Is 2 prime in $\mathbb{Z}+\mathbb{Z} ω+\mathbb{Z} ω^2+...+\mathbb{Z} ω^{21}$ where $ω$ is 23rd root of unity I believe it is, because I did a similar problem for $\mathbb{Z}+\mathbb{Z}\omega$ where $\omega=e^{2\pi i/3}$ is 3rd root of unity and proved the element $1-\omega$ is prime.
I did this by showing $\mathbb{Z}+\mathbb{Z}\omega$ is Euclidean with respect to $N(a+b\omega)=(a+b\omega)(a+b\omega^2)=a^2+b^2-ab$, then showing $1-\omega$ is irreducible using properties of the norm.
I don't see how I would be able to do this for the 23rd root of unity since I don't know what the norm should be. It was already pretty tedious for then 3rd root of unity so I imagine this line of thought would be even more tedious for the 23rd root of unity.
Any help is greatly appreciated!
 A: All the important ideas have already been expressed in the comments, but let me flesh it out in a full answer.
Let $p=23$, let $\omega=e^{\frac{2\pi i}{p}}$ and let $R$ be the ring $R={\mathbb Z}[\omega]$ (this is simpler to write than your sum!). The square residues modulo $p$ are $S=\lbrace 1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18 \rbrace$.
By the theory of quadratic Gauss sums, we know that
$$
\sqrt{-p}= \sum_{j=1}^{p-1} \big(\frac{j}{p}\big)\omega ^ j = 1+2\sum_{s\in S} \omega^s
$$
and hence, if we put $x=\frac{1-\sqrt{-p}}{2}, y = \frac{1+\sqrt{-p}}{2}$, we have
$$
x=-\sum_{s\in S} \omega^s \in R, y=1+\sum_{s\in S} \omega^s \in R.
$$
Now, $xy=\frac{1+p}{4}=6$ is divisible by $2$ in $R$, but neither $x$ nor $y$ is divisible by $2$ in $R$. So $2$ is not a prime in $R$.
A: Let $R=\mathbb{Z}+\mathbb{Z}\omega+\mathbb{Z}\omega^2+...+\mathbb{Z}\omega^{21}$
Let $$\alpha=1+\omega^2+\omega^4+\omega^5+\omega^6+\omega^{10}+\omega^{11}\in R$$
$$\beta=1+\omega+\omega^5+\omega^6+\omega^7+\omega^9+\omega^{11}\in R$$
then,
$$\alpha\beta=1+\omega+\omega^5+\omega^6+\omega^7+\omega^9+\omega^{11}+\\+\omega^2+\omega^3+\omega^7+\omega^8+\omega^9+\omega^{11}+\omega^{13}+\\+\omega^4+\omega^5+\omega^9+\omega^{10}+\omega^{11}+\omega^{13}+\omega^{15}+\\+\omega^5+\omega^6+\omega^{10}+\omega^{11}+\omega^{12}+\omega^{14}+\omega^{16}+\\+\omega^6+\omega^7+\omega^{11}+\omega^{12}+\omega^{13}+\omega^{15}+\omega^{17}+\\+\omega^{10}+\omega^{11}+\omega^{15}+\omega^{16}+\omega^{17}+\omega^{19}+\omega^{21}+\\+\omega^{11}+\omega^{12}+\omega^{16}+\omega^{17}+\omega^{18}+\omega^{20}+\omega^{22}\\$$
Using the fact that $\omega^{22}+\omega^{21}+\omega^{20}+...+\omega^2+\omega+1=0$ we get,
$$\alpha\beta=2(\omega^5+\omega^6+\omega^7+\omega^9+\omega^{10}+3\omega^{11}+\omega^{12}+\omega^{13}+\omega^{15}+\omega^{16}+\omega^{17})$$
Thus $2\mid(\alpha\beta)$ but $2\nmid\alpha$ because if it did, then,
$$\alpha=1+\omega^2+\omega^4+\omega^5+\omega^6+\omega^{10}+\omega^{11}=2z$$
for some $z\in R\implies\exists n_0,n_1,n_2,...,n_{21}\in\mathbb{Z}$ such that,
$$1+\omega^2+\omega^4+\omega^5+\omega^6+\omega^{10}+\omega^{11}=2(n_0+n_1\omega + n_2\omega^2+...+n_{21}\omega^{21})$$
Thus $n_0=1/2$ which is a contradiction. Thus, $2\nmid\alpha$. Similarly, $2\nmid\beta$.
Therefore, $2$ is not prime in $R$
