Math 254A. Valued Rings and Valued Fields (Preliminary)

This handout gives some basic definitions and results concerning valued rings and valued fields.

Definition 1. A valued ring is an ordered pair (R, | · |) , where R is a ring and | · | is an absolute value on R. A homomomorphism ϕ: (R, | · |_{R}) → (S, | · |_{S}) of valued rings is a ring homomorphism ϕ: R → S such that |ϕ(r)|_{S} = |r|_{R} for all r ∈ R.

Note that all homomorphisms ϕ: (R, |·|_{R}) → (S, |·|_{S}) of valued rings are injective, since if r ∈ ker ϕ then |r|_{R} = |ϕ(r)|_{S} = 0 , which implies r = 0 . As such, we may at times use the word embedding instead of homomorphism.

Definition 2. A valued ring (R, | · |) is archimedean or non-archimedean if its absolute value is archimedean or non-archimedean, respectively.

A valued ring (R, | · |) gives rise to a metric space, with metric d(x, y) = |x − y|. This in turn, provides a topology on R.

Definition 3. A valued ring (R, |·|) is complete if its associated metric space is complete (i.e., all Cauchy sequences in R converge to a limit in R).

Definition 4. A valued field is a valued ring (K, | · |) in which K is a field. A homomorphism of valued fields is a homomorphism of valued rings (whose domain and codomain are valued fields). A valued field is archimedean, non-archimedean, or complete if it has that property as a valued ring.

From now on, we will often write (R, | · |) as just R.

Theorem 5. Let R = (R, | · |) be a valued ring.

(a). The valued ring R can be embedded as a dense subring of a valued ring

(R̂, | · | ) . (b). The complete valued ring (R̂, | · | ) from part (a) satisfies the following universal property. Every homomorphism ψ: R → S, with S complete, factors uniquely through a homomorphism ψ̂: R̂ → S of valued rings. In particular, R̂ (with the embedding R → R̂) is unique up to unique isomorphism.

Proof. If R is the trivial ring, then the result is also trivial (and is left as an exercise). Therefore we assume from now on that R ≠ (0) .

(a). Let R_{0} = R^{ℕ} be the ring R × R × … of ℕ-indexed sequences in R, and let R_{1} ⊆ R_{0} be the subset of Cauchy sequences.

Claim. R_{1} is a subring of R_{0} .

Proof. A sum of two Cauchy sequences is Cauchy, and so is the negative of a Cauchy sequence; therefore R_{1} is an additive subgroup of R_{0} . The constant sequence (1, 1, …) is Cauchy; therefore R_{1} contains a unity element. Finally, the fact that R_{1} is closed under multiplication follows from (i) all Cauchy sequences are bounded, and (ii)

|a_{n}b_{n} − a_{m}b_{m}| ≤ |a_{n}||b_{n} − b_{m}| + |b_{m}||a_{n} − a_{m}| .

This proves the claim. □

Claim. Let 𝔮 = {(a_{n}) ∈ R_{1} : |a_{n}| → 0 as n → ∞ }. Then 𝔮 is a prime ideal in R_{1} .

Proof. It is easy to see that 𝔮 is an additive subgroup of R_{1} . To see that it is an ideal, let a = (a_{n})_{n∈ℕ} ∈ 𝔮 and b = (b_{n})_{n∈ℕ} ∈ R_{1} . Then the sequence (a_{n}b_{n}) converges to zero because the sequence (|b_{n}|) is bounded and a_{n} → 0 . Therefore 𝔮 is an ideal in R_{1} .

Finally, the ideal 𝔮 is prime because the sequence (1, 1, …) is not in 𝔮 (trivially), and if (a_{n}), (b_{n}) ∉ q then the sequences (|a_{n}|) and (|b_{n}|) converge to nonzero real numbers, and therefore so does the sequence (|a_{n}b_{n}|) . □

Now let R̂ = R_{1}/𝔮. It is an entire ring.

Let a = (a_{n}) ∈ R_{1} . Then lim_{n→∞} |a_{n}| converges (by the triangle inequality the sequence (|a_{n}|) is a Cauchy sequence of real numbers). Also, if a = (a_{n}) and b = (b_{n}) are in R_{1} and a − b ∈ 𝔮, then lim_{n→∞} |a_{n}| = lim_{n→∞} |b_{n}|, so |a| := lim_{n→∞} |a_{n}| is a well-defined function on R̂. It is easy to check that this satisfies the conditions for an absolute value on R̂, so (R̂, | · | ) is a valued ring.

Since constant sequences are Cauchy, a ↦ (a, a, …) is a well-defined ring homomorphism ϕ_{0} : R → R_{1} . Let ϕ: R → R̂ be the map corresponding to ϕ_{0} . Then it is clear that |ϕ(a)| = |a| for all a ∈ R, so ϕ is a homomorphism of valued rings.

Next, the image of ϕ is dense in R̂. Indeed, let α ∈ R̂ be any element. We need to show that for all ϵ > 0 there is an a = a_{ϵ} ∈ R such that |α − ϕ(a)| ≤ ϵ. To see this, lift α to a Cauchy sequence (a_{0}, a_{1}, …) ∈ R_{1} . Pick N = N_{ϵ} ∈ ℕ such that |a_{i} − a_{j}| ≤ ϵ for all i, j ≥ N . In particular, |a_{i} − a_{N}| ≤ ϵ for all i ≥ N , so |α − ϕ(a_{N})| = lim_{i→∞} |a_{i} − a_{N}| ≤ ϵ, as was to be shown.

Finally, we claim that R̂ is complete. Indeed, let (α_{0}, α_{1}, …) be a Cauchy sequence in R̂. Given n ∈ ℕ, pick N_{n} ∈ ℕ such that |α_{i} − α_{j}| ≤ 2^{−n} for all i, j ≥ N_{n}, and pick a_{n} ∈ R such that |α_{N_{n}} − ϕ(a_{n})| ≤ 2^{−n}. Then the sequence (a_{0}, a_{1}, …) is Cauchy, because for all n ≤ m in ℕ,

\begin{aligned}|a_{n} − a_{m}| = |ϕ(a_{n}) − ϕ(a_{m})| & ≤ |ϕ(a_{n}) − α_{N_{n}}| + |α_{N_{n}} − α_{N_{m}}| + |α_{N_{m}} − ϕ(a_{m})| \\ & ≤ 2^{−n} + 2^{−n} + 2^{−m} \end{aligned}

≤ 3 · 2^{−n} .

Therefore this sequence defines an element α ∈ R̂. Also lim_{i→∞} α_{i} = α (in R̂), because for all n ≥ 0 and all i ≥ max{n, N_{n}},

|α_{i} − α| ≤ |α_{i} − α_{N_{m}}| + |α_{N_{m}} − ϕ(a_{n})| + lim_{j→∞}(|ϕ(a_{n}) − ϕ(a_{j})| + |ϕ(a_{j}) − α)| )

≤ 2^{−n} + 2^{−n} + lim_{j→∞} |ϕ(a_{n}) − ϕ(a_{j})| + lim_{j→∞} |ϕ(a_{j}) − α)|

≤ 2^{−n} + 2^{−n} + 3 · 2^{−n} + 0 = 5 · 2^{−n} .

Thus (R̂, | · | ) is complete, so (a) is proved.

(b). Exercise (for now). □