What does the stalk perceive?

Let $\mathcal{F}$ be a sheaf (or a separated presheaf) of abelian groups on a scheme $X$. Let $s$ be a global section of $\mathcal{F}$.


Notice that even $s(p) = 0$, $s_p$ could still be nonzero.

Example. We give an example to illustrate the fact that two sections become the same element in the stalk at $p$ iff they restrict to the same section on a neighborhood of $p$ (but not that they agree pointwise) (Of course, this is simply by definition of stalk).

 Let $X = \text{Spec} A$. Two global sections $f, g \in A$ satisfies $f_p = g_p$ iff $f_p - g_p$ is $0$ in $A_p$. So let's look at the kernel of $A \to A_p$.

These consist of $s \in A$ such that $h s = 0$ for some $h \not \in p$. Thus $s $ is $0$ on $D(h)$, which is a neighborhood of $p$. Thus if $s_p = 0$ then $s$ vanishes on a neighborhood of $p$.

The converse is not true.

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We see that if $s_p = 0$ then $s$ is $0$ on a neighborhood of $p$ and thus $s(p) = 0$. Is the converse true?

The answer is No. Warning:
 

$$s(p) = 0 \not \implies  s_p = 0.$$

Example. Let $X = \text{Spec} k[x]/(x^2)$. Then $X$ has only one point, but there are extra first derivative vectors at the single point of $X$ also. As in, two global sections of $X$ are identical iff they agree on points of $X$ and also have the same first derivative. Now let $s = x$. At the point $p= (x)$, we have $s(p) $ is $0$ in $ A/(x)$ (or equivalent, in $\text{Frac} A/(x) = A_{(x)}/(x)$). However $s_p$ is not $0$ in $A_{(x)} = (k[x])_{(x)}/(x^2)$. This is because $s= x$ is  $0$ in the latter iff $x f = x^2 g$ for some $f \not \in (x)$, which is impossible as $k[x]$ is a UFD. Thus $\text{Supp} s = X$.

Intuitively, what happened is that for $s_p$ to be $0$, $s$ needs to be $0$ on a neighborhood of $p$. In our case, a neighborhood of $p$ would include the extra derivatives at $p$.   So since $s$ only vanish at $p$ with order $1$, its stalk at $p$ is not $0$.

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