Find the prime decomposition of 2 in the cyclotomic field $\mathbb{Q} ( \xi_5 ) $. How do I do that?

Question

I am listening to a course of Algebraic number theory and just found a non-constructive proof for this. Have you any hints for me how to tackle this problem?

Answer 1

Let $\zeta$ be a $p$-th root of unity, $p$ prime, and $q$ be a prime $\ne p$. Here $p=5$, $q=2$. Then the ring of integers of $\Bbb Q(\zeta)$ is $\Bbb Z[\zeta]$ and $\zeta$ has minimum polynomial $X^{p-1}+X^{p-2}+\cdots+X+1=\Phi_p(X)$ say. Then $(q)$ splits into distinct prime ideal factors in $\Bbb Z[\zeta]$. A typical factor is $(p,f_j(\zeta)$ where $$\Phi_p(X)\equiv f_1(X)f_2(X)\cdots f_r(X)\pmod q$$ and the $f_i$ are monic polynomials irreducible modulo $q$. Actually we may produce the $f_i$ is a uniform way. Let $\xi$ be some non-trivial $p$-th root of unity in a finite field containing $\Bbb F_q$. Then consider the sequence $\xi$, $\xi^q$, $\xi^{q^2},\ldots$. Eventually one hits a $t$ with $\xi^{q^t}=\xi$. Consider the polynomial $\prod_{j=0}^{t-1}(X-\xi^{q^i})$. Its coefficients lie in $\Bbb F_q$ and so one can think of it as the reduction modulo $q$ of a polynomial over $\Bbb Z$. This polynomial is one of the $f_i$.

In the example at hand, the sequence $\xi$, $\xi^q$, $\xi^{q^2},\ldots$ goes $\xi$, $\xi^2$, $\xi^4$, $\xi^8=\xi^3$, $\xi^{16}=\xi$, so $t=4$ and the $f_i$ is $(X-\xi)(X-\xi^2)(X-\xi^4)(X-\xi^3)$ which is congruent to $X^4+X^3+X^2+X+1$. The prime ideal factor of $(2)$ that arises is $(2,\zeta^4+\zeta^3+\zeta^2+\zeta+1)=(2,0)=(2)$, that is $(2)$ is a prime ideal in $\Bbb Z[\zeta]$.