Let ${\bf a}(x)\in (\mathbb{Z}/q\mathbb{Z})[x]$, where $q$ is a prime. Suppose that ${\bf a}(1)\equiv 0\pmod{q}$.
Question: How to show that ${\bf a}(x)$ is not invertible in $\frac{\mathbb{Z}/q\mathbb{Z}}{x^N-1}$ where $N$ is prime.
try: Using assumptions we get $(x-1)\mid {\bf a}(x)$ in $(\mathbb{Z}/q\mathbb{Z})[x]$.
But I don't know how to continue the proof.
Thanks for any help
We have $x-1\mid a(x)$.
On the other hand, $x-1\mid x^N-1$ and so the gcd between $a(x)$ and $x^N-1$ is a multiple of $x-1$.
It follows that $a(x)$ is not invertible in the quotient ring $\Bbb Z_q[x]/\langle x^N-1\rangle$, since the invertible elements in this ring have $gcd(a(x),x^N-1)= 1$. All other elements are zero divisors (bar $0$ which is absorbing).
Let $gcd(a(x),x^N-1)=d$. Then the extended Euclidean algorithm gives $d = c(x)(x^N-1) + b(x)a(x)$.
If $d=1$, then $b(x)$ will be the inverse of $a(x)$ in the quotient ring.
The situation in $\Bbb Z_n$ is similar.