TITLE: What are the units of cyclotomic integers?
QUESTION [31 upvotes]: This question made me realize I had a misconception about the cyclotomic integers: I thought the units were exactly the roots of unity. There are only finitely many units but infinitely many integers so the question is impossible to solve unless there are more units. So what are the units of cyclotomic integers?

REPLY [2 votes]: We take the $ p^{\text{th}} $ cyclotomic ring of integers $ \mathbb{Z}[\zeta] $, $ p $ an odd prime, a primitive root $ \gamma\pmod{p} $ and the homomorphism $ \sigma\zeta=\zeta^\gamma $. Kummer took the units
$$\tag{1} \varepsilon_{j}=\dfrac{\sigma^j\zeta-\sigma^{j}\zeta^{-1}}{\sigma^{j-1}\zeta-\sigma^{j-1}\zeta^{-1}}=\sigma^{j-1}\left(\dfrac{\sigma\zeta-\sigma\zeta^{-1}}{\zeta-\zeta^{-1}}\right),\quad 1\le j\le\mu-1, $$
with $ \mu=(p-1)/2 $. These are real units and Kummer also proved that every unit can be expressed as a real unit times a root of unity $ (-\zeta)^n $. We denote the fundamental system of units with $ \hat\varepsilon_j $, $ 1\le j\le\mu-1 $. A unit of the fundamental system of units can also be expressed as a real unit times a root of unity. We choose real units $ \hat\varepsilon_j $ for the fundamental system of units.
Assume that the multiplicative quotient module
$$ [\hat\varepsilon_1,\dots,\hat\varepsilon_{\mu-1}]/[\pm1,\varepsilon_1,\dots,\varepsilon_{\mu-1}] $$
is not trivial or has an order $ N $ greater than $ 1 $. Let the prime $ 2 $ divide the order $ N $. Then there exists a real unit $ E(\zeta) $ of order $ 2 $ in the quotient module and exponents $ x_k $ such that
$$ (-1)^s\prod_{j=1}^{\mu-1}\varepsilon_j^{x_j}=E^2(\zeta),\quad x_j\in\mathbb{Z},\quad s\in\{0,1\}. $$
Kummer observed that the left hand side is a positive number and it must also be a positive number for all conjugates $ \sigma^kE^2(\zeta) $. This allows us to search for possible exponents for which this criterion holds. We just solve the linear system of congruences:
$$ \left(\dfrac{1-\text{sign}\;\sigma^i\varepsilon_j}{2}\right)_{i\times j}\cdot\begin{pmatrix}x_1 \\ \vdots \\ x_{\mu-1}\end{pmatrix}\equiv\begin{pmatrix}s \\ \vdots \\ s\end{pmatrix}\pmod{2}. $$
In the $ 163^{\text{rd}} $ cyclotomic ring of integers we obtain three candidates
$$ E_{0}^2=-\prod_{j=0}^{26}\varepsilon_{1+3j},\quad E_{1}^2=-\prod_{j=0}^{26}\varepsilon_{2+3j},\quad E_{2}^2=E_{0}^2\cdot E_{1}^2, $$
with the primitive root $ \gamma=2\pmod{163} $. Let $ p-1=ef $. The units are invariant under the homomorphism $ \sigma^e $, $ e=3 $, with $ \sigma^{ef/2}\varepsilon_k=\sigma^{\mu}\varepsilon_k=\varepsilon_k $ for a real unit $ \varepsilon_k\equiv\varepsilon_k(\zeta+\zeta^{-1}) $ because $ \sigma^\mu\zeta=\zeta^{-1} $ and we should assume that the units can be expressed by the Gaussian periods
$$ \eta_j=\sigma^j\zeta+\sigma^{j+e}\zeta+\dots+\sigma^{j+(f-1)e}\zeta=\sum_{k=0}^{f-1}\sigma^{ke+j}\zeta,\quad j=0,\dots,e-1, $$
thus, we assume that $ E_k=a_0\eta_0+a_1\eta_1+a_2\eta_2 $ holds for some integers $ a_j $. This gives with $ \sigma\eta_2=\eta_0 $ for periods $ \eta_j $ with length $ f=54 $
\begin{align}
\pm\sigma^0 E_i&=a_0\eta_0+a_1\eta_1+a_2\eta_2 \\ \tag{2}
\pm\sigma^1 E_i&=a_0\eta_1+a_1\eta_2+a_2\eta_0 \\
\pm\sigma^2 E_i&=a_0\eta_2+a_1\eta_0+a_2\eta_1
\end{align}
Writing the units and periods as complex numbers, we can easily solve the linear system of equations and obtain $ E_{0}^2=(5+\eta_2)^2 $ and $ E_{1}^2=(5+\eta_1)^2 $ with $ 1+\eta_0+\eta_1+\eta_2=0 $. There is no unit with order $ 4 $ in the quotient module. The $ 349^{\text{th}} $ cyclotomic ring of integers has the four linearly independent units
\begin{align*}
E(1,3)&=(30 \eta_{0}+30 \eta_{1}+36 \eta_{2}+30 \eta_{3}+42 \eta_{4}+37 \eta_{5})^2, f=58 \\
E(2,4)&=(37 \eta_{0}+30 \eta_{1}+30 \eta_{2}+36 \eta_{3}+30 \eta_{4}+42 \eta_{5})^2, f=58 \\
-E(2,3)&=(8 \eta_{0}+7 \eta_{1}+6 \eta_{2}+6 \eta_{3}+7 \eta_{4}+6 \eta_{5})^2, f=58 \\
-E(2,5)&=(7 \eta_{0}+7 \eta_{1}+6 \eta_{2})^2, f=116
\end{align*}
with
$$ E(a,b)=\prod_{j=0}^{28}\varepsilon_{a+6j}\varepsilon_{b+6j} $$
The periods were built with the primitive root $ \gamma=2\pmod{349} $. There is no unit with order $ 4 $ in the quotient module.
