Abstract Nonsense

Localization (Pt. I)

Point of Post: In this post we discuss the notion of localization of rings at multiplicatively closed subsets which subsumes the discussion of the field of fractions for an integral domain.

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Motivation

Often times when doing ring theory one likes to pass from a ring $R$ to some quotient ring $R/\mathfrak{a}$ since the ideal theory of $R/\mathfrak{a}$ is simpler (just being the ideal theory of $R$ ‘chopped off’ as made precise by the fourth isomorphism theorem) and hopefully we can pull facts back from $R/\mathfrak{a}$ to facts about $R$. Localization can be thought of as being similarly minded. Namely, taking a ring $R$ and trying to put it ‘inside’ a bigger ring where things may be more clear. A sort of contrived example of this desire to go from murky to clear via enlarging the ring (which, incidentally has nothing to do with localization) is the following. Suppose you wanted to prove that $x^2+1$ could not be factored in the polynomial ring $\mathbb{R}[x]$ in a non-trivial way. I mean, there is an obvious way to do this by assuming you have such a factorization and finding what the coefficients may be. But, there is a more clever way. Namely, we know that $x^2+1$ factors as $(x-i)(x+i)$ in $\mathbb{C}[x]$. But, $\mathbb{C}[x]$ has unique factorization (in a sense not yet defined) and so any factorization of $x^2+1$ in $\mathbb{R}[x]$ would be a factorization in $\mathbb{C}[x]$ and so equal to $(x-i)(x+i)$ but this is impossible since $x+i\notin\mathbb{R}[x]$. So, now let’s think of a more apropos example (one that actually involves localization). One can recall that I claimed (or a corollary of what I claimed) was that the natural map $\mathbb{Z}[x]\to\mathbb{Z}^{\mathbb{Z}}$ by taking a polynomial to its associated function was an embedding, so that thinking of polynomials as functions or sums of formal symbols was the same. This isn’t obvious, but for reasons not obvious now but eventually obvious (mostly because $\mathbb{Q}$ is a field), it’s clear that the mapping $\mathbb{Q}[x]\to\mathbb{Q}^\mathbb{Q}$ is an embedding. But, we can then conclude since the restriction of  the map $\mathbb{Q}[x]\to\mathbb{Q}^\mathbb{Q}$ to $\mathbb{Z}[x]\to\mathbb{Z}^\mathbb{Z}$ must also be an embedding. This is the characteristic type of use (in the immediate future) for localization. Namely, we put the structure $\mathbb{Z}$ into the nicer structure $\mathbb{Q}$ and then prove the result there, but then we were able to conclude about the case for $\mathbb{Z}$. How exactly are we making the ring bigger? Well, taking the passage $\mathbb{Z}\leadsto\mathbb{Q}$ as a prime example again, what we are doing to a ring $R$ when we localize it, is that we are adding in inverses. Namely, you could think of $\mathbb{Q}$ as precisely the smallest ring containing $\mathbb{Z}$ in which all the non-zero elements have inverses. To be more precise we fixed a multiplicatively closed subset $D$ of $\mathbb{Z}$ (namely $\mathbb{Z}-\{0\}$) and created a bigger ring $\mathbb{Q}$ containing $\mathbb{Z}$ for which we precisely added inverses for the elements of $D$.

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So, the only thing that I feel like remains to be discussed is…why localization? Namely, why would someone call localization localization? As many things in math, it’s a ‘shut up and you’ll see’ concept, but I’ll see if I can explain it properly. One of the best examples of  localization can be seen with smooth manifolds. If we have some smooth manifold $M$ we can fix some point $p\in M$. We can then try to study the properties of $M$ by studying the ‘smooth function germs’ at $p$. Intuitively, these consist of equivalence classes $[f]$ of smooth functions $f:M\to \mathbb{R}$ in the set of all such functions $C^\infty(M)$ where $f\sim g$ if and only if there exists a neighborhood $U$ of $p$ for which $f_{\mid U}=g_{\mid U}$. So, the smooth function germs are precisely classes of ‘non-equivalent near $p$‘ functions. Clearly the set of germs $\mathcal{G}_p$ gives us a lot of information about how $M$ acts near $p$, namely it gives us local data. That said, it’s easy to see that we can turn $\mathcal{G}_p$ into a ring by multiplying and adding equivalence class wide. Well, and here’s where this is all relevant, the set $\mathfrak{p}=\left\{f\in C^\infty(M):f(p)=0\right\}$ is a prime ideal in $C^\infty(M)$ and if we take the localization of $C^\infty(M)$ with respect to the multipicative set $C^\infty(M)-\mathfrak{p}$ we get a ring $C^\infty(M)$. Well, long story short, $C^\infty(M)_\mathfrak{p}\cong\mathcal{G}_p$. Thus, the localization $C^\infty(M)_\mathfrak{p}$ gives us local information about $M$ at $p$.

