# Free modules internal to a nice category

Let $\mathcal{C}$ be a cocomplete category with finite products such that for all $X \in \mathrm{Ob}(\mathcal{C})$ the functor $X \times - : \mathcal{C} \to \mathcal{C}$ is cocontinuous; perhaps we will even need that $\mathcal{C}$ is cartesian closed, and/or locally finitely presentable. Let $R$ be a ring object in $\mathcal{C}$ and consider the category ${}_R \mathrm{Mod}$ of left $R$-module objects internal to $\mathcal{C}$ (these are objects $X \in \mathrm{Ob}(\mathcal{C})$ equipped with morphisms $+ : X \times X \to X$, $\cdot : R \times X \to X$, $0 : 1 \to X$ and $- : X \to X$, such that certain diagrams commute). We have a forgetful functor ${}_R \mathrm{Mod} \to \mathcal{C}$. How can we describe its left adjoint (if it exists)? That is, how does the free left $R$-module object on $X \in \mathrm{Ob}(\mathcal{C})$ look like?

We are familiar with the case $\mathcal{C}=\mathsf{Set}$. What's special in this situation is that $X$ is a coproduct of copies of $1$, and in that case the free left $R$-module is just a direct sum of copies of "$R$". But I don't directly see what to do with arbitrary objects $X$.

• Are you asserting that the left adjoint exists, or are you asking whether it exists? – Zhen Lin Sep 30 '15 at 10:46
• Such an module $F(X)$ would be an initial object in the category $X/\mathcal{U}$, where $\mathcal{U}$ is the forgetful functor, right? I.e. an object $F(X)$ such that there is a 1-1 correspondence between morphisms $X\rightarrow \mathcal{U}(M)$ and module homomorphisms $F(X)\rightarrow M$. It's not obvious to me that such an object exists in such a general setting. – Espen Nielsen Sep 30 '15 at 11:17
• @ZhenLin: I was pretty sure that it exists, but meanwhile I'm not so sure anymore. So my question is (a) does it exist, (b) how can we construct it. – Martin Brandenburg Sep 30 '15 at 13:21
• @EspenNielsen: It's not obvious to me either. That's why I asked. ;) – Martin Brandenburg Sep 30 '15 at 13:22
• Actually, come to think of it, can we even define the "obvious" morphism $X \to R^X$ without assuming $X$ is decidable? So maybe we won't always be able to define $F (X)$ as a submodule of $R^X$. – Zhen Lin Sep 30 '15 at 13:56

Let $\mathcal{S}$ be a category with finite products and let $R$ be a ring in $\mathcal{S}$. There is an $\mathcal{S}$-enriched Lawvere theory $\mathcal{T}_R$ where $\mathcal{T}_R (n, m) = R^{m \times n}$ with composition defined by matrix multiplcation. We can define models of $\mathcal{T}_R$ in $\mathcal{S}$ to be an object $M$ together with morphisms $\alpha_n : \mathcal{T}_R (n, 1) \times M^n \to M$ making the obvious diagrams of the form below commute: $$\require{AMScd} \begin{CD} \mathcal{T}_R (m, 1) \times \mathcal{T}_R (n, m) \times M^n @>>> \mathcal{T}_R (n, 1) \times M^n \\ @VVV @VV{\alpha_n}V \\ \mathcal{T}_R (m, 1) \times M^m @>>{\alpha_m}> M \end{CD}$$ It is straightforward to check that the category of models of $\mathcal{T}_R$ is equivalent to the category of $R$-modules (as concrete categories over $\mathcal{S}$).

Now, suppose $\mathcal{S}$ has colimits of countable diagrams. Then we can define the coend $$F (X) = \int^{n : \mathbf{FinSet}} \mathcal{T}_R (n, 1) \times X^n$$ and if $\times$ preserves colimits in each variable, then there is an induced $R$-module structure on $F (X)$. There is also an evident morphism $X \to F (X)$, and it exhibits $F (X)$ as the free $R$-module generated by $X$. Thus, very concretely, the free $R$-module generated by $X$ is the quotient of $$\coprod_{n \ge 0} R^n \times X^n$$ by a certain countable family of parallel pairs: writing an element of $R^n \times X^n$ suggestively as $$\begin{pmatrix} r_1 & \cdots & r_n \end{pmatrix} \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix}$$ we are imposing the relations $$\begin{pmatrix} a_1 & \cdots & a_m \end{pmatrix} \begin{pmatrix} x_{f (1)} \\ \vdots \\ x_{f (m)} \end{pmatrix} = \begin{pmatrix} b_1 & \cdots & b_n \end{pmatrix} \begin{pmatrix} x_{1} \\ \vdots \\ x_{n} \end{pmatrix}$$ for every map $f : \{ 1, \ldots, m \} \to \{ 1, \ldots, n \}$, where $b_1, \ldots, b_n$ are defined as follows, $$b_j = \sum_{f (i) = j} a_{i}$$ i.e. $\vec{b} = \vec{a} F^\mathsf{T}$, where $F$ is the 0-1 matrix corresponding to the map $f$ and $\vec{a}$ and $\vec{b}$ are the obvious row vectors.

Of course, we can run the same argument whenever we have an $\mathcal{S}$-enriched Lawvere theory.

Also, as mentioned in the comments, I believe we can construct $F (X)$ when $\mathcal{S}$ is only a locally presentable category, without assumptions on $\times$. I suspect the cost is that we will have to iterate the coend construction infinitely many times before we obtain an $R$-module.

• So we take $\coprod_{n \geq 0} (R \times X)^n$ and mod out the relations $(\dotsc,(a,x),(b,x),\dotsc) \sim (\dotsc,(a+b,x),\dotsc)$ and $(\dotsc,(a,x),\dotsc,(b,y),\dotsc) \sim (\dotsc,(b,y),\dotsc,(a,x),\dotsc)$, maybe more? (I would like to have a very explicit description of the free module.) – Martin Brandenburg Sep 30 '15 at 16:04
• You also need one to handle $0$. Essentially, you are using the fact that maps of sets are generated by deletion, duplication, and permutation of pairs. – Zhen Lin Sep 30 '15 at 17:27
• I can take $b=0$ in my first relation. Is there anything else I missed? – Martin Brandenburg Sep 30 '15 at 22:24
• No, you need one that says that e.g. $((a, x))$ is equal to $((a, x), (0, y))$. – Zhen Lin Sep 30 '15 at 22:32