# 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.

• The terminal $R$-module is the zero module. Based on your recent questions about computation of terminal objects, you seem to be missing some of the prerequisites for topos theory. You would probably benefit from spending some time with a general category theory book such as Leinster's Basic Category Theory, Riehl's Category Theory in Context, or Mac Lane's Categories for the Working Mathematician before continuing with toposes. – Kevin Arlin Feb 2 at 15:11

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 $$X$$.
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).