Let $\zeta_\lambda$ be a $\lambda$th root of unity, where $\lambda$ is a rational prime. Let $h(\zeta_\lambda) \in \Bbb Z[\zeta_\lambda]$ be a cyclotomic prime and let $x, y \in \Bbb Z$ be coprime integers such that $h(\zeta_\lambda) \mid x+\zeta_\lambda^jy$, with $\lambda \nmid j$.
I wish to prove that $h(\zeta_\lambda) \mid \lambda$ or $h(\zeta_\lambda) \mid p$ where $p$ is a rational prime and $p \equiv 1 \bmod \lambda$. I was given the following "preparation" to prove:
Suppose $h(\zeta_\lambda) \mid x+\zeta_\lambda^jy$. Then $h(\zeta_\lambda)\mid N(x+\zeta_\lambda^jy)$, the norm of $x + \zeta_\lambda^jy$. Since $N(x+\zeta_\lambda^jy) \in \Bbb Z$, it has a prime factorisation, and since $h(\zeta_\lambda)$ is a cyclotomic prime, it divides exactly one of these prime factors, so $h(\zeta_\lambda) \mid p$ for some rational prime $p$. Now, $p \nmid y$, since if $p\mid y$ then, as $h(\zeta_\lambda) \mid p$, we have $\zeta_\lambda^jy \equiv 0 \bmod p$ and hence $\zeta_\lambda^j \equiv 0 \bmod h(\zeta_\lambda)$, from which it follows that $x \equiv 0 \bmod h(\zeta_\lambda)$, contradicting the way that $x$ and $y$ were chosen. So $\zeta_\lambda^jy \equiv k \bmod p$ for some $k \in \Bbb Z$. Since $y \neq 0$, it is invertible modulo $p$, so that $\zeta_\lambda^j \equiv k^\prime \bmod p$, and hence $\zeta_\lambda^j \equiv k^\prime \bmod h(\zeta_\lambda)$. Finally, since $\zeta_\lambda$ is conjugate to its powers, we apply the conjugation that sends $\zeta_\lambda^j \mapsto \zeta_\lambda$ and conclude that $\zeta_\lambda \equiv k^\prime \bmod h(\zeta_\lambda)$ for some $k^\prime \in \Bbb Z$.
I'm struggling to see the relevance of this to the problem statement though. If $k \in \Bbb Z$ is the integer to which $\zeta_\lambda$ is congruent modulo $h(\zeta_\lambda)$ then I guess $$x + \zeta_\lambda^jy \equiv 0 \bmod h(\zeta_\lambda) \iff x + k^jy \equiv 0 \bmod p$$
but again I'm not sure how to develop this idea.
$\newcommand{\la}{\lambda}\newcommand{\ze}{\zeta}$Consider the quotient ring $R=\Bbb Z[\la]/(h(\ze))$. As $h(\ze)$ is a prime element it divides a rational prime $p$, and so $R$ is a finite field of characteristic $p$. But $x\equiv -\ze^j y\pmod{h(\ze)}$. This means that $\ze^j\equiv -x/y\pmod p$ (here $1/y$ is a modulo $p$ reciprocal of $y$). Taking $\la$-th powers gives $1\equiv(-x/y)^\la$, so some integer has order $1$ or $\la$ modulo $p$. In the latter case we must have $p\equiv1\pmod\la$. In the former case $-x/y\equiv1\pmod p$, that means that $\zeta\equiv1\pmod{h(\ze)}$. In this case $h(\ze)\mid(\ze-1)$ and as $\ze-1$ has norm $p$, then $\la=p$.