# Abstract Nonsense

## 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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and

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

$\blacksquare$

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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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References:

[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 -

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