For a number field \(K\) with ring of integers \(\OO_K\), the norm \(N_{K/\QQ}\) is multiplicative, an element of \(\OO_K\) is a unit if and only if its norm is \(\pm1\), and \(a\divides b\) in \(\OO_K\) implies \(N_{K/\QQ}(a)\divides N_{K/\QQ}(b)\) in \(\ZZ\). The unit group \((\ZZ/n)^\times\) has order \(\phi(n)\), where \(\phi\) is Euler's totient function; Euler’s theorem \(a^{\phi(n)}\equiv 1\pmod n\) for \(\gcd(a,n)=1\) is Lagrange’s theorem applied to the class of \(a\) in \((\ZZ/n)^\times\), and Fermat’s little theorem is the case \(n=p\) prime.
Norms in number fields
Multiplicativity of the norm $N_{K/\QQ}$ and units of $\OO_K$ (Tag PR-2ZW5Z)
Let \(K\) be a number field with ring of integers \(\OO_K\). For \(a \in K\), let \(N(a) \coloneqq \det(m_a)\), where \(m_a\colon K \to K\) is the \(\QQ\)-linear map \(x \mapsto ax\); this defines the norm \(N = N_{K/\QQ}\colon K \to \QQ\).
Totients and modular powers
Euler's totient of a prime power (Tag FF-SOE5P)
For a prime \(p\) and an integer \(k\ge1\), the Euler totient of \(p^k\) is \[ \phi(p^k)=p^{k-1}(p-1)=p^k\qty{1-\frac1p}. \]
An integer \(m\) with \(1\le m\le p^k\) fails to be coprime to \(p^k\) exactly when \(p\divides m\), and there are \(p^{k-1}\) such multiples of \(p\). So \(\phi(p^k)=p^k-p^{k-1}\).
Euler's theorem (Tag FT-BAV4D)
Let \(n\ge1\) and let \(a\in\ZZ\) with \(\gcd(a,n)=1\). Then \[ a^{\phi(n)}\equiv1\pmod n, \] where \(\phi\) is the Euler totient function.
The class of \(a\) lies in the group \((\ZZ/n\ZZ)^\times\) of order \(\phi(n)\), so its order divides \(\phi(n)\) by Lagrange’s theorem.
If \(n>1\) and \(d=\gcd(a,n)>1\), then \(a^k\not\equiv1\pmod n\) for every \(k\ge1\), since \(d\) divides both \(a^k\) and \(n\) but not \(1\). For \(a=2\) and \(n=4\), \(2^{\phi(4)}=2^2=4\equiv0\pmod4\).
Fermat's little theorem (Tag FT-RVIGS)
Let \(p\) be a prime and let \(a\in\ZZ\). Then \[ a^p\equiv a\pmod p, \] and if \(p\nmid a\), then \(a^{p-1}\equiv1\pmod p\).
If \(p\nmid a\), the class of \(a\) lies in the group \(\FF_p^\times\) of order \(p-1\), so \(a^{p-1}\equiv1\pmod p\) by Lagrange’s theorem, and multiplying by \(a\) gives \(a^p\equiv a\pmod p\). If \(p\divides a\), both sides of \(a^p\equiv a\) are \(0\) modulo \(p\).