Let $\alpha$ be an element of $\mathbb{Q}(\zeta)$. We denote by $N(\alpha)$ the norm of $\alpha$ with respect to $\mathbb{Q}(\zeta)/\mathbb{Q}$.
Let $\alpha$ and $\beta$ be elements of $\mathbb{Z}(\zeta)$. We denote $\alpha \equiv \beta$ (mod $(1 - \zeta)$) if $\alpha - \beta$ belongs to $(1 - \zeta)\mathbb{Z}(\zeta)$.
Lemma 1 $N(1 - \zeta) = l$
Proof: This is immediate by the following formula(replacing $X$ by $1$). $1 + X + \cdots + X^{l-1} = (X - \zeta)(X - \zeta^2)\cdots (X - \zeta^{l-1})$.
Lemma 2 Let $\alpha = f(\zeta)$ be an element of $\mathbb{Z}(\zeta)$, where $f(X)$ is a polynomial in $\mathbb{Z}[X]$. Then $\alpha \equiv 0$ (mod $(1 - \zeta)$) if and only if $f(1) \equiv 0$ (mod $l$).
Proof: Since $\zeta \equiv 1$ (mod $(1 - \zeta))$, $f(\zeta) \equiv f(1)$ (mod $(1 - \zeta))$.
Suppose $\alpha \equiv 0$ (mod $(1 - \zeta)$). Then $f(1) \equiv 0$ (mod $(1 - \zeta)$) by the above congruence. Taking norms of $f(1)$ and $1 - \zeta$, we get $f(1)^{l-1} \equiv 0$ (mod $l$) by the Lemma 1. Hence $f(1) \equiv 0$ (mod $l$)
Conversely suppose $f(1) \equiv 0$ (mod $l$). By the Lemma 1, $f(1) \equiv 0$ (mod $(1 - \zeta)$). Hence $\alpha \equiv 0$ (mod $(1 - \zeta)$) by the above congruence. This completes the proof.
Proposition Let $\alpha$ and $\beta$ be elements of $\mathbb{Z}(\zeta)$. Suppose $\alpha\beta \equiv 0$ (mod $(1 - \zeta)$). Then $\alpha \equiv 0$ (mod $(1 - \zeta)$) or $\beta \equiv 0$ (mod $(1 - \zeta)$).
Proof: This follows immediately from Lemma 2