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 :)

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 $X$.
