What is the motivation behind definition of localization of rings?

From where does the term "localization" come from?

Why is the equivalence relation between the ordered pairs $(m,u),(m',u')$ with $ m,m' \in M$ and $u,u' \in U$ is defined as $(m,u)\sim (m',u')$ if there exists $v\in U$ such that $v(u'm-um')=0$?

Here $M$ is an $R$-module and $U$ is multiplicatively closed subset of the ring $R$.

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    $\begingroup$ Think fractions. $\endgroup$ Commented Oct 13, 2011 at 8:23
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    $\begingroup$ The word "localization" comes from applications in geometry. By localizing at, say, a prime $p$, we "forget" everything but what happens at the point p. $\endgroup$ Commented Oct 13, 2011 at 9:30
  • $\begingroup$ Related on MO: mathoverflow.net/questions/21903 $\endgroup$
    – Watson
    Commented Sep 25, 2016 at 16:19

2 Answers 2


The term "localization" is indeed taken from geometry. Consider the ring $R = C(\mathbb{R})$ (continuous real valued function on the real numbers) and the set $U$ of function that don't vanish at the origin. Clearly $U$ is multiplicative. Now, what is the geometric interpretation of $U^{-1}R$ ? If we are only interested in the behavior of functions near the origin, a function that doesn't vanish at the origin doesn't vanish in some open neighborhood of the origin. Therefore, by restricting such function to a small enough neighborhood, we get a non-vanishing function and therefore it is possible to take its (multiplicative) inverse. Thus, $U^{-1}R$ can be thought of as the result of concentrating attention to small neighborhoods of the origin and hence the name "localization".

In a general commutative ring we adopt the intuition of the geometric case and think of elements of the ring as "functions" on some "geometric object" whose points are the prime ideals of the ring. The full picture of this is given by the modern algebro-geometric concept of a scheme.


We want $U^{-1}M$, the localization of $M$ at $U$, to consist of "fractions of $M$ over $U$"; a good start toward this definition is to consider $M\times U$, writing $m/u$ for $(m,u)$. However, this is not enough: as with ordinary fractions, we expect there to be multiple ways of writing the same fraction ($1/2 = 2/4$). Thus we have to impose some equivalence relation. Let's try to reverse-engineer such an equivalence relation. Begin with two fractions that are equal: $$\frac mu = \frac{m'}{u'}$$ and "cross-multiply" to get $$mu' = mu.$$ This is a relation that can be checked in the original module $M$, and we might naively try to use this as an equivalence relation. However, we have forgotten that elements of $U$ can be cancelled; i.e. for $v\in U$, we have $$\frac mu = \frac{m'}{u'} = \frac{vm'}{vu'}$$ giving the equation $$mvu' = vm'u,$$ or $v(mu' - m'u)=0$. Thus we must weaken our equivalence relation on $M\times U$ to $(m,u)\sim(m',u')\iff v(m'u-mu')=0$ for some $v\in U$. It turns out that this is sufficient to make $M\times U/\sim$ a module in the "right" way. (satisfies some universal property, is equal to $M\otimes_R U^{-1}R$, etc.)

Remark: The "naive" definition suffices when $vn=0$ implies $v=0$ or $n=0$ for $v\in U$, $n\in M$; for example, when $M=R$ and $R$ is an integral domain. In general, though, I'm not sure that the "naive" definition even gives an equivalence relation (I'll check later, but I'm pretty sure that it doesn't).

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    $\begingroup$ That existence of $v \in U$ part is required for the transitivity, but not for cancellation. $\endgroup$ Commented Oct 14, 2011 at 8:02

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