# Why does the congruence hold?

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 .

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

## 2 Answers

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 $$S$$