# Abstract Nonsense

## Direct Limit of Rings (Pt. I)

Point of Post: In this post we discuss the direct limit of rings, prove their universal characterization, and give some examples.

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Motivation

Unsurprisingly, we can take the direct limit of rings in much the same way we can take the direct limit of modules. This reflects that the notion of a direct limit can truly be defined in any category, though we shall not take that view (yet). All the intuition is going to be the same for that of modules, and so as to not just repeat it verbatim, we omit it in the below material. We shall see that many, many interesting constructions can be described as certain direct limits.

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Directed Systems

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Let $\left(\mathcal{A},\leqslant\right)$ be a preordered set, then a directed system of rings over $\left(\mathcal{A},\leqslant\right)$ is an ordered pair $\left(\{R_\alpha\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}:R_\alpha\to R_\beta\}_{\alpha,\beta\in\mathcal{A},\; \alpha\leqslant\beta}\right)$ where $R_\alpha$ is a ring for each $\alpha\in\mathcal{A}$ and the $f_{\alpha,\beta}$ are ring homomorphisms satisfying $f_{\beta,\gamma}\circ f_{\alpha,\beta}=f_{\alpha,\gamma}$ whenever $\alpha\leqslant\beta\leqslant\gamma$ and $f_{\alpha,\alpha}=\text{id}_{\alpha}$.

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Let’s now take a look at some examples of directed systems of rings:

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Let $R$ be some ring and define $f_{i,j}:R^i\to R^j$ to be the natural inclusion of $R^i$ into the first $i$ coordinates of $R^j$. Clearly then $\left(\{R^i\}_{i\in\mathbb{N}},\{f_{i,j}:R^i\to R^j\}_{i,j\in\mathbb{N},\; i\leqslant j}\right)$ forms a directed system over the directed set $\left(\mathbb{N},\leqslant\right)$.

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Consider a set $\{R_\alpha\}_{\alpha\in\mathcal{A}}$ of rings and consider the trivial preorder on $\mathcal{A}$. We then define the morphism $f_{\alpha,\alpha}:R_\alpha\to R_\alpha$ to be $f_{\alpha,\alpha}=\text{id}_{R_\alpha}$. Evidently then $\left(\{R_\alpha\},\{f_{\alpha,\alpha}\}_{\alpha\in\mathcal{A}}\right)$  forms a trivial directed system over the preordered set $\left(\mathcal{A},\leqslant_{\text{triv}}\right)$.

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Fix some $z\in\mathbb{C}$. Let $\mathcal{U}_z$ be the neighborhood system of $z$ (i.e. all open sets containing $z$). Define an ordering on $\mathcal{U}_z$ by the opposite inclusion ordering, i.e. $U\leqslant V$ if and only if $V\subseteq U$. Define then for $U\leqslant V$ in $\mathcal{U}_z$ the maps $\text{res}_{U,V}:\mathcal{H}(U)\to\mathcal{H}(V)$ (where $\mathcal{H}$ denotes the ring of holomorphic functions on the set) to be the restriction map $\mathcal{H}(U)\ni f\mapsto f_{\mid V}\in\mathcal{H}(V)$. Evidently then (this really is easy to check, the restriction of a restriction is a restriction of the smaller) $\left(\{\mathcal{H}(U)\}_{U\in\mathcal{U}_z},\{\text{res}_{U,V}:\mathcal{H}(U)\to\mathcal{H}(V)\}_{U,V\in\mathcal{U}_z,\; U\leqslant V}\right)$ forms a directed system over the directed set $\left(\mathcal{U}_z,\leqslant\right)$.

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Let $R$ be a commutative unital ring and let $S\subseteq R$ be a unitally multiplicative subset of $R$. Define the relation $\leqslant$ on $S$ by $a\leqslant b$ if and only if $a\mid b$ (in the usual notion of divisibility). For each $s\in S$ let $R_s$ denote the localization of $R$ at $\{1,s,s^2,\cdots\}$. There are obvious maps $f_{s,t}:R_s\to R_t$ when $s\mid t$. Namely, since $s\mid t$ we know that $t=sr$ for some $r\in R$, fixing such an $r$ we define $\displaystyle f_{s,t}\left(\frac{x}{s^n}\right)=\frac{r^n x}{t^n}$. Note that this is in fact independent of which divisor we take. Indeed, suppose that $sr_1=t=sr_2$ then $\displaystyle \frac{r_1^n x}{t^n}=\frac{r_2^nx}{t^n}$ indeed it suffices by definition to show that $r_1^nxt^n=r_2^nxt^n$ but $r_1^nxt^n=r_1x(r_2s)^n=r_1^nr_2^nxs^n$ and $r_2^nxt^n=r_2^nx(r_1s)^n=r_1^nr_2^nxs^n$ from where the equality follows.  We claim that $f_{t,u}\circ f_{s,t}=f_{s,u}$ when $s\leqslant t\leqslant u$. Indeed, suppose that $t=rs$ and $u=r't$ then $u=rr's$ and

