Questions: Local Rings

Question 1

Which of the following is a local ring?

• A. ℤ, because (0) is its only prime ideal
• B. k[x]/(x²), because its only maximal ideal is (x̄) and every element not in (x̄) is a unit
• C. ℤ/6ℤ, because it has only finitely many ideals
• D. k[x, y], because (x, y) is a maximal ideal

Answer: (B)

In k[x]/(x²), an element a + bx̄ is a unit if and only if a ≠ 0 (since (a + bx̄)(a⁻¹ - ba⁻²x̄) = 1). The non-units are exactly multiples of x̄, which form the ideal (x̄). Since (x̄) is the unique maximal ideal, k[x]/(x²) is local. ℤ has infinitely many maximal ideals (2), (3), (5), .... ℤ/6ℤ has two maximal ideals (2̄) and (3̄). k[x,y] has maximal ideals (x-a, y-b) for each (a,b), so it is not local.

Question 2

A commutative ring R is local if and only if the set of non-units forms an ideal.

• A. This is true and the ideal of non-units is automatically the unique maximal ideal
• B. This is false — the non-units can form an ideal without the ring being local
• C. This is true but only for Noetherian rings
• D. This is false — in any ring, the non-units form an ideal

Answer: (A)

If the non-units form an ideal 𝔪, then 𝔪 is the unique maximal ideal: any ideal consisting entirely of non-units is contained in 𝔪, and 𝔪 itself is proper (it doesn't contain 1). Conversely, if R has a unique maximal ideal 𝔪, every non-unit lies in some maximal ideal, which must be 𝔪. So the non-units are exactly 𝔪. This clean characterization is often the most useful definition of a local ring.

Question 3

The localization of ℤ at the prime ideal (p) is a local ring with residue field 𝔽_p.

• T. True
• F. False

Answer: True

ℤ₍ₚ₎ consists of fractions a/b with p ∤ b. The unique maximal ideal is pℤ₍ₚ₎ = {a/b : p | a, p ∤ b}. The residue field is ℤ₍ₚ₎/pℤ₍ₚ₎ ≅ ℤ/pℤ = 𝔽_p, because modding out by pℤ₍ₚ₎ kills exactly multiples of p and inverts everything else, leaving a copy of the finite field with p elements.

Question 4

Every field is a local ring.

• T. True
• F. False

Answer: True

In a field, every nonzero element is a unit, so the only non-unit is 0. The set of non-units is {0} = (0), which is the unique maximal ideal. The residue field is k/(0) ≅ k itself. Fields are the 'trivial' local rings — their unique maximal ideal is as small as possible.

Question 5

Explain why localization at a prime ideal always produces a local ring, and what geometric intuition this corresponds to.

Model answer: If 𝔭 is a prime of R and S = R \ 𝔭, then S⁻¹R = R_𝔭 has prime ideals corresponding to primes of R contained in 𝔭. The ideal 𝔭R_𝔭 is the unique maximal ideal because 𝔭 is the largest prime contained in itself. Any element outside 𝔭R_𝔭 has the form a/s with a ∉ 𝔭, so a ∈ S and a/s is a unit in R_𝔭.

Geometrically, Spec(R) is a space whose points are prime ideals. Localizing at 𝔭 'zooms in' to an infinitesimal neighborhood of 𝔭, discarding all geometric information about other points. The resulting local ring R_𝔭 captures the local geometry at 𝔭 — smoothness, singularity type, tangent directions — while ignoring the global structure. This is why local rings are the algebraic analog of germs of functions in differential geometry.