Abstract Nonsense

Crushing one theorem at a time

Localization (Pt. II)

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

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So, as stated, one of the main reasons localizations is important is the following:

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Theorem: Let R be a commutative unital ring and D a unital multiplicatively closed subset. Then, D\subseteq\left(D^{-1}R\right)^\times

Proof: Let d\in D, then we see that \displaystyle d\frac{1}{d}=1 and so d\in \left(D^{-1}R\right). \blacksquare

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In particular, if we do take R to be an integral domain and D=R-\{0\} then D^{-1}R is a field, which we call the field of fractions of R and denote it \text{Frac}(R).

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Now, what we’d like to say is that \text{Frac}(R)  is the ‘smallest’ field containing R. In fact, we make the more precise following universal characterization:

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Theorem(Universal Characterization of Localizations): Let R be a commutative unital ring and D a multiplicatively closed subset.  Then, given any commutative unital ring S and a unital homomorphism f:R\to S such that f(D)\subseteq S^\times then f factors through \ell_D. Said differently, there exists a unique  morphism g:D^{-1}R\to S such that g\circ\ell_D=f.

Proof: Clearly such a map, if it exists, is unique. So, to produce such a map we define g:D^{-1}R\to S given by \displaystyle \frac{r}{d}\mapsto f(r)f(d)^{-1}. Let’s first show that this is actually a well-defined map. Note that if \displaystyle \frac{r}{d}=\frac{r'}{d'} then xrd'=xr'd for some x\in D. We see then that f(xrd')=f(xr'd) and so f(x)f(r)f(d')=f(x)f(r')f(d) noting that since x,d\in D we have that f(x),f(d) are units we may conclude from this that f(r)f(d)^{-1}=f(r')f(d')^{-1} and so \text{ }

 \displaystyle g\left(\frac{r}{d}\right)=f(r)f(d)^{-1}=f(r')f(d')^{-1}=g\left(\frac{r'}{d'}\right)

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To prove that g is a morphism we note that evidently 1\mapsto 1, and that

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\displaystyle \begin{aligned}g\left(\frac{r}{d}+\frac{r'}{d'}\right) &=g\left(\frac{rd'+r'd}{dd'}\right)\\ &=f(rd'+r'd)f(dd')^{-1}\\ &=(f(r)f(d')+f(r')f(d))f(d)^{-1}f(d')^{-1}\\ &=f(r)f(d)^{-1}+f(r')f(d')^{-1}\\ &=g\left(\frac{r}{d}\right)+g\left(\frac{r'}{d'}\right)\end{aligned}

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\displaystyle g\left(\frac{r}{d}\frac{r'}{d'}\right)=g\left(\frac{rr'}{dd'}\right)=(f(rr'))f(dd')^{-1}=f(r)f(d)^{-1}f(r')f(d')^{-1}=g\left(\frac{r}{d}\right)g\left(\frac{r'}{d'}\right)

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The conclusion follows by noticing that

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\displaystyle g(\ell_D(r))=g\left(\frac{r}{1}\right)=f(r)f(1)^{-1}=f(r)1^{-1}=f(r)


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This gives us that the localization of a ring at a multiplicative subset is ‘unique’ for the properties it possesses, namely:

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Theorem: Let R and S be commutative unital rings and D\subseteq R multiplicative. Then, S\cong D^{-1}R if and only if there exists a unital homomorphism f:R\to S such that g(R)\subseteq S^\times, f(r)=0 if and only if dr=0 for some d\in D, and every element of S is of the form f(r)f(d)^{-1} for some r\in R and d\in D.

Proof: By the first property and the previous universal characterization of D^{-1}R we have that there exists a unital homomorphism f:D^{-1}R\to S, the second property says it’s injective since \displaystyle g\left(\frac{r}{d}\right)=0 implies f(r)=0 which implies that d'r=0 for some d'\in D and so \displaystyle \frac{r}{d}=0, and clearly the last characteristic says precisely that g is a surjection. It thus follows that g is an isomorphism. \blacksquare

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As a corollary to this we get the following:

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Theorem: Let R be an integral domain and k a field containing R. Then, the subbfield (intersection of all fields) containing R is isomorphic to \text{Frac}(R).

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Ideal Theory of Localizations

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So, we want to figure out, given a commutative unital ring R and a unital multiplicative subset D what the ideals of the localization D^{-1}R look like. Well, as always we have natural maps from \mathfrak{L}(R) to \mathfrak{L}(D^{-1}R) given by \mathfrak{a}\mapsto \langle \ell_D(\mathfrak{a})\rangle and \mathfrak{L}(D^{-1}R)\to\mathfrak{L}(R) given by \mathfrak{a}\mapsto \ell_D^{-1}(\mathfrak{a}). While this doesn’t give us the whole story it’s a start. That said, while it is, at least theoretically, apparent what \ell_D^{-1}(\mathfrak{a}) looks like, it’s not immediately as obvious what \langle\ell_D(\mathfrak{a})\rangle looks like. Luckily, and not altogether unexpectedly, there is a nice presentation for this set. Namely

: \text{ }

Theorem: Let R be a commutative unital ring and D a unital multiplicative subset. Then, \langle \ell_D(\mathfrak{a})\rangle is equal to

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\displaystyle \left\{\frac{a}{d}:a\in\mathfrak{a}\text{ and }d\in D\right\}

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Proof: Evidently this set is an ideal containing \ell_D(\mathfrak{a}) and so it suffices to prove that any ideal containing \ell_D(\mathfrak{a}) contains this set. To see this we merely note that if \mathfrak{b} is an ideal in D^{-1}R containing \ell_D(\mathfrak{a}) then it contains \displaystyle \frac{a}{1}\frac{1}{d}=\frac{a}{d} for every a\in\mathfrak{a} and d\in D, from where the conclusion follows. \blacksquare

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

[2] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.


October 7, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , , ,


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