Isomorphism of sheaves of rings

Question

Suppose we have locally ringed spaces $(X, F_1), (X, F_2)$.

If $F_1(X)$ and $F_2(X)$ are isomorphic as rings, can we conclude that sheaves $F_1$ and $F_2$ are isomorphic?

I think this is true because isomorphism of sheaves $\phi : F_1 \rightarrow F_2$ are the set of isomorphisms of rings $\phi(U) : F_1(U) \rightarrow F_2(U)$ for any open set $U$ in $X$.

Furthermore, if we have locally ringed spaces $(X, F_1), (Y, F_2)$ where X and Y are homeomorphic and $F_1(X)$ and $F_2(Y)$ are isomorphic as rings, can we still conclude that sheaves $F_1$ and $F_2$ are isomorphic as sheaves?

Thank you.

Answer 1

This is not true.

For any local ring $A$, the constant sheaf $\underline A$ on $X$ yields a locally ringed space on $X$. Hence, if for any given nonconstant sheaf $F$ on $X$, its global sections $A = F(X)$ happens to be a local ring, then $F ≠ \underline A$, but $F(X) = \underline A(X)$.

This gives a lot of counterexamples. For example any projective space $\mathbb P^n_A$ or most affine spectra $\operatorname{Spec} A$ for a local ring $A$. For the affine spectra, just make sure that they have more than one point.