# Abstract Nonsense

## Inverse Limit of Rings (Pt. I)

Point of Post: In this post we discuss the notion of the inverse limit of rings, defining inverse systems, defining inverse limits in terms of universal characterizations, and then showing that inverse limits always exist.

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Motivation

As we have already seen it’s easy to go from constructions involving limits of modules to constructions involving limits of rings by just changing “$R$-map” to “ring homomorphism” and “submodule” to “ideal”. This continues in this post where we discuss the notion of inverse limits of inverse systems of rings. The motivation, and the key ideas are the same as they were when we discussed the inverse limits of modules, and so we shall omit motivating remarks and the finer details of proofs since they are almost verbatim, the same.

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Inverse System of Rings

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

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Now, let’s take a look at some examples of inverse systems of rings:

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Let $\{\alpha,\beta,\gamma\}=\mathcal{A}$ and define a preorder $\leqslant$ on $\mathcal{A}$ subject to $\alpha,\beta\leqslant\gamma$. Then, an inverse system of rings over $\left(\mathcal{A},\leqslant\right)$ is a set $\{R_\alpha,R_\beta,R_\gamma\}$ of rings and a set of maps described in the following diagram:

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$\underset{\circlearrowright \text{id}_{R_\alpha}}{R_\alpha}\xrightarrow{f_{\alpha,\gamma}}\underset{\circlearrowright\text{id}_{R_\gamma}}{R_\gamma}\overset{f_{\beta,\gamma}}{\longleftarrow}\underset{\circlearrowright \text{id}_{R_\beta}}{R_\beta}$

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Let $R$ be be a ring (most often a UFD or a PID) we can then define a relation on the set $\mathcal{A}=R$ by $a\leqslant b$ if and only if $a\mid b$. Then, we have canonical maps $f_{a,b}:R/(b)\to R/(a)$ whenever $a\leqslant b$ given by reduction modulo $(a)$. Said more directly, we map $x+(b)$ to $x+(a)$–this is (not to insult anyone’s intelligence) well-defined for if $x+(b)=y+(b)$ then $x-y\in(b)\subseteq (a)$ so that $x+(a)=y+(a)$. Evidently, $f_{a,a}=\text{id}_{R/(a)}$ and $f_{a,c}=f_{a,b}\circ f_{b,c}$ for all $a\leqslant b\leqslant c$ in $R$. Thus, we see that $\left\{\{R/(a)\}_{a\in\mathcal{A}},\{f_{a,b}:R/(b)\to R/(a)\}_{a,b\in \mathcal{A},\; a\leqslant b}\right)$ is an inverse system over $\left(\mathcal{A},\leqslant\right)$

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Note that the previous example can easily be generalized. Namely, given any subset $\mathcal{A}\subseteq R$ we can define the divisibility relation on $\mathcal{A}$ and create a similar inverse system over $\left(\mathcal{A},\leqslant\right)$. An interesting example is when we take $\mathcal{A}$ to be a chain. In particular, fix some $a\in R$ (where, for the sake of niceness, we assume that $R$ is unital, but of course this is unnecessary) and consider $\mathcal{A}=\{1,a,a^2,a^3,\cdots\}$ we then have natural maps $f_{n,m}:R/(a^m)\to R/(a^n)$ whenever $n\leqslant m$.

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In a slightly different (although clearly a more general case of the previous paragraph) direction we consider, given a ring $R$ and some ideal $\mathfrak{a}$ or $R$ the rings $R/\mathfrak{a}^n$ for each $n\in\mathbb{N}$. This evidently gives us for each $n\leqslant m$ a ring homomorphism $f_{n,m}:R/\mathfrak{a}^m\to R/\mathfrak{a}^n$ given by reduction modulo $\mathfrak{a}^n$ (recall that $\mathfrak{a}\supseteq\mathfrak{a}^2\supseteq\cdots$). Clearly one has that $f_{n,n}=\text{id}_{R/\mathfrak{a}^n}$ and $f_{n,k}=f_{n,m}\circ f_{m,k}$ whenenver $n\leqslant m\leqslant k$. Thus, we see that we have constructed an inverse system $\left(\{R/\mathfrak{a}^n\}_{n\in\mathbb{N}}:\{f_{n,m}:R/\mathfrak{a}^m\to R/\mathfrak{a}^n\}_{n,m\in\mathbb{N},\; n\leqslant m}\right)$ over the directed set $\left(\mathbb{N},\leqslant\right)$.

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Let $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$and define the trivial order on $\mathcal{A}$. We then get the trivial inverse system over $\left(\mathcal{A},\leqslant\right)$ with just the maps $f_{\alpha,\alpha}=\text{id}_{R_\alpha}$

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Inverse Limit of Rings

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We now attempt to define the inverse limit of an inverse system of rings. In particular, given a preoredered set $\left(\mathcal{A},\leqslant\right)$ and an inverse system $\left(\{R_\alpha\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}:R_\beta\to R_\alpha\}_{\alpha,\beta\in\mathcal{A},\; \alpha\leqslant\beta}\right)$ over $\left(\mathcal{A},\leqslant\right)$ we define an inverse limit of this system to be an ordered pair $\left(R,\{\pi_\alpha\}\right)$ where $R$ is a ring and $\pi_\alpha:R\to R_\alpha$ are ring homomorphisms such that $f_{\alpha,\beta}\circ\pi_\beta=f_\alpha$ whenever $\alpha\leqslant\beta$ and which satisfy the following universal property: given any ring $S$ and any set $g_\alpha:S\to R_\alpha$ of ring homomorphisms such that $f_{\alpha,\beta}\circ g_\beta=g_\alpha$ there exists unique ring homomorphism $j:S\to R$ such that $\pi_\alpha \circ j=g_\alpha$.

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As always, the inverse limit of an inverse system of rings is unique up to unique isomorphism:

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

Proof: By the existence of the maps $\pi_\alpha:R\to R_\alpha$ and $\eta_\alpha:S\to R_\alpha$ we are afforded maps $j:R\to S$ and $k:S\to R$ such that $\eta_\alpha\circ j=\pi_\alpha$ and $\pi_\alpha\circ k=\eta_\alpha$. From this we may conclude that $\eta_\alpha\circ (j\circ k)=\pi_\alpha\circ k=\eta_\alpha$ and $\pi_\alpha\circ (k\circ j)=\eta_\alpha\circ j=\pi_\alpha$–but since $\text{id}_S$ and $\text{id}_R$ also satisfy this we may conclude by uniqueness that $j\circ k=\text{id}_S$ and $k\circ j=\text{id}_R$ and so $j:R\to S$ is an isomorphism. $\blacksquare$

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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.