advanced30 min studyLesson 523 in Mathematics

Localization and local rings

Invert a controlled set of scalars, track which elements disappear and isolate the algebra near a prime ideal.

On this page

What you’ll understand

  • Use the correct fraction-equivalence rule when zero divisors are present.
  • Describe primes and units in a localization at a prime.
  • Explain exactness of localization and what a single localization cannot detect globally.

Localization lets selected scalars become invertible while changing as little else as possible. In a domain it resembles familiar fractions, but zero divisors make the equality rule more subtle: multiplying by a newly invertible scalar can force a previously nonzero element to vanish. Localizing away from a prime then produces a local ring, concentrating attention on that prime and the primes below it. This construction is an algebraic microscope rather than a quotient that simply sets every outside element equal to one.

Fractions must allow newly killed elements

Let S be a multiplicatively closed subset of a commutative unital ring R containing one. Fractions r/s and r′/s′ are identified if some t in S kills rs′−r′s. The extra multiplier is essential when R has zero divisors. In particular, r/1 is zero precisely when some element of S annihilates r. If zero belongs to S, the localization is the zero ring; otherwise it can still kill nonzero zero divisors. For a module M the same rule gives m/s, and m/1 vanishes exactly when a denominator-set element kills m.

The universal property says that a ring map from R inverting every element of S factors uniquely through S^{-1}R. This explains why different denominator descriptions can yield the same localization. It also shows why elements of S become units rather than become equal to one. When R is a domain and S excludes zero, the map embeds R in its fraction field. Without the domain assumption, injectivity needs the separate condition that no nonzero element is annihilated by a member of S.

rs=r′s′  ⟺  ∃t∈S: t(rs′−r′s)=0,m1=0 in S−1M  ⟺  ∃s∈S: sm=0\frac r s=\frac{r'}{s'}\iff\exists t\in S:\ t(rs'-r's)=0,\qquad\frac m1=0\text{ in }S^{-1}M\iff\exists s\in S:\ sm=0
The multiplier t repairs ordinary cross multiplication in the presence of zero divisors. Localization can erase elements rather than merely add formal inverses.↔ Scroll wide equations horizontally.

A prime determines a local microscope

Prime ideals of S^{-1}R correspond to primes of R disjoint from S, by extension and contraction. For a prime p choose S=R∖pS=R\setminus\mathfrak p. The resulting R_p has unique maximal ideal pR_p, so it is a local ring. Its residue field is the fraction field of R/p, not generally R/p itself. For Z localized at (3), fractions have denominators not divisible by three, and the residue field is F₃. At the zero prime of a domain, localization gives the entire fraction field instead of a small neighborhood with a nonzero maximal ideal.

A local ring’s elements outside its unique maximal ideal are units. This follows because a nonunit lies in some maximal ideal, which must be the unique one. In R_p this unit criterion makes statements such as a matrix being invertible depend on whether its determinant lies outside p. Localization can therefore turn a globally noninvertible map into an invertible one. That is a feature of the microscope, not proof that the original map was an isomorphism. To detect a finite or arbitrary module’s vanishing globally, test all prime localizations, or all maximal localizations, rather than just one.

Spec⁡(S−1R)↔{q∈Spec⁡R:q∩S=∅},κ(p)=Rp/pRp≅Frac⁡(R/p)\operatorname{Spec}(S^{-1}R)\leftrightarrow\{\mathfrak q\in\operatorname{Spec}R:\mathfrak q\cap S=\varnothing\},\qquad\kappa(\mathfrak p)=R_\mathfrak p/\mathfrak pR_\mathfrak p\cong\operatorname{Frac}(R/\mathfrak p)
The correspondence retains precisely the primes avoiding the inverted scalars. The residue field is formed after localization.↔ Scroll wide equations horizontally.

Exactness survives, faithfulness may not

Localization is exact on modules. A concrete kernel argument clears denominators: if a localized element maps to zero, some denominator multiplier sends its numerator into the original kernel, providing a localized kernel representative. Surjectivity survives by lifting a numerator. This works without a Noetherian hypothesis. It is one of the basic examples of flatness, introduced later. Exactness means relations remain valid after this operation; it does not mean every nonzero module remains visible. A torsion module can be annihilated when its annihilating scalar is inverted.

For example, Q⊗_Z Z/6Z is zero, even though Z/6Z is nonzero. Conversely, if M is nonzero, choose a nonzero element m and a maximal ideal containing its proper annihilator. The element m remains nonzero after localization there, since no outside scalar can annihilate it. This proves that all local microscopes together detect module vanishing. The distinction between exactness and detection will return as flatness versus faithful flatness, and it prevents local computations from being promoted to global conclusions without checking which primes were discarded.

Inverting a zero divisor erases another one

Let R=k[x,y]/(xy) and invert x. Describe R_x and identify a nonzero element of R that becomes zero.

  1. The class of y is nonzero in R, since y is not a multiple of xy in k[x,y]. But x·y=0 in R.

  2. Once x is invertible, multiplication of xy=0 by x^{-1} forces y=0. Every localized polynomial therefore reduces to a Laurent polynomial in x.

  3. The maps sending y to zero and x to the Laurent variable give an isomorphism R_x≅k[x,x^{-1}]. This ring is a domain even though the original ring had two zero-divisor branches.

The result

Localization can remove an entire component: the nonzero class of y vanishes because an inverted scalar annihilates it. Ordinary injective fraction-field intuition would miss this.

Explore the model

Interactive lab

Interactive lab · modular

A scalar can annihilate a nonzero class

Pause and Reset, then set modulus m=12 and additive step=4. Predict the nonzero class 4 becomes zero after three repeated additions, exhibiting 3·4=0. Step to confirm. For a fresh comparison Reset and set m=12,step=3; predict 4·3=0. Algebraically, inverting an annihilating scalar would kill the displayed class; the instrument itself performs only additive residue arithmetic and does not construct a localization.

A scalar can annihilate a nonzero classAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forNewly invertible annihilators can force nonzero module elements to disappear.

Check your understanding

Why can m/1 vanish in S⁻¹M even when m≠0?

Try it yourself

Describe Z_(5), decide whether 2 and 5 are units there, and compute the localization of Z/6Z at the same prime.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Denominators avoid multiples of five. In Z/6Z, consider whether the annihilator six is among those denominators.

Compare with the solution

Z_(5) consists of rational fractions a/b with 5 not dividing b. The element 2 is a unit with inverse 1/2; the element 5 lies in the maximal ideal and is not a unit. Since six is not divisible by five, it becomes invertible, but it annihilates Z/6Z. Therefore (Z/6Z)_(5)=0. This single localization cannot detect that the original module is nonzero; primes dividing its torsion remain relevant.

Take it with you

  • Localization equality allows a denominator multiplier to kill a difference.
  • Localizing at a prime gives a local ring and a residue field.
  • Localization preserves exactness but can lose nonzero modules; all prime tests together recover detection.

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