Rank($U$) = Rank($U^2$) for group of units $U$

Question

I am reading the paper "Algebraic Integers on the Unit Circle" by Ryan C. Daileda (http://www.sciencedirect.com/science/article/pii/S0022314X05002027). I am confused about how he concludes that the rank of $U$ is equal to the rank of $U^2$ in the proof of Theorem 1.

Can anyone please explain?

(Edit to add the theorem and proof): Let $K$ be an algebraic number field, $U$ be its group of units, $V$ be the subgroup of units of modulus $1$, and $R$ be the group of real units in $K$ $(R = U \cap \mathbb{R})$.

Then the theorem states: "Let $K$ be a number field closed under complex conjugation, and let $U, V,$ and $R$ be as above. Then $\operatorname{rank}(V)+\operatorname{rank}(R) = \operatorname{rank}(U)$." The proof is:

"By hypothesis, if $u \in U$, then $\bar{u} \in U$ as well. Therefore, $u\bar{u} = |u|^2 \in R$, and $u/\bar{u} \in V$. This means that $u^2 = |u|^2 \frac{u}{\bar{u}} \in RV$.

Hence $U^2 \subset RV \subset U$, so that $\operatorname{rank}(U) = \operatorname{rank}(U^2)= \operatorname{rank}(RV)$. It is clear that $R\cap V = \{\pm 1\}$, giving $\operatorname{rank}(RV) = \operatorname{rank}(R) + \operatorname{rank}(V)$, so we're done."

Answer 1

All right, well first of all Dirichlet's unit theorem tells you that the unit group (modulo roots of unity, which are irrelevant for computing rank) is a finitely generated free group on $r+s-1$ generators with $r$ the number of real embeddings and $s$ the number of complex ones.

If we use additive notation

$$U/\text{Tor}(U)\cong \bigoplus_{i=1}^{r+s-1}\Bbb Z$$

it is easier to see that the squares here are a subgroup of index $2^{r+s-1}$ since $U^2/\text{Tor}(U^2)$ is generated by squares of generators of the torsion-free part of $U$. But then if the subgroup is of finite index, they have the same rank.

Edit: In case you've not seen the proof of finite index explicitly, it's quite direct:

If $x\in U^2$ and $x_1,\ldots, x_{r+s-1}$ are generators for $U/\text{Tor}(U)$ then $x=(x_1^{e_1}\ldots x_{r+s-1}^{e_{r+s-1}})^2$ so that the powers are all even, hence $x_1^2,\ldots, x_{r+s-1}^2$ generate $U^2/\text{Tor}(U^2)$. Consider the projection map

$$U/\text{Tor}(U)\to U^2/\text{Tor}(U^2)$$

which is given on the additive structure by the multiplication by $2$ map, you get an isomorphism

$$U^2/\text{Tor}(U^2)\cong (2\Bbb Z)^{r+s-1}$$

for the image.

Edit 2: Let's just do this explicitly. In this particular case it's not even hard. You can see that $x_1^{2e_1}\cdot\ldots\cdot x_{r+s-1}^{2e_{r+s-1}}=1$ (i.e. the identity element of the group) implies that all the $e_i$ are zero because we know that this holds in $U$ which is a bigger group. Hence $x_1,\ldots ,x_{r+s-1}$ are linearly independent elements of $U^2$ hence it has rank $\ge r+s-1$, but it also has rank $\le r+s-1$ because all elements of $U^2$ are also elements of $U$ and it has rank $n$.

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I think what confused you is thinking that the things in the proof were intended to prove $\text{rank}(U^2)=\text{rank}(U)$, but in reality that's known from this argument via basic algebra. The real reason it's in there is because you want to trap the group $RV$ which is between them to compute its rank, which is part of what was being shown.

Answer 2

$U/U^2$ is a finitely generated (multiplicative) abelian group every element of which has order $2$. Thus it is finite.