Points of affine scheme and points of topos Let $X =$Spec$(R)$ be an affine scheme and $\mathscr{E}_{X}$ denote the topos of sheaves of sets on $X$. Is it true that the (geometric) points of $\mathscr{E}_{X}$ are in bijection with the prime ideals of Spec$(R)$? That is, that the points of this topos are in bijection with the points of the scheme $X$?
To each prime ideal $p \in $Spec$(R)$, one can form a morphism to $\varphi: R \rightarrow \overline{k}_{p}$ from $R$ to an algebraically closed field. This gives a geometric morphism from sets to $\mathscr{E}_{X}$ i.e. a point of the topos $\mathscr{E}_{X}$. 
However, I am unsure whether or not each such point of the topos must be of this form. If anyone can shed any light on this, or just point me to a reference where this is discussed, that would be great. Thanks. 
Edit: I have conflated two ideas simply because they are both called points --- This could just be a coincidence.  
 A: Yes, @user45878 is right.
Assuming the Boolean Prime Ideal Theorem (which is slightly weaker than the axiom of choice, but still wildly unconstructive), the spectrum of a ring is a sober space, such that there is no difference between topological points and set-theoretical points.
In constructive mathematics, the classical definition of the spectrum as the topological space of prime ideals isn't very useful -- the definition can be made, but the resulting space won't satisfy the expected universal property. (There are models of constructive mathematics in which there are nontrivial rings without any prime ideals.) But there is a nice twist in this case, which is why I'm making this extended comment.
In constructive mathematics, you usually define the spectrum in a different way -- as a locale, as a topos or as a formal space. Such gadgets still have a notion of points, but they're not fundamental to their definition and it can happen that such a gadget doesn't have any points and still is nontrivial. The points of these constructive versions of the spectrum are in one-to-one correspondence with the "prime filters" of the ring.
The notion of "prime filter" is a direct axiomatization of what's classically is the complement of a prime ideal. For instance, where we have the axiom "$xy \in \mathfrak{p} \Rightarrow (x \in \mathfrak{p}) \vee (y \in \mathfrak{p})$", a prime filter satisfies "$(x \in F) \wedge (y \in F) \Rightarrow xy \in F$".
Thus, even in constructive mathematics, the answer to your question is "yes" -- modulo the passage from prime ideals to prime filters.
