ON UNIT GROUPS OF EXTENSION RINGS: AN EXAMPLE

K. Alan Loper Lawrence University

The following is a well known result of algebraic number theory [1] or [2].

\underline{Theorem (Dirichlet)}. Suppose Q[α] is a finite degree extension field over the field Q of rational numbers and R is the integral closure in Q[α] of the ring Z of integers. Let r and 2s be the numbers of real and nonreal embeddings, respectively, of Q[α] into the field of complex numbers. Then U(R), the group of units of R, can be written as a direct product U(R) ≅ G × H where G is a finite group and H is a free abelian group of rank r + s − 1.

One peculiarity of the above result is that while it appears to be entirely algebraic, it seems that all known proofs involve some analysis. In fact, there does not seem to exist an entirely algebraic proof of the weaker conclusion that U(R) has finite rank. Following this same line of questioning, a friend of the author recently posed the following problem.

\underline{Conjecture}. Let R be an integral domain with field of fractions K and suppose that U(R), the unit group of R, has finite rank. Suppose that F is a finite degree extension field over K and that S is the integral closure in F of R. Then U(S), the unit group of S, has finite rank.

The purpose of this note is to give a counterexample to the above conjecture, thus illustrating the difficulty of giving an algebraic proof of Dirichlet’s theorem.

\underline{Example}. Let Z be the ring of integers. We will define an infinite sequence of rings as follows:

R_{0} = Z , R_{1} = R_{0}[x_{1}, \sqrt{2x_{1} + 3}, \sqrt{3x_{1} + 5}] ,

and for i ≥ 1,

R_{i} = R_{i−1}[x_{i}, \sqrt{2x_{i} + 3}, \sqrt{3x_{i} + 5}] .

Now, let

R_{∞} = ⋃_{i=1}^{∞} R_{i} and let S = R_{∞}[\sqrt{2}, \sqrt{3}] .

We make two claims.

\underline{Claim 1}. U(R_{∞}) = \{1, −1\}.

\underline{Proof}. Suppose not. Then U(R_{t}) ≠ \{1, −1\} for some t ≥ 1. Choose t to be minimal with respect to this property and let u_{1} ∈ U(R_{t}) with u_{1} ∉ {1, −1}. Then we can write

u_{1} = f + g\sqrt{2x_{t} + 3} + h\sqrt{3x_{t} + 5} + l\sqrt{2x_{t} + 3}\sqrt{3x_{t} + 5}

with f, g, h, l ∈ R_{t−1}[x_{t}]. We are then led naturally to define several conjugates of u_{1} as follows:

u_{1} = f + g\sqrt{2x_{t} + 3} + h\sqrt{3x_{t} + 5} + l\sqrt{2x_{t} + 3}\sqrt{3x_{t} + 5}
u_{2} = f − g\sqrt{2x_{t} + 3} − h\sqrt{3x_{t} + 5} + l\sqrt{2x_{t} + 3}\sqrt{3x_{t} + 5}
u_{3} = f − g\sqrt{2x_{t} + 3} + h\sqrt{3x_{t} + 5} − l\sqrt{2x_{t} + 3}\sqrt{3x_{t} + 5}
u_{4} = f + g\sqrt{2x_{t} + 3} − h\sqrt{3x_{t} + 5} − l\sqrt{2x_{t} + 3}\sqrt{3x_{t} + 5}

Clearly, u_{i} ∈ U(R_{t}) and u_{i} ∉ {1, −1} for each i. Now, let u_{∗} = u_{1}u_{2}u_{3}u_{4}. Then u_{∗} ∈ U(R_{t}) and tedious computation yields

u_{∗} = (f^{2} − g^{2}(2x_{t} + 3) − h^{2}(3x_{t} + 5) + l^{2}(2x_{t} + 3)(3x_{t} + 5))^{2} − 4(fl − gh)^{2}(2x_{t} + 3)(3x_{t} + 5)

