For example, 1-(-1) is in the ideal <2>, whenever $n$ is even.
Suppose $R=\Bbb Z[\zeta_n]$, and (by the above), that $n$ is odd. We know 1+$\zeta_n$ can be multiplied by $1+\zeta_n^2+\zeta_n^4+\cdots+\zeta_n^{n-1}$ to get 1+0=1, so $1+\zeta_n$ is a unit of $R$.
However, $1-\zeta_n$ seems to fall into an ideal of $R$ generated by $\{\zeta_n^i-\zeta_n^{i+1}\}_{0\leq i<n}$. Is this true? If so, how does this ideal intersect with $\Bbb Z$? (e.g., if $k=\Bbb Q(\zeta_n)$, then $R$ is the algebraic integers of $k$, and so every ideal of $R$ should be liftable to an ideal of $\Bbb Z$).
Let $\Phi_n(x)$ be the cyclotomic polynomial.
Claim: If $n$ is divisible by two distinct primes, then $1-\zeta_n$ is a unit in $\mathbb Z[\zeta_n]$.
Proof: We only need to show that if $p\neq q$ are primes then $\zeta_{pq}$ is a unit in $\mathbb Z[\zeta_{pq}]$, because $1-\zeta_n$ is a divisor of $1-\zeta_{pq}$ if $pq\mid n$, and $\mathbb Z[\zeta_{pq}]\subseteq \mathbb Z[\zeta_{n}]$.
Now, consider the polynomial $$\begin{align} \Phi_{pq}(x) &=\frac{(x^{pq}-1)(x-1)}{(x^p-1)(x^q-1)}\\ &=\frac{(1+x+x^2+\dots x^{pq-1})}{(1+x+\dots x^{p-1})(1+x+\dots+x^{q-1})}\tag{1} \end{align} $$
It has as roots all are exactly the primitive $pq$th roots of unity, which are all in $\mathbb Z[\zeta_{pq}]$. So it splits as:
$$\prod_{(k,pq)=1} \left(x-\zeta_{pq}^k\right)$$
Now, $\Phi_{pq}(1)= 1$ by (1), so $1$ is a multiple of $1-\zeta_{pq}$ in $\mathbb Z[\zeta_{pq}]$, so $1-\zeta_{pq}$ is a unit.
Note that $$\Phi_n(x)=\prod_{(n,k)=1} (x-\zeta_n^k)$$ So if $1-\zeta_n$ is a unit, then so is $1-\zeta_n^k$ for $(k,n)=1$, and therefore so would be $\Phi_n(1)$.
But when $n=p^m$ for prime $p$, we can easily show that $\Phi_n(1)=p$. So $1-\zeta_n$ is not a unit, and is a divisor of $p$. So the answer is $p\mathbb Z$.