If $\mathcal{F}$ is quasi-coherent on $X=\operatorname{Spec}(A)$, then do we have $\mathcal{F}(X)^\sim = \mathcal{F}$? It is a well known fact that on any affine scheme $X= \operatorname{Spec}(A)$ the functor $M \mapsto M^\sim$ gives an equivalence of the category of $A$-modules and the category of quasi-coherent $\mathcal{O}_X$-modules. See for instance Hartshorne, II.5.2 or Görtz/Wedhorn 7.17.
In particular, for every quasi-coherent $\mathcal{F}$ on $X$ there is some $A$-module $M$ such that $M^\sim$ is isomorphic to $\mathcal{F}$ as $\mathcal{O}_X$-modules.
Since $\mathcal{F}(X)$ is an $A$-module, this at least suggests that $\mathcal{F}(X)^\sim \neq \mathcal{F}$ for general quasi-coherent $\mathcal{F}$. I was wondering why this should be true and hence I would like to see an example of this!
I am grateful for any kind of help!
 A: In most of modern algebra, the notion of a real literal equality of sets or objects is usually too restrictive to be useful: who cares if we use $\{0,-1\}$ with addition/multiplication mod 2 as $\Bbb F_2$ instead of $\{0,1\}$? All the algebraic properties we would care about are satisfied by one just as well as the other.
This is especially true when working with sheaves. Sheaves are in general a gigantic amount of information: we require a module over the appropriate ring for each open subset and then restriction maps between these modules for each inclusion of open sets.
For instance, if we pick $A=\Bbb F_2[x]_{(x)}$ then $\operatorname{Spec} A$ is $\{(0),(x)\}$ as a set with the topology $\{\emptyset,\{(0)\},\{(0),(x)\}\}$ and the structure sheaf is given as follows:


*

*$\mathcal{O}_X(\emptyset)=\{0\}$

*$\mathcal{O}_X(\{(0)\}) = \Bbb F_2(x)$

*$\mathcal{O}_X(\{(0),(x)\}) = \Bbb F_2[x]_{(x)}$
where the only nonzero restriction map is the canonical embedding of $\Bbb F_2[x]_{(x)}$ in to its fraction field $\Bbb F_2(x)$.
Now define a new sheaf $\mathcal{P}$ by (almost) the same data:


*

*$\mathcal{O}_X(\emptyset)=\{\}$

*$\mathcal{O}_X(\{(0)\}) = \Bbb F_2(x)$

*$\mathcal{O}_X(\{(0),(x)\}) = \Bbb F_2[x]_{(x)}$
where by $\{\}$ I mean the ring which has one element $$ and we have $+=$ and $\cdot =$. We keep the only nontrivial restriction map as the same as in $\mathcal{O}_X$ above, and all other restriction maps are forced since we're mapping in to a ring with one element. Now, $\mathcal{P}$ has the same global sections as $\mathcal{O}_X$ constructed above (literally the same set!) but is not equal as a sheaf because $0\neq $.
This is really silly! Replacing one module in the definition of our sheaf by an isomorphic one which doesn't have the same underlying set breaks the relation of equality, but not isomorphism.

In category theory, it turns out that even isomorphisms of categories are inappropriate, and we want to talk about equivalences instead. To see why, consider the following example. Let $\mathcal{C}$ be the category of all finite-dimensional vector spaces with $k$-linear maps over some fixed base field $k$, and let $\mathcal{D}$ be the category which has as objects $k^n$ for all integers $n\geq 0$, and morphisms the $k$-linear maps. These categories are not isomorphic: there are no functors $\mathcal{C}\to\mathcal{D}$ and $\mathcal{D}\to\mathcal{C}$ which we can define which compose both ways to the identity functor on $\mathcal{C}$ and $\mathcal{D}$. This is because $\mathcal{C}$ has many, many, many, many more objects that $\mathcal{D}$: the objects of $\mathcal{C}$ form a set, but the objects of $\mathcal{D}$ form a proper class! On the other hand, these two categories are "essentially the same" - they're both just finite dimensional vector spaces over our field. So equivalence comes to the rescue and lets us consider these categories as equivalent (where we only require that the composition of functors be isomorphic to the identity, not literally equal to the identity), instead of equal or isomorphic.
Moral of the story: use the correct notion of isomorphism. In this case (and in many others), literal equality is inappropriate.