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$\displaystyle f_{t,u}\left(f_{s,t}\left(\frac{x}{s^n}\right)\right)=f_{u,t}\left(\frac{r^n x}{t^n}\right)=\frac{r^nr'^n x}{u^n}=\frac{(rr')^nx}{u^n}=f_{s,u}\left(\frac{x}{s^n}\right)$

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since evidently $f_{s,s}=\text{id}_{R_s}$ for each $s\in S$ (since $s=1s$, we may take our representative complementary divisor to be $1$) we thus have that $\left(\{R_s\}_{s\in S},\{f_{s,t}:R_s\to R_t\}_{s,t\in S,\; s\leqslant t}\right)$ is a directed system over the (easily verified) directed set $\left(S,\leqslant\right)$.

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Direct Limits

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Now that we have some examples of directed systems of rings, we can define the directed limits over such systems. Namely, given a preordered set $\left(\mathcal{A},\leqslant\right)$ and a directed system $\left(\{R_\alpha\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}:R_\alpha\to R_\beta\}_{\alpha,\beta\in\mathcal{A},\; \alpha\leqslant \beta}\right)$ over this preordered set, we define a direct limit of this system to be an ordered pair $\left(R,\{\varphi_\alpha\}\right)$ where $R$ is some ring and $\{\varphi_\alpha:R_\alpha\to R\}$ is a set, called a limit cone, of ring homomorphisms satisfying $\varphi_\beta\circ f_{\alpha,\beta}=\varphi_\alpha$ and universal with respect to this property: given any set of ring homomorphisms $\{g_\alpha:R_\alpha\to S\}_{\alpha\in\mathcal{A}}$ (where $S$ is some ring) such that $g_\beta\circ f_{\alpha,\beta}=g_\alpha$ there exists a unique ring homomorphism $j:R\to S$ satisfying $j\circ \varphi_\alpha=g_\alpha$.

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Shocking as this may be (not shocking at all, anything defined in terms of universal characterizations ‘morally’ should be) unique up to (unique) isomorphism. Indeed:

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Theorem: Let $\left(\mathcal{A},\leqslant\right)$ a preordered set, and $\left(\{R_\alpha\}_{\alpha\in\mathcal{A}},\left\{f_{\alpha,\beta}:R_\alpha\to R_\beta\right\}_{\alpha,\beta\in\mathcal{A},\; \alpha\leqslant\beta}\right)$ a directed system over $\left(\mathcal{A},\leqslant\right)$. Then, if $(R,\{\varphi_\alpha\})$ and $(S,\{\psi_\alpha\})$ are two direct limits of this direct system, then $R\cong S$.

Proof: By virtue of the existence of the maps $\varphi_\alpha:R_\alpha\to R$ and $\psi_\alpha:R_\alpha\to S$ satisfying $\varphi_\beta\circ f_{\alpha,\beta}=\varphi_\alpha$ and $\psi_\beta\circ f_{\alpha,\beta}=\psi_\beta$ we afforded by the universal properties of the two direct products maps $j:S\to R$ and $k:R\to S$ such that $j\circ\psi_\alpha=\varphi_\alpha$ and $k\circ\varphi_\alpha=\psi_\alpha$. From this we deduce that $(j\circ k)\circ\varphi_\alpha=\varphi_\alpha$ and $(k\circ j)\circ\psi_\alpha=\psi_\alpha$. But, since the identity maps $\text{id}_R$ and $\text{id}_S$ satisfy these two identities respectively we have by uniqueness that $j\circ k=\text{id}_R$ and $k\circ j=\text{id}_S$ respectively. Thus, $j$ is an isomorphism as desired. $\blacksquare$

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

December 4, 2011 -