Note that u_{∗} ∈ R_{t−1}[x_{t}]. Since each u_{i} ∈ U(R_{t}) then u_{i}^{−1} ∈ U(R_{t}) for each i and a similar computation to that done above will yield u_{∗}^{−1} ∈ R_{t−1}[x_{t}]. Since R_{t−1} is a domain and u_{∗} is a unit in R_{t−1}[x_{t}] then the degree of u_{∗} as a polynomial in x_{t} must be zero. Thus, u_{∗} ∈ R_{t−1} and then by the minimality of t we must have either u_{∗} = 1 or u_{∗} = −1. We now define

ϕ = f^{2} − g^{2}(2x_{t} + 3) − h^{2}(3x_{t} + 5) + l^{2}(2x_{t} + 3)(3x_{t} + 5)

and

ψ = 2(fl − gh) .

Then

u_{∗} = ϕ^{2} − ψ^{2}(2x_{t} + 3)(3x_{t} + 5)

with

ϕ, ψ ∈ R_{t−1}[x_{t}] .

We now have two cases to consider. Either ψ = 0 or ψ ≠ 0. We will see that both cases lead to contradictions.

Suppose ψ = 0. Then either ϕ = 1 or ϕ = −1 and so

f^{2} + l^{2}(2x_{t} + 3)(3x_{t} + 5) = g^{2}(2x_{t} + 3) + h^{2}(3x_{t} + 5) + a

where

a ∈ {1, −1} .

By comparing degrees of the polynomials in x_{t} on the right and on the left hand sides of this equation, we see that f^{2} = 1 and g = h = l = 0. This implies that u_{1} = 1 which is a contradiction.

Suppose ψ ≠ 0. Since u_{∗} has degree zero in x_{t}, then ϕ^{2} and ψ^{2}(2x_{t} + 3)(3x_{t} + 5) have the same degree in x_{t}. Let 2n be this degree. Then we can write ϕ = ϕ_{n}x_{t}^{n} + ⋯ + ϕ_{0} and ψ = ψ_{n−1}x_{t}^{n−1}+⋯+ψ_{0} where each ϕ_{i} and ψ_{i} is in R_{t−1}. Then ϕ_{n}/ψ_{n−1} = \sqrt{6} is in the field of fractions of R_{t−1}. Let F_{t−1} and K be the fields of fractions of R_{t−1} and Z[x_{1}, x_{2}, …, x_{t−1}], respectively. Since F_{t−1} is the composition of 2t − 2 quadratic extensions of K, the Galois group of F_{t−1} over K is an elementary abelian 2–group of order 2^{2t−2}. Such a group contains 2^{2t−2} − 1 maximal subgroups and so F_{t−1} contains exactly 2^{2t−2} − 1 quadratic extensions of K. We can construct 2^{2t−2} − 1 such extensions by adjoining to K expressions of the form

∏_{i=1}^{t−1}(\sqrt{2x_{i} + 3})^{d_{i}}(\sqrt{3x_{i} + 5})^{e_{i}}

where d_{i}, e_{i} = 0 or 1 for each i. Clearly, none of these extensions is equal to K[\sqrt{6}]. Thus, \sqrt{6} ∉ F_{t−1} and we have reached a contradiction.

Since neither ψ = 0 nor ψ ≠ 0 is possible, we conclude that U(R_{∞}) = {1, −1}.

\underline{Claim 2}. U(S) has infinite rank.

\underline{Proof}. The equation

(\sqrt{2}\sqrt{3x_{i} + 5} + \sqrt{3}\sqrt{2x_{i} + 3})(\sqrt{2}\sqrt{3x_{i} + 5} − \sqrt{3}\sqrt{2x_{i} + 3}) = 1

demonstrates that u_{i} = (\sqrt{2}\sqrt{3x_{i} + 5} + \sqrt{3}\sqrt{2x_{i} + 3}) is a unit in S for each i. Then the set {u_{i}|i ≥ 1} generates a group of units in S with infinite rank. Thus U(S) has infinite rank.

\underline{References}

1. D. A. Marcus, Number Fields, Springer–Verlag, New York/Berlin, 1977.

2. G. Karpilovsky, Unit Groups of Classical Rings, Oxford University Press, New York, 1988.