Kummer also had a method in store for computing units that have an odd order $ q\ne p $ in the multiplicative quotient module. In this case we have
$$\tag{3} \varepsilon_q(\zeta)=\prod_{j=1}^{\mu-1}\varepsilon_j^{x_j}=E^q(\zeta),\quad x_j\in\mathbb{Z}. $$
If the sign on the left hand side is negative, we take the unit $ -E(\zeta) $ so that we could leave this sign out. Now, $ E^q(\zeta)=E(\zeta^q)+q\omega_q(\zeta) $ with some cyclotomic integer $ \omega_q(\zeta) $. This gives
$$ \varepsilon_q^q(\zeta)=\left\lbrace E(\zeta^q)+q\omega(\zeta)\right\rbrace^q\equiv E^q(\zeta^q)=\varepsilon_q(\zeta^q)\pmod{q^2} $$
or $ \varepsilon_q^q(\zeta)\equiv\varepsilon_q(\zeta^q)\pmod{q^2} $. With $ \varepsilon_j\equiv\varepsilon_j(\zeta) $ for the units $ (1) $ we also have $ \varepsilon_j^q(\zeta)=\varepsilon_j(\zeta^q)+q\omega_j(\zeta) $ with some cyclotomic integer $ \omega_j(\zeta) $ and this gives
$$ {\varepsilon_q^q(\zeta)}={\left\lbrace \prod_{j=1}^{\mu-1}\varepsilon_j^{x_j}(\zeta) \right\rbrace^q}
={\prod_{j=1}^{\mu-1}\left\lbrace \varepsilon_j(\zeta^q)+q\omega_j(\zeta) \right\rbrace^{x_j}}
={\prod_{j=1}^{\mu-1}\varepsilon_j^{x_j}(\zeta^q)\left\lbrace1+q\dfrac{\omega_j(\zeta)}{\varepsilon_j(\zeta^q)} \right\rbrace^{x_j}}
\equiv{\left\lbrace \prod_{j=1}^{\mu-1}\varepsilon_j^{x_j}(\zeta^q) \right\rbrace\cdot\prod_{j=1}^{\mu-1}\left\lbrace1+qx_j\dfrac{\omega_j(\zeta)}{\varepsilon_j(\zeta^q)} \right\rbrace}
\equiv{\left\lbrace \prod_{j=1}^{\mu-1}\varepsilon_j^{x_j}(\zeta^q) \right\rbrace\cdot\left\lbrace1+\sum_{j=1}^{\mu-1}qx_j\dfrac{\omega_j(\zeta)}{\varepsilon_j(\zeta^q)} \right\rbrace}
\equiv{\varepsilon_q(\zeta^q)+\varepsilon_q(\zeta^q)\sum_{j=1}^{\mu-1}qx_j\dfrac{\omega_j(\zeta)}{\varepsilon_j(\zeta^q)}}
\pmod{q^2} $$
or
$$ \sum_{j=1}^{\mu-1}x_j\dfrac{\omega_j(\zeta)}{\varepsilon_j(\zeta^q)}\equiv0\pmod{q} $$
with $ \varepsilon_q^q(\zeta)\equiv\varepsilon_q(\zeta^q)\pmod{q^2} $ and $ \varepsilon_q(\zeta^q)\not\equiv0\pmod{q} $. This is a linear system of congruences so that we can search for possible exponents $ x_j $ and compute the units similar to the system of equations $ (2) $. In the $ 401^{\text{st}} $ cyclotomic ring of integers, we obtain two linearly independent units
$$ {\prod_{j=0}^{24}\varepsilon_{1+8j}^{2}\cdot\varepsilon_{3+8j}\cdot\varepsilon_{4+8j}\cdot\varepsilon_{5+8j}^{2}\cdot\varepsilon_{6+8j}}
={(3836\eta_0+3637\eta_1+2718\eta_2+3877\eta_3+2915\eta_4+2835\eta_5+3116\eta_6+3559\eta_7)^3} $$
and
$$ {\prod_{j=0}^{24}\varepsilon_{1+8j}^{2}\cdot\varepsilon_{2+8j}^{2}\cdot\varepsilon_{3+8j}\cdot\varepsilon_{4+8j}^{2}\cdot\varepsilon_{7+8j}}={(-85\eta_0-118\eta_1-89\eta_2-89\eta_3-95\eta_4-107\eta_5-116\eta_6-111\eta_7)^3} $$
with periods $ \eta_k $ of length $ f=50 $ and the primitive root $ \gamma=3\pmod{401} $. A more detailed approach to finding these units can be taken from the sections $ 16.5 $ and $ 16.6 $, here.
René Schoof states in his paper Class numbers of real cyclotomic fields of prime conductor, $ 2002 $, that he had found all factors dividing the number $ N $ (or the class number of the $ p^{\text{th}} $ real cyclotomic field) for primes $ p<10000 $ with a likelihood of $ 98\% $. If we could make it for the remaining $2\% $ we would be able to compute many fundamental systems of units!