Understanding the notation $a \equiv b \bmod p \mathbb{Z} [\zeta]$

Question

Im trying to follow Keith Conrad's notes on Fermats Last Theorem for regular primes but I'm having trouble with some unfamiliar notation. Half way down page three he begins talking about a congruence which I'm not sure about. Namely, $y(\zeta - \zeta^{-1})$$ \equiv 0 $ mod p$\mathbb Z[\zeta]$.

My initial thought was that perhaps it's just reminding me what ring we are working in and I just just treat it as (mod p) but he uses (mod p) many times so there must be some subtle difference I'm not grasping. (He also uses mod p$\mathbb Z$)

Yes I am, just never seen it written like that before. Dietrich Burde's answer below is more than sufficient :) — J.Plaister, Nov 06 '16 at 21:31
Presumably it is a typo that you have no term on the RHS of the conrgrience. right? The definition of congruences extends to general rings in the obvious way. You might find this answer illuminating. — Bill Dubuque, Nov 06 '16 at 21:42
Oh yes, sorry my mistake. There is supposed to be a 0 there. And thank you! I'll check that out now — J.Plaister, Nov 06 '16 at 21:54

score 1 · Accepted Answer · answered Nov 06 '16 at 21:11

1

The notation is $a\equiv b\bmod I$ for elements $a,b$ of a ring $R$ and an ideal $I$ of $R$. It means that $b-a\in I$. For $R=\mathbb{Z}$ we would have $I=n\mathbb{Z}$ for some integer $n$, and then we obtain the usual congruence.

In the text on Fermat the ring is $R=\mathbb{Z}[\zeta]$, the ring of integers in the cyclotomic field $\mathbb{Q}[\zeta]$, with the ideal $I=(p)=p\mathbb{Z}[\zeta]$.

answered Nov 06 '16 at 21:11

Dietrich Burde

130,978

Oh i see, that's simple enough then. Perfect explanation, thank you. – J.Plaister Nov 06 '16 at 21:33
You are welcome! – Dietrich Burde Nov 06 '16 at 21:35

Understanding the notation $a \equiv b \bmod p \mathbb{Z} [\zeta]$

1 Answers1