What is the localization of ℤ at the multiplicative set S = {1, 2, 4, 8, ...} = {2ⁿ : n ≥ 0}?
AThe ring ℤ[1/2] of fractions with denominator a power of 2
BThe field ℚ of all rational numbers
CThe ring ℤ/(2) of integers modulo 2
DThe ring of 2-adic integers ℤ₂
Localizing ℤ at S = {2ⁿ} inverts exactly the powers of 2, producing fractions a/2ⁿ — the ring ℤ[1/2]. This ring sits strictly between ℤ and ℚ: it contains 3/4 and 7/8 but not 1/3. To get all of ℚ, you would localize at S = ℤ \ {0}, inverting all nonzero integers. Localization at a single element (or powers of it) is the simplest non-trivial example and illustrates how localization selectively enlarges a ring.
Question 2 Multiple Choice
When localizing a ring R at a prime ideal 𝔭, you form S⁻¹R where S = R \ 𝔭. Why must S be multiplicatively closed?
AS is not always multiplicatively closed — it works only because prime ideals are maximal
BS = R \ 𝔭 is multiplicatively closed precisely because 𝔭 is prime: if a, b ∉ 𝔭, then ab ∉ 𝔭 (otherwise 𝔭 would contain a or b by the prime property)
CS is multiplicatively closed because R \ 𝔭 is always a subring
DMultiplicative closure is not actually required for localization to work
The definition of a prime ideal says: if ab ∈ 𝔭 then a ∈ 𝔭 or b ∈ 𝔭. Contrapositively: if a ∉ 𝔭 and b ∉ 𝔭, then ab ∉ 𝔭. This is exactly the statement that S = R \ 𝔭 is closed under multiplication. Since 1 ∉ 𝔭 (prime ideals are proper), S also contains 1. This is one of the key reasons prime ideals, rather than arbitrary ideals, play a central role in commutative algebra — their complements are the right sets to localize at.
Question 3 True / False
If R is an integral domain, the localization of R at S = R \ {0} is the field of fractions of R.
TTrue
FFalse
Answer: True
When S = R \ {0}, you invert every nonzero element, producing all fractions a/b with b ≠ 0. For R = ℤ, this gives ℚ. For R = k[x], this gives k(x), the field of rational functions. The field of fractions is the 'most localized' version of an integral domain — you have made every nonzero element invertible, which is exactly what it means to be a field. Every other localization of R (at a smaller multiplicative set) sits between R and its fraction field.
Question 4 True / False
Localization is an exact functor on modules.
TTrue
FFalse
Answer: True
If 0 → M' → M → M'' → 0 is an exact sequence of R-modules, then 0 → S⁻¹M' → S⁻¹M → S⁻¹M'' → 0 is exact as S⁻¹R-modules. This exactness — localization preserves kernels and cokernels — is fundamental. It means properties like 'an element is zero' or 'a map is surjective' can be checked locally at each prime ideal. This is the algebraic foundation of the local-global principle.
Question 5 Short Answer
Explain what it means to 'localize ℤ at the prime ideal (5)' and describe the resulting ring.
Think about your answer, then reveal below.
Model answer: Localizing ℤ at (5) means forming S⁻¹ℤ where S = ℤ \ (5) — all integers not divisible by 5. The result is the ring ℤ₍₅₎ of fractions a/b where b is not divisible by 5. This ring contains ℤ but not all of ℚ: 1/3 ∈ ℤ₍₅₎ but 1/5 ∉ ℤ₍₅₎. It is a local ring with unique maximal ideal 5ℤ₍₅₎, consisting of fractions a/b where 5 | a but 5 ∤ b. Every element not in this maximal ideal is a unit.
Localization at a prime 'zooms in' on divisibility by that prime, forgetting all other primes. In ℤ₍₅₎, the primes 2, 3, 7, etc. become invertible (harmless units), and only the prime 5 retains its special status. This is why the resulting ring is local — there is exactly one prime left to worry about. This construction is the algebraic analog of looking at a geometric object near a specific point.