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Localization

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Let $R$ be a commutative unital ring and $D$ a unitally mulitplicative subset.  We define an equivalence relation on $R\times D$ given by $(a,b)\sim (c,d)$ if and only if there exists $s\in D$ with $sad=scb$. This is clearly an equivalence relation since $1ab=1ab$ so that $(a,b)\sim (a,b)$, it’s symmetric since if $sad=scb$ then evidently $scb=sad$, and it’s transitive because if $(a,b)\sim (c,d)$, say $sad=sbc$ and $(c,d)\sim(e,f)$ say $s'cf=s'ed$ then

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$ss'daf=s'fsad=s'fsbc=sbs'cf=ssbs'ed=ss'deb$

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We then wish to define operations on the set $(R\times D)/\sim$ of equivalence classes which makes it into a ring. At this point, there may be some curiosity about what is the point of this construction? Roughly, the ordered pair $(a,b)$ could be equally well denoted $\displaystyle \frac{a}{b}$ and so in this case that $S\subseteq R^\times$ the relation $\displaystyle \frac{a}{b}=\frac{c}{d}$ if and only if $ad=bc$ and so this mirror’s precisely the construction of $\mathbb{Q}$ from $\mathbb{Z}$. To this end, we’ll find it more convenient to denote the equivalence class $[(a,b)]$  in $(R\times D)/\sim$ as $\displaystyle \frac{a}{b}$. This perspective then indicates us how we should define the operations on $(R\times D)/\sim$ namely we add and multiply them just as we added and multiplied fractions:

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$\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}\qquad \frac{a}{b}\frac{c}{d}=\frac{ac}{bd}$

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where it becomes clear (although we have already implicitly used this) why it is vital to have $D$ closed, since otherwise $bd\notin D$ and so this doesn’t make sense. It is an exercise in routine tedium to show that these operations are well-defined and make $(R\times D)/\sim$ into a commutative unital ring, which we denote by $D^{-1}R$ and call the localization of $R$ at $D$. It has unit equal to $\displaystyle \frac{1}{1}$ and zero element equal to $\displaystyle \frac{0}{1}$ which we denote by $1$ and $0$ to little confusion.

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Note, we shall see a specific case of localization when $R$ is an integral domain and $D=R-\{0\}$ for which $D^{-1}R$ will be a field, this is commonly encountered in most more basic algebra courses. From this we may get the wrong impression, namely that $D^{-1}R$ is a ‘bigger, nicer’ ring than $R$, and in particular $R$ naturally embeds into $D^{-1}R$. This is not an unreasonable assumption, and in most well-behaved cases it will be true. That said, one should be aware that this could be totally wrong. For example, let $R$ be the matrix ring $\text{Mat}_2(\mathbb{R})$ and let $D$ be the set of all non-negative powers of the matrix $\left(\begin{smallmatrix}0 & 1\\ 0 & 0\end{smallmatrix}\right)=A$, then we claim that $D^{-1}R$ is the zero ring. Indeed, one merely notes that

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$\displaystyle \frac{1}{1}=\frac{A^2}{A^2}=\frac{0}{A^2}=\frac{0}{1}$

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from where it follows that $D^{-1}R=\{0\}$. In fact, it’s easy to see that $D^{-1}R=\{0\}$ precisely when $D$ contains a nilpotent element, or since $D$ is closed, equivalently if $0\in D$. Clearly from this we can see that hoping to embed $R$ into $D^{-1}R$ is, in general, doomed. That said, there is a natural homomorphism $R\to D^{-1}R$ given by $\displaystyle r\mapsto \frac{r}{1}$, we call this map the localization map associated to $D$ and often denote it $\ell_D$. It’s pretty easy to see that $\ell_D(x)=0$ and only if $dx=0$ for some $d\in D$. In other words, $\ker\ell_D$ is the set of all ring elements of $R$ which are annihilated by at least one element of $D$. So $\ker \ell_D$ is trivial if and only if $D$ contains no zero divisors of $R$, and so in particular $\ell_D$ is an embedding if and only if $D$ contains no zero divisors of $R$.

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Remark: It will become very convenient to identify the image $\ell_D(R)\subseteq D^{-1}R$ with $R$ itself, so that we denote $\ell_D(x)$ as $x$. This coincides with our notations for $1$ and $0$. So, when we say something to the effect of ‘let $R\cap X\subseteq D^{-1}R$‘ we know what we mean is that  ‘let $\ell_D(R)\cap X\subseteq D^{-1}R$‘.

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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.

September 29, 2011 -

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