How can we show that $x^N - 1$ is an ideal of the ring $\mathbb{Z}[x]$?
I understand that $I$ is an ideal if $fg$ is in I for all $f$ in $\mathbb{Z}[x]$, all $g \in I$. But how do we see this for a value like $x^N - 1$? I feel that I am missing an insight here short of multiplying everything out to see why it is an ideal.
Given any subset $S \subseteq R$ of any (commutative) ring, we can define the generated ideal of $S$, denoted by $\langle S\rangle$ or sometimes also $(S)$, to be smallest ideal, which contains $S$. If $S = \{r\}$ contains only a single element we sometimes also write $\langle r\rangle$ instead of $\langle \{r\}\rangle$.
So $\langle x^N - 1\rangle$ is an ideal simply by the definition. If you want to express it as a set, you have $\langle x^N - 1\rangle = \{ f \cdot (x^N - 1) | f \in \mathbb Z [x]\}$. One can show that the right side is an ideal by showing that it is an additive subgroup and closed under multiplication by arbitrary ring elements.