Eisenstein reciprocity

A number ${\displaystyle \alpha \in {\mathcal {O}}_{m}}$ is called primary[2][3] if it is not a unit, is relatively prime to ${\displaystyle m}$, and is congruent to a rational (i.e. in ${\displaystyle \mathbb {Z} }$) integer ${\displaystyle {\pmod {(1-\zeta _{m})^{2}}}.}$

The following lemma[4][5] shows that primary numbers in ${\displaystyle {\mathcal {O}}_{m}}$ are analogous to positive integers in ${\displaystyle \mathbb {Z} .}$

Suppose that ${\displaystyle \alpha ,\beta \in {\mathcal {O}}_{m}}$ and that both ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are relatively prime to ${\displaystyle m.}$ Then

• There is an integer ${\displaystyle c}$ making ${\displaystyle \zeta _{m}^{c}\alpha }$ primary. This integer is unique ${\displaystyle {\pmod {m}}.}$
• if ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are primary then ${\displaystyle \alpha \pm \beta }$ is primary, provided that ${\displaystyle \alpha \pm \beta }$ is coprime with ${\displaystyle m}$.
• if ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are primary then ${\displaystyle \alpha \beta }$ is primary.
• ${\displaystyle \alpha ^{m}}$ is primary.

The significance of   ${\displaystyle 1-\zeta _{m}}$  which appears in the definition is most easily seen when   ${\displaystyle m=l}$   is a prime.  In that case   ${\displaystyle l=(1-\zeta _{l})(1-\zeta _{l}^{2})\dots (1-\zeta _{l}^{l-1}).}$   Furthermore, the prime ideal   ${\displaystyle (l)}$   of   ${\displaystyle \mathbb {Z} }$   is totally ramified in  ${\displaystyle \mathbb {Q} (\zeta _{l})}$

${\displaystyle (l)=(1-\zeta _{l})^{l-1},}$

  and the ideal   ${\displaystyle (1-\zeta _{l})}$   is prime of degree 1.[6][7]

For ${\displaystyle \alpha ,\beta \in {\mathcal {O}}_{m},}$ the m-th power residue symbol for ${\displaystyle {\mathcal {O}}_{m}}$ is either zero or an m-th root of unity:

${\displaystyle \left({\frac {\alpha }{\beta }}\right)_{m}={\begin{cases}\zeta {\mbox{ where }}\zeta ^{m}=1&{\mbox{ if }}\alpha {\mbox{ and }}\beta {\mbox{ are relatively prime}}\\0&{\mbox{ otherwise}}.\\\end{cases}}}$

It is the m-th power version of the classical (quadratic, m = 2) Jacobi symbol (assuming ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are relatively prime):

• If ${\displaystyle \eta \in {\mathcal {O}}_{m}}$ and ${\displaystyle \alpha \equiv \eta ^{m}{\pmod {\beta }}}$ then ${\displaystyle \left({\frac {\alpha }{\beta }}\right)_{m}=1.}$
• If ${\displaystyle \left({\frac {\alpha }{\beta }}\right)_{m}\neq 1}$ then ${\displaystyle \alpha }$ is not an m-th power ${\displaystyle {\pmod {\beta }}.}$
• If ${\displaystyle \left({\frac {\alpha }{\beta }}\right)_{m}=1}$ then ${\displaystyle \alpha }$ may or may not be an m-th power ${\displaystyle {\pmod {\beta }}.}$

${\displaystyle \left({\frac {\zeta _{m}}{a}}\right)_{m}=\zeta _{m}^{\frac {a^{m-1}-1}{m}}.}$  [8]

${\displaystyle \left({\frac {1-\zeta _{m}}{a}}\right)_{m}=\left({\frac {\zeta _{m}}{a}}\right)_{m}^{\frac {m-1}{2}}.}$  [8]

Let  ${\displaystyle \alpha \in {\mathcal {O}}_{m}}$ be primary (and therefore relatively prime to   ${\displaystyle m}$), and assume that  ${\displaystyle \alpha }$  is also relatively prime to  ${\displaystyle a}$. Then[8][9]

${\displaystyle \left({\frac {\alpha }{a}}\right)_{m}=\left({\frac {a}{\alpha }}\right)_{m}.}$

Assume that ${\displaystyle p}$ is an odd prime, that ${\displaystyle x^{p}+y^{p}+z^{p}=0\;\;}$   for pairwise relatively prime integers (i.e. in ${\displaystyle \mathbb {Z} }$ )   ${\displaystyle x,y,z}$ and that ${\displaystyle p\nmid xyz.\;\;}$

This is the first case of Fermat's last theorem. (The second case is when ${\displaystyle p\mid xyz.\;}$)   Eisenstein reciprocity can be used to prove the following theorems

(Wieferich 1909)[13][14] Under the above assumptions,   ${\displaystyle 2^{p-1}\equiv 1{\pmod {p^{2}}}.\;\;}$

The only primes below 6.7×10¹⁵ that satisfy this are 1093 and 3511. See Wieferich primes for details and current records.

(Mirimanoff 1911)[15] Under the above assumptions   ${\displaystyle 3^{p-1}\equiv 1{\pmod {p^{2}}}.}$

Analogous results are true for all primes ≤ 113, but the proof does not use Eisenstein's law. See Wieferich prime#Connection with Fermat's last theorem.

(Furtwängler 1912)[16][17] Under the above assumptions, for every prime   ${\displaystyle r\mid x,\;\;\;r^{p-1}\equiv 1{\pmod {p^{2}}}.}$

(Furtwängler 1912)[18] Under the above assumptions, for every prime   ${\displaystyle r\mid (x-y),\;\;\;r^{p-1}\equiv 1{\pmod {p^{2}}}.}$

(Vandiver)[19] Under the above assumptions, if in addition   ${\displaystyle p>3,}$   then   ${\displaystyle x^{p}\equiv x,\;y^{p}\equiv y}$   and   ${\displaystyle z^{p}\equiv z{\pmod {p^{3}}}.}$

Eisenstein's law can be used to prove the following theorem (Trost, Ankeny, Rogers).[20]   Suppose   ${\displaystyle a\in \mathbb {Z} }$   and that   ${\displaystyle l\nmid a}$   where   ${\displaystyle l}$   is an odd prime. If   ${\displaystyle x^{l}\equiv a{\pmod {p}}}$   is solvable for all but finitely many primes   ${\displaystyle p}$   then   ${\displaystyle a=b^{l}.}$