Abstract Nonsense

Crushing one theorem at a time

Direct Limit of Rings (Pt. III)


Point of Post: This is a continuation of this post.

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Examples of Direct Limits (cont.)

We now come to our final example. What we claim is that the localization S^{-1}R is the direct limit of the R_s‘s over this directed system. The put-in  maps \varphi_s:R_s\to S^{-1}R are just the inclusions (which are trivially well-defined). What we now need to prove is that \varphi_t\circ f_{s,t}=\varphi_s whenever s\leqslant t\in S. To see this we merely note that if t=rs then

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\displaystyle \varphi_t\left(f_{s,t}\left(\frac{x}{s^n}\right)\right)=\varphi_t\underbrace{\left(\frac{r^n x}{t^n}\right)}_{\in R_t}=\underbrace{\frac{r^n x}{t^n}}_{\in S^{-1}R}\overset{(\ast)}{=}\underbrace{\frac{x}{s^n}}_{\in S^{-1}R}=\varphi_s\left(\frac{x}{s^n}\right)

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where the (\ast) step is true since, now being in the full S^{-1}R, we may “cancel” the r^n whereas we couldn’t before (you can’t have denominators of the form s^n in R_t!). What we must now show is that S^{-1}R is universal with respect to this property. Namely, suppose that g_s:R_s\to X is some set of ring homomorphisms (where X is some given ring) such that g_t\circ f_{s,t}=g_s for all s\leqslant t\in S. We then define j:S^{-1}R\to X by the rule \displaystyle j\left(\frac{x}{s}\right)=g_s\left(\frac{x}{s}\right). To prove that this map is well-defined we merely note that if \displaystyle \frac{x}{s}=\frac{y}{t} then there exists some u\in S such that uxt=usy thus

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\displaystyle \begin{aligned}j\left(\frac{x}{s}\right) &=g_s\left(\frac{x}{s}\right)\\ &=g_{usy}\left(f_{s,usy}\left(\frac{x}{s}\right)\right)\\ &=g_{uxy}\left(\frac{uxy}{usy}\right)=g_{usy}\left(\frac{t}{uxt}\right)\\ &=g_{usy}\left(f_{t,utx}\left(\frac{x}{t}\right)\right)\\ &=g_t\left(\frac{x}{t}\right)\\ &=j\left(\frac{x}{t}\right)\end{aligned}

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Clearly this is a ring homomorphism and j\circ \varphi_s=g_s Moreover, it’s clear that this was the only such map that would have worked since if k:S^{-1}R\to X is another such map then we see that for any \displaystyle \frac{x}{s}\in S^{-1} one has that

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\displaystyle j\left(\frac{x}{s}\right)=j\left(\varphi_s\left(\frac{x}{s}\right)\right)=g_s\left(\frac{x}{s}\right)=k\left(\varphi_s\left(\frac{x}{s}\right)\right)=k\left(\frac{x}{s}\right)

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from this we may conclude that S^{-1}R is the direct limit of this directed system.
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General Construction
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What we would now like to do is prove that given any directed set of rings we can always construct a direct limit for the directed sequence. The methodology for doing this, for making this construction, is precisely the same as it was in the case of modules. The proof that this construction actually is a direct limit is exactly the same in both cases as well, and so we just state the fact that this construction works as the following theorem:
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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_\alpha\to R_\beta\}_{\alpha,\beta\in\mathcal{A},\; \alpha\leqslant\beta}\right) be a directed system of rings over (\mathcal{A},\leqslant). Then, if \mathfrak{a} is the ideal
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\mathfrak{a}=\left\langle \rho_\alpha(f_{\alpha,\beta}(x_\alpha))-\rho_\alpha(x_\alpha):\alpha,\beta\in\mathcal{A},\; \alpha\leqslant\beta\right\rangle
\text{ }
of \displaystyle \bigoplus_{\alpha\in\mathcal{A}}R_\alpha (where \displaystyle \rho_\alpha:R_\alpha\to\bigoplus_{\alpha\in\mathcal{A}}R_\alpha are the canonical injections) then \displaystyle \left(\left(\bigoplus_{\alpha\in\mathcal{A}}R_\alpha\right)/\mathfrak{a},\{\rho_\alpha\}\right) is a direct limit of this directed system.
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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.

[4] Lang, Serge. Algebra. Reading, MA: Addison-Wesley Pub., 1965. Print.

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December 4, 2011 - Posted by | Algebra, Ring Theory | , , , , , ,

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