I have always studied arithmetical definitions of the most common algebraic structures. Basically we see algebraic structures as set with operations, and in general we ca think them as models for specific first order theories in some specified langauges.
Are there more categorical definitions for the most common algebraic structures? I mean definitions which exploit arrows and diagrams?

For instance, I think we can define a field F, inside the category of rings by asking that, for any nonzero ring $B$ (which i think we can translate by saying $B$ is not terminalin $\operatorname{Ring}$) $\operatorname{Ring}(F,B)$ is only made of monomorphisms.

But then what about other important structures defined from rings as ideals or modules? And what about rings themselves (inside abelian groups)? Is there some general framework to develop such a treatment of algebraic structures?

Any reference tackling the issue with rather acessible tools (basic category theory) would be great too! Thanks for your help.

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    $\begingroup$ Kernels are possible to define categorically. And I'm sure you know that monomorphisms and epimorphisms are categorical versions of injective and surjective functions. $\endgroup$
    – Arthur
    Commented Mar 19, 2020 at 16:51
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    $\begingroup$ The standard approach to defining algebraic structure through category theory is through Lawvere theories (also simply called algebraic theories), which allows you to define algebraic structures and their models in categories with sufficient structure in a presentation-free way. $\endgroup$
    – varkor
    Commented Mar 19, 2020 at 17:32
  • $\begingroup$ If your rings require a unity and unital morphisms (as seems to be the case if you have $\mathrm{Ring}(F,B)$ consist only of monomophisms with $F$ a field), do you disallow the $0$ ring somehow? Because otherwise, (i) the $0$ ring satisfies the given property, but is not a field; and (ii) $\mathrm{Ring}(F,\mathbf{0})$ consists only of monomoprhisms if and only if $F=\mathbf{0}$. So I'm not sure your characterization works. $\endgroup$ Commented Mar 19, 2020 at 18:40
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    $\begingroup$ Note however that fields are not algebras in the sense of universal algebra, since "multiplicative inverse" is not an operation and cannot be defined via identities. They are partial algebras, though. $\endgroup$ Commented Mar 19, 2020 at 18:47
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    $\begingroup$ This recent mathoverflow question of mine is related. It turns out that you can define sets (and hence everything else) categorically inside the category of groups. $\endgroup$ Commented Mar 20, 2020 at 12:00

2 Answers 2


There are a lot of ways of talking about what a algebraic theory is, in a categorical fashion. As was mentioned, Lawvere theories are one of them, but you might also be interested in the notion of monad.

To put it roughly, a Lawvere theory is a category $C$ whose objects are generated by iterated coproducts of a specific object (and an initial object $0$). I will denote $1$ the specific object, $2$ the coproduct $1+1$, $3$ the coproduct $1+1+1$ and so on. We think of the category $C$ as a collection of operations in all arities, together with equalities between these operations. That is exactly what an algebraic theory is.

To be clearer, we will call a model of the Lawvere theory $C$ a contravariant functor $C\to \mathbf{Set}$ that sends the colimits on limits. That might not yet mean a lot, but if we actually unfold what a model is : there is an image for the object $1$, that I will call $M$, then there is an image for the object $2$, but since I have to send coproducts on products, it has to $M^2$, and so on... So the images of the objects are $1,M,M^2,M^3,\ldots$ (where $1$ is the singleton set), in terms of data, that's really just the data of a set $M$.

Suppose now that the theory $C$ has a map $f : 1 \to 2$, then the functor has to produce an arrow $\mu_f:M^2\to M$. Well that's just a binary operation! So you see you really can encode operations as morphisms $C$. But that's not all : since $3 = 2+1$, the arrow $1\to 2$ together with the identity induce automatically an arrow $f+\mathrm{id} : 2 \to 3$, which translate as an operation $(\mu_f,\mathrm{id}) M^3\to M^2$, obtained by multiplying by $\mu_f$ the first two components of a triple in $M^3$ and leaving the third one untouched, and similarly $3 = 1+2$, so there is an induced arrow $\mathrm{id}+f : 2 \to 3$ which turns into an arrow $(\mathrm{id},\mu_f):M^3\to M^2$ which leaves the first component of a triple untouched and multiply the last two. So you see that the arrows of $C$ not only encode composition law, but also encode all other operations these composition laws induce.

