For any ring $R$, prove that $R$-$\mathbf{Mod}$ has no subobject classifier. This is Exercise I.3 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]".
The Question:

For any ring $R$, prove that the category $R$-$\mathbf{Mod}$ of all left $R$-modules has no subobject classifier.

I assume that the morphisms of $R$-$\mathbf{Mod}$ are module homomorphisms; that is, $M\stackrel{f}{\rightarrow} N$ is given by, for all $x,y\in M$ and all $r\in R$,
$$\begin{align}
f(x+y)&=f(x)+f(y)\\
f(rx)&=rf(x).
\end{align}$$
I'm guessing that rings are intended to have a $1$ and are not necessarily commutative.
A definition of a subobject classifier is given on page 32, ibid.

Definition: In a category $\mathbf{C}$ with finite limits, a subobject classifier is a monic, ${\rm true}:1\to\Omega$, such that to every monic $S\rightarrowtail X$ in $\mathbf{C}$ there is a unique arrow $\phi$ which, with the given monic, forms a pullback square
$$\begin{array}{ccc}
S & \to & 1 \\
\downarrow & \, & \downarrow {\rm true}\\
X & \stackrel{\dashrightarrow}{\phi} & \Omega.
\end{array}$$

Thoughts:
Following the answers to my previous question on the nonexistence of a subobject classifier in $\mathbf{FinSets}^{\mathbf{N}}$, I have considered using the Yoneda Lemma; however, I'm not sure how or whether it applies: the "target category," so to speak, for the Lemma is $\mathbf{Sets}$.
Also, I ask myself, "what would a subobject classifier in $R$-$\mathbf{Mod}$ look like?"
To answer this, I considered first the existence of a terminal object in the category. My guess is that it's $I=(\{0_R, 1_R\}, \times_R, +_R)$, since, for any $R$-module $M$, we have $!: M\to I$ given by
$$!(m)=\begin{cases}
0_R &: m=0_M, \\
1_R &: \text{ otherwise}.
\end{cases}$$
But I don't think this is right. Perhaps my problem is my understanding of left $R$-modules.
Please help :)
 A: The terminal and initial object is the $0$-module, $\{0\}$. Addition/multiplication with elements in $R$ given bu the only way possible. Consider $S = 0$. Then we get that $\ker (\phi) = 0$ and every $X$ embeds into $\Omega$. A morphism with zero kernel in $R$-$\mathbf{Mod}$ has to be a monomorphism and because right adjoints preserve monomorphisms and the forgetful functor $R$-$\mathbf{Mod} \rightarrow \mathbf{Set}$ is a right adjoint $\phi$ has to be injective on set level. There are no size restrictions on $R$ modules thus we arrive at a contradiction, there can't be an injection $X \rightarrow \Omega$ for every $X$.
A: To expand on a Yoneda Lemma based approach: suppose we have a subobject classifier $\Omega \in R{-}\mathbf{Mod}$.  Then the first thing we might wonder about $\Omega$ is what its underlying set would be.  Now, the underlying set functor $U : R{-}\mathbf{Mod} \to \mathbf{Set}$ is representable by the $R$-module $R$, so we must have
$$U(\Omega) \simeq \operatorname{Hom}_{R{-}\mathbf{Mod}}(R, \Omega) \simeq \operatorname{Sub}_{R{-}\mathbf{Mod}}(R).$$
But the submodules of $R$ are exactly the left ideals of $R$.  (So at this point, we might already suspect that there will be no subclassifier object, at least in general, since there is no obvious $R$-module structure on the set of left ideals of $R$.  But it remains to find a precise contradiction.)
Next, we can look at what the action of $R$ on this set would be.  This would be induced by the morphism $r\cdot : R \to R$, so we would get that the action has to be:
$$r \boxdot I = \{ x \in R \mid rx \in I \}.$$
Thus, for example, if $R$ is a nontrivial ring in which 2 is a unit, then we already have a contradiction since this gives $(-1) \boxdot I = I = 1 \boxdot I$ for every ideal $I$, so $I = 0 \boxdot I = \langle 1 \rangle$; thus the zero ideal would also be equal to the unit ideal, contradicting the assumption that $R$ is nontrivial.
As for the sum operation (let us call that $\boxplus$ to distinguish it from the usual ideal sum), let us consider the diagonal morphism $\Delta : R \to R \oplus R$ along with the inclusion morphisms $i_1, i_2$; we then see that since $\Delta = i_1 + i_2$, we would have to have
$$I \boxplus J = i_1^*(I \oplus J) \boxplus i_2^*(I \oplus J) = \Delta^*(I \oplus J) = I \cap J.$$
This now gives a contradiction in the general case: for every ideal $I$, we must have $\langle 1 \rangle = 0 \boxdot I = I \boxplus (-1) \boxdot I = I \boxplus I = I \cap I = I$.  Thus, again, the implication that $\langle 0 \rangle = \langle 1 \rangle$ would give that $R$ must be trivial.
(And in fact, if $R$ is the trivial ring in which $0 = 1$, then the category of $R$-modules is equivalent to the one-object, one-morphism category since every $R$-module has exactly one element; and this does give a (degenerate) topos with subobject classifier $\{ 0 \}$.)

For another point of view, let's say we were at the point where we know $U(\Omega)$ must be the set of left ideals of $R$.  Then, we can calculate that given a submodule $N \subseteq M$, the underlying function $U(M) \to U(\Omega)$ would need to send $x \mapsto \{ \lambda \in R \mid \lambda x \in N \}$, i.e. it sends $x$ to the annihilator ideal of $x + N \in M / N$.  However, what we find is that just knowing the annihilator ideals of $x + N, y + N \in M / N$ is not enough information to conclude what the annihilator ideal of $x + y + N \in M / N$ must be.  From here, the rest of the proof would consist of coming up with a counterexample which works for any nontrivial ring $R$, and drawing a contradiction from that (the upshot being that the sum of the two corresponding ideals under the $R$-module structure of $\Omega$ would have to be equal to two distinct ideals at the same time).
