Good catch! Feel free to edit it in, I've made my answer community wiki –  Dylan Wilson Feb 11 '13 at 4:25
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