Now the grand final : Imagine that in the category $C$ the two arrows $(\mathrm{id}+f)\circ f : 1 \to 3$ and $(f+\mathrm{id})\circ f : 1 \to 3$ coincide. Then it implies that for the model $M$, the following equality is satisfied : $$ \mu_f\circ (\mu_f,\mathrm{id}) = \mu_f(\mathrm{id},\mu_f)$$ More concretely, for all $x,y,z \in M$, evaluating on both sides on the triple $(x,y,z)$, yields $\mu_f(\mu_f(x,y),z) = \mu_f(x,\mu_f(y,z))$. That's called the associativity of $\mu_f$. So equalities between compositions of arrows in the theory $C$ translate to axioms of a theory. To sum this up : a Lawvere theory encodes operations and axioms between them as arrows in a category.

Doing so I can write down explicitly a Lawvere theory for monoids, commutative monoids, groups, abelian groups,... As it has previously been mentioned, I chose models to be a functor to $\mathbf{Set}$, but I could have chosen any category with (finite) products, and made sense of a reasonable notion of monoids, groups... in this category. Topological groups (models for the L. theory of groups in the category of (nice) topological spaces), Lie groups (models for the L. theory of groups in the category of manifolds) are particular examples of this. Someone with more knowledge than me in algebraic geometry could give thousands more examples.

That is nice and all, but this is not the only way to do such tricks. You can also understand an algebraic theory as what we call a monad. This is an endofunctor $T$, together with natural transformations $\eta : 1 \to T$ and $\mu : T\circ T \to T$ satisfying some axioms that I won't spell out here. Note that now you need to restrict yourself in a given category. I will try and explain how these also correspond to algebraic theories by taking a specific example : Consider the monad on $\mathbf{Set}$ which associates to each set $X$ the set $X^*$ of lists made out of elements of $X$. The unit $\eta$ is given by associating the function which associates to each element $x$ the one element list $(x)$, and the multiplication $\mu$ associates to a list of list of elements of $X$ the big list obtained by concatenating all the lists.

I will say that an algebra for this monad is a set $X$ equipped with a map $X^*\to X$. Now what is that : A map $X^* \to X$ is something that associates to every list of element of $X$ another element of $X$. For instance given two elements $x,y$ of $X$, I can build the list $(x,y)$, and get an element of $x$ out of it, that I can call the product of $x$ and $y$. Now this map is required to respect the maps $\mu$ and $\eta$, and spelling out explicitly what that means will show that an algebra for this monad is a monoid. So in a monad the operations are encoded by the image of the object, and the axioms of the theory are encoded by the $\eta$ and the $\mu$.

So the point now is to compare both :

  • Lawvere theory are abstract and can span the same notion across various different categories (groups, topological groups, Lie groups,...) whereas monads take place in a specific category
  • All Lawvere theories give rise to a monad in set, such that the algebra of this monads are exactly the same as the models of the Lawvere theory
  • There are monads that do not correspond to a Lawvere theory, but finitary monads do. So monads in the sets are a bit more general than Lawvere theory

You might also want to look at operads, which are yet another way to present a theory (pretty similar to Lawvere theory), and which relate to both Lawvere theory and monads

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    $\begingroup$ Definitely thanks for this wonderful answer! $\endgroup$ Commented Mar 21, 2020 at 8:01

The most general answer I am aware of is that of @Varkor‘s comment: Lawvere theories.

A more elementary special case is that of. monoid objects and group objects in a category and of corresponding algebras/modules/$G$-objects. For example (commutative) rings are nothing more than (commutative) monoid objects in the category of abelian groups. Modules over a fixed ring are nothing more than objects in the category of modules for this monoid etc. This is rather straightforward, but requires (at least I did) some time to getting used to. The cool thing about this is that it allows to consider algebra in completely different categories than Set, for example topological groups and topological vector spaces can be obtained using this general nonsense.

The following can be done in any category with finite products. A group object is an object $G$ together with morphisms $m:G\times G \rightarrow G$, $i:G\cong G$ and $e:1\rightarrow G$ subject to the usual axioms of group theory, e.g. associativity $m(m\times G) = m(G\times m): G \times G \times G$ neutral laws like $m(e\times G) = 1_G$ and inverse laws like $m(i\times G)\Delta = e!_G$, where $\Delta$ denotes the diagonal and $!_G$ is the unique arrow to the terminal object. The usual definition of a topological group just states that multiplication and inversion are continuous. If you take the definition of a group object and consider it in the category Top this is trivially satisfied, so the only thing to worry about is that all the group axioms are there (I gave only an excerpt).

Wikipedia lists more examples such as Lie-groups.

  • $\begingroup$ Thanks for your answer! Could you espand the final a bit? I am very courious. Are you saying that if we consider the categorical definitions of groups and vector spaces in Top we get automatically topological groups and topological vector spaces? (with all the required compatibility conditions realozed?) $\endgroup$ Commented Mar 20, 2020 at 9:33
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    $\begingroup$ I added some more details. $\endgroup$ Commented Mar 20, 2020 at 10:29

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