Let $p$ be a prime number and $\mathbb{Q}(\zeta)$ be the pth cyclotomic number field where $\zeta$ is any primitive pth root of unity.

Writing $$t=b_0+b_1\zeta+...+b_{p-2}\zeta^{p-2} $$ with $b_j \in \mathbb{Z}$ , we get $$t^p \equiv b_0^p+b_1^p+...+b_{p-2}^p \pmod{p\mathbb{Z}[\zeta]}$$

I understand the congruence if I consider this modulo p .

  • $\begingroup$ But that is a congruence modulo $p$. $\endgroup$ Jun 23 '19 at 17:22
  • $\begingroup$ I do not really understand it because there you consider the ideal of p , (p) . $\endgroup$ Jun 23 '19 at 17:25
  • $\begingroup$ Instead of what? What else would a congruence “modulo $p$” be in your eyes? $\endgroup$
    – k.stm
    Jun 23 '19 at 17:32

First prove that $t^p = b_0^p + (b_1\zeta)^p + \cdots + (b_{p-2}\zeta^{p-2})^p$.

Hint: write it out and note what terms get a coefficient divisible by $p$, then note that all those coefficients are in your ideal $(p)$.

Another hint: Maybe start small and show $(a + b)^p = a^p + p(\cdots) + b^p$.

Next, use $\zeta^p = 1$ to arrive at your final answer.


The Binomial Theorem based proof of the Freshmans's Dream $\, (a+b)^p = a^p + b^p\,$ in $\,\Bbb F_p\,$ uses only the ring axioms plus $\,p = 0\,$ and $\,a,b\,$ commute. So the proof will work in any such ring.

Said universally, $\,(x+y)^p = x^p + y^p$ in $\Bbb F_p[x,y]\,$ so it holds in any ring containing $\,\Bbb F_p\,$ by the universal mapping property of polynomial rings, i.e. simply evaluate the polynomoial dream at $\,x = a, y = b\,$ to map it into any ring where $\,p=0\,$ and $\,a,b\,$ commute.

In particular the dream is true in your $\,S = R/pR\,$ by $\,p \in pR\,\Rightarrow\, p=0\,$ in $S$


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