Math 593 Assignment # 3 Solutions

Bryden R. Cais

October 2, 2002

1. (a) ℤ(x), the field of rational functions in x.

(b) ℚ[\sqrt{2}] = ℚ(\sqrt{2}) is a field.

(c) ℂ(x, y) = ℂ(x)(y).

2. Let F denote the field of fractions of R = k[[x]] and for any f ∈ k((x)) let v(f) be the unique integer such that x^{v(f)}f = c_{0} + c_{1}x + · · ·. Define

ϕ : k((x)) ⟶ F

by

ϕ(f) = \begin{cases}\frac{f}{1} & if f ∈ k[[x]] \\ x^{v(f)}f/x^{v(f)} & otherwise \end{cases} .

It is clear that ϕ is an injective homomorphism. To show surjectivity, let f/g ∈ F. Then x^{v(g)}g is a unit in k[[x]] (from lecture) so there exists h ∈ k[[x]] such that gh = 1. Thus,

\frac{f}{g} = x^{v(g)}fh.

Clearly fh ∈ k[[x]] and since v(g) ∈ ℤ, f/g is in the image of ϕ.

3. For any maximal ideal 𝔐, it is clear that 1 ∉ 𝔐 and hence a = a/1 ∈ A_{𝔐} for any a ∈ A and any 𝔐.

It follows that

A ⊂ ⋂_{\begin{matrix}𝔐⊆A \\ 𝔐 maximal \end{matrix}} A_{𝔐}.

For the reverse inclusion, let x ∈ A_{𝔐} for all maximal 𝔐 ⊂ A. Then we can write x = a_{𝔐}/b_{𝔐} for each 𝔐 with a_{𝔐} ∈ A, b_{𝔐} ∈ A \ 𝔐. Observe that the ideal B generated by the b_{𝔐} is all of A, for if not, then B ⊂ 𝔐_{0} for some maximal ideal 𝔐_{0}, but b_{𝔐_{0}} ∈ B and b_{𝔐_{0}} ∉ 𝔐_{0} ⊃ B, a contradiction. Hence we can write

1 = ∑_{𝔐∈T} c_{𝔐}b_{𝔐}

for some finite set T where c_{𝔐} ∈ A. It follows that

x = ∑_{𝔐∈T} xc_{𝔐}b_{𝔐} = ∑_{𝔐∈T} c_{𝔐}a_{𝔐} ∈ A,

and we are done.

4. (a) We use the result from class that a ring R is local with unique maximal ideal 𝔐 if any f ∈ R \ 𝔐 is a unit. So let 𝔐 ⊂ C_{0} be the ideal of germs of functions vanishing at 0 (this is obviously an ideal). If [(U, f)] ∈ C_{0} \ 𝔐, then f(0) ≠ 0 so that by continuity, there exists a neighborhood V ⊂ U of 0 with f nonvanishing on V. Then [(V, 1/f)] ∈ C_{0} \ 𝔐 and

[V, 1/f][U, f] = [V, 1] = id ∈ C_{0},

so f is invertable, as required.

(b) Now let C(ℝ) denote the ring of all continuous functions on ℝ and let 𝔐 ⊂ C(ℝ) be the maximal ideal of all functions vanishing at 0. Define

φ : C(R)_{𝔐} ⟶ C_{0}

by

φ([f/g]) = [U_{g}, f/g],

where [f/g] denotes the equivalence class of f/g and U_{g} is a neighborhood of 0 on which g does not vanish (such a neighborhood always exists since g(0) ≠ 0 and g is continuous). We claim that φ is an isomorphism. There are several things to check:

i. The map φ is well defined: That is, we must show that it does not depend on the choice of representative f/g. So suppose that f/g, h/k are any two representatives of the same equivalence class in C(ℝ). Then there exists a continuous function s ∉ 𝔐 such that s(fk − gh) ≡ 0 on ℝ. Since s ∉ 𝔐, there exists a neighborhood U of 0 such that s does not vanish on U. It follows that fk − gh must vanish on U and hence that f/g = k/h on U ∩ U_{g} ∩ U_{h} so that [(U_{g}, f/g)] = [(U_{h}, k/h)], so φ is well defined.

ii. Injectivity: Suppose that φ([f/g]) = φ([k/h]). Then there exists some neighborhood U of 0 such that f/g − k/h ≡ 0 on U. Now there exists some ϵ > 0 with (−ϵ, ϵ) ⊂ U. Let

χ_{ϵ}(t) = \begin{cases}0 & if |t| > ϵ \\ 1 & if |t| < ϵ/2 \\ 2 (1 − \frac{|t|}{ϵ}) & otherwise \end{cases} .

Then clearly χ_{ϵ} ∈ C(ℝ) \ 𝔐 and

χ_{ϵ}(fh − gk) ≡ 0

on all of ℝ, in other words, [f/g] = [k/h] in C(ℝ)_{𝔐}, as required. iii. Surjectivity: Let [(U, f)] ∈ C_{0}. Then for some ϵ > 0 we have (−ϵ, ϵ) ⊊ U. Now define

f̃(t) = \begin{cases}f(t) & if t ∈ (−ϵ, ϵ) \\ f(ϵ) & if t ≥ ϵ \\ f(−ϵ) & if t ≤ −ϵ \end{cases} .

Then it is not hard to see that f̃ ∈ C(ℝ) ⊂ C(ℝ)_{𝔐} and that φ([f̃]) = [(U, f)]. iv. The map φ is a homomorphism: This follows directly from the definition of addition and

multiplication in C_{0}.