I've been going through Emily Riehl's textbook on categories, and struggle with the exercise 1.5.xi.

Consider the functors $Ab \to Grp$ (inclusion), $Ring \to Ab$ (forgetting the multiplication), $(-)^\times: Ring \to Grp$ (taking the group of units), $Ring \to Rng$ (inclusion), $Fld \to Ring$ (inclusion) and $Mod_R \to Ab$ (forgetful). Determine which functors are full, which are faithful, and which are essentially surjective. Do any define an equivalence of categories? (Warning: A few of these questions conceal research-level problems, but they can be fun to think about even if full solutions are hard to come by.)

As mentioned, some of the problems are pretty hard, so I skipped some of them. For now, I've determined that properties only for $Ab \to Grp$, $Ring \to Ab$ and $Fld \to Ring$. This are my solutions:

  • $Ab \to Grp$: almost obviously, this functor is not essentially surjective, whileas it's fully faithful: if two groups are abelian, then any abelian homomorphism between them is just a group homomorphism, so it's faithful, and if two groups in $Grp$ are in fact abelian groups, then they have the same set of morphisms as objects in $Ab$, so it's fully faithful, and really $Ab$ is a full subcategory of $Grp$.
  • $Ring \to Ab$: this functor is not essentially surjective(because e.g. $\mathbb{Q}/\mathbb{Z}$ is not a ring); it's faithful, since different ring homomorphisms are going to be different group homomorphisms under functor action. But this functor is not full: consider rings $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}$. If $U: Ring \to Ab%$, then consider $\varphi: U(\mathbb{Z}/n\mathbb{Z}) \to U(\mathbb{Z}): x \mapsto 0$. There's no $\psi: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}$ such that $U(\psi) = \varphi$, so this functor is not full.
  • $Fld \to Ring$: in some sense same as for $Ab \to Grp$: $Fld$ is full subcategory of $Ring$ and this functor is not obviously essentially surjective.

I'd like to know whether I am correct. Seems like there was nothing hard, but I have doubts on my solutions.

Also I'm interested in the properties of another functors. I was thinking about $Mod_R \to Ab$: since it's forgetful, it's faithful. But is it full? Or essentially surjective? Obviously, not both(otherwise, this is an equivalence of categories), and the answer is somehow depends on the ring(because if $R = \mathbb{Z}$, then this is the equivalence of categories). Probably I can show that such functor is essentially surjective, since I kinda can create module via tensor product $G \otimes_\mathbb{Z} R$(which is kinda left-adjoint to $Mod_R \to Ab$), but I'm not sure in it.

So, can anyone elaborate the situation with $Mod_R \to Ab$ and the other functors? Or maybe give a tip on it. Concretely, I have no clue on:

  • $(-)^\times: Ring \to Grp$ (taking the group of units)
  • $Ring \to Rng$ (inclusion)



2 Answers 2


It's not so clear why Ring to Ab is not essentially surjective. For example, the Rng to Ab is obviously essentially surjective: For any abelian group $G$, it can be lifted to be a Rng by defining $a\cdot b=0$ for all $a,b\in G$. While $\mathbb Q/\mathbb Z$ isn't equipped with an obvious ring structure, it's unclear why there couldn't be some way to define the multiplication. This needs a proof: Does every Abelian group admit a ring structure?

Mod$_R$ to Ab is not necessarily essentially surjective: Let $R=\mathbb F_2$, then any module $M$ over $R$ is just vector spaces over $\mathbb F_2$, in particular every element of $M$ has order $2$, which isn't the case for general abelian groups.

It's not necessarily full either: Let $R=\mathbb C[T]$ and $M\simeq\mathbb C^n$ is a $R$-module with $T$ acts on $M$ through a matrix $A$. Then End$_R(M)$ consists of the set of matrices that commute with $A$, hence End$_R(M)$ is not necessarily End$_{\mathbb Z}(M)$. Or even simpler, we can have $\mathbb R$-linear transformations between $\mathbb C^n$ and $\mathbb C^m$ that are not $\mathbb C$-linear.

  • $\begingroup$ Thank you for the response! There's also a proof on why there's no ring structure here. Thank you for the clarification on $Mod_R \to Ab$ as well. So basically it's not full because any abelian group is $\mathbb{Z}$-module, so if the functor is full, then $End_R(M) \simeq End_\mathbb{Z}(UM)$, but this is not true in general, right? $\endgroup$ Nov 28, 2021 at 19:53
  • $\begingroup$ Yes. This is also related to Morita equivalence. $\endgroup$ Nov 29, 2021 at 2:35
  • $\begingroup$ Oh, could you explain the relation to Morita equivalence? I don't know much about it, where I can read something on it? $\endgroup$ Nov 29, 2021 at 12:19
  • $\begingroup$ I read about it in Jacobson's Basic Algebra II, but there are many resources on this topic. Rings $R$ and $S$ are Morita equivalent if their categories of modules are equivalent, however this equivalence is in general given by tensor product, not forgetful functors. Clearly this is related to this problem, and details are left for further reading. $\endgroup$ Nov 30, 2021 at 0:18
  • $\begingroup$ As an example of the last sentence, we can use complex conjugation $\mathbb{C} \to \mathbb{C}$, which is a homomorphism of abelian groups but is not $\mathbb{C}$-linear. $\endgroup$
    – Anakhand
    Jan 19 at 9:23

Regarding the two other functors you ask about:

The functor $\text{Ring} \to \text{Group}$ that takes the group of units is neither faithful, full nor essentially surjective. These two threads take care of this:

The inclusion functor $\text{Ring} \to \text{Rng}$ is faithful, but neither full nor essentially surjective.

  • Faithful. Indeed, the data that defines the identity (equality) of a rng homomorphism and that of a ring homomorphism is exactly the same (a set function), the only difference is the properties it is required to have, namely unital or not (which we forget with this functor).
  • Not full. Consider the rng homomorphism $\mathbb{Z} \hookrightarrow \mathbb{Z} \times \mathbb{Z}$ defined as $n \mapsto (n, 0)$.
  • Not essentially surjective. Suppose that a rng $R$ is isomorphic to a ring $S$ through the isomorphism $\phi$. Let $x \in R$ be the element such that $\phi(x) = 1_S$. Then $\phi(x^2) = 1_S$, so by injectivity $x = x^2$. Therefore, if a rng is isomorphic to a ring, it must have an idempotent. But there are plenty of examples of rngs without idempotents, e.g. $2\mathbb{Z}$.

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