If sections of certain sheaves agree on an open neighbourhood does it agree globally? Let $X$ be a non-reduced scheme whose associated reduced scheme is smooth. Assume as well that $X$ is irreducible. Suppose that two global sections of the tangent sheaf of $X$ agree on a non-empty open set in $X$. Does this mean that the two global sections are the same? Does the same statement hold true if we replace the tangent sheaf by the normal sheaf of some closed subscheme of $X$?
 A: Firstly, "smooth" doesn't make sense for a scheme (it is a property of a morphism between schemes, not of an individual scheme); do you mean regular? Or does your scheme lie over a field $k$, so that smoothness is being discussed relative to $k$?
Also, what do you mean by the tangent sheaf of $X$?  If $X$ is smooth over a field $k$, then I would think of the tangent sheaf as being the dual to the sheaf $\Omega^1_{X/k}$, which is locally free by smoothness.  But your $X$ is not smooth over $k$ (being non-reduced); are you defining the tangent sheaf as a sheaf of derivations, or in some other way?

In any case, in general on a non-reduced scheme functions (and hence also sections of locally free sheaves) are not determined by their restriction to a dense open subset.  
Considering the difference of the two functions/sections, it is enough to consider
whether there is a function/section supported on proper closed subset.
Since sections of a locally free sheaf of rank $n$ are locally $n$-tuples of
functions, I will restrict to the case of functions, i.e. sections of $\mathcal O_X$.
If we work locally, on some affine open Spec $A$ (and let me assume $A$ is Noetherian, for safety), the following are equivalent:


*

*Every element of $A$ that vanishes on a dense open subset of Spec $A$
actually vanishes.

*Every associated prime of $A$ is minimal.

*$A$ is $S_1$.
(The last two conditions are more-or-less equivalent by definition;
the equivalence of the first two uses the basic theory of associated
primes.)
Now a Noetherian ring if reduced if and only if its is $R_0$ (i.e. generically reduced) and $S_1$.
So if we take a ring that is generically reduced but non-reduced, it is not
$S_1$, and hence admits non-zero elements that vanish on a dense open subset.
E.g. $A = k[x,y]/(xy, y^2).$   Note that $A_{\mathrm{red}}= k[x]$, 
so Spec $A$ is irreducible (the underlying reduced subscheme is just the affine line); that $A[1/x] = k[x,1/x],$ so 
Spec $A[1/x]$ is reduced; and $y$ vanishes when restricted to Spec $A[1/x]$.
But $y$ does not actually vanish in $A$; it is a nilpotent element supported
just at the point $x = 0$.
