Polynomial ideals

Question

I got stuck with an exercise while preparing for my exam, and could use a hint or two to move on...

Let $f(X) = a_n X^n+a_{n-1} X^{n-1}+ \cdots +a_0 \in \mathbb{Z}[X]$ with $a_0\neq 0$

Assuming that $X\in \langle f(X)\rangle$ then prove: $n \leq 1$ and either $\langle f(X)\rangle = \langle X\rangle$ or $f(X) = \mathbb{Z}[X]$

I have shown that if $5 \in \langle f(X)\rangle$ then $n = 0 $ and $a_0\mid 5$, and got to at point where I have:

$$X\in \langle f(X)\rangle \Rightarrow X = \lambda f(x), \quad \lambda \in \mathbb{Z}[X]$$

Which tells me that $\deg(f(x))$ is either one or zero.

If it is $0$ then $f(X)$ is a unit

If $\deg(f(X))$ is 1 then $f(X)$ is irreducible and $\langle f(X)\rangle$ is a maximal ideal

Answer 1

$I=\langle f(X)\rangle$ is a principal ideal, which means that every element of $I$ is divisible by $f(X)$. The only divisors of $X$ are the units and associates of $X$ itself, so if $X\in \langle f(X)\rangle $ then $f(X)$ is a unit or $f(X)$ is an associate of $X$. In the first case $I=\mathbb{Z}[X]$ because there exists a $g$ such that $fg=1$, and since $I$ is an ideal this means that for any $h\in\mathbb{Z}[X]$ we have $h\cdot 1=h\in I$. In the second case we have $I=\langle X\rangle$ since $I$ consists precisely of the multiples of $X$.

Answer 2

$X=f(X)g(X) \Rightarrow deg f(X) + deg g(X)=1 \Rightarrow deg f(X)=0,1.$ If $deg f(X)=0,$ then $f(X)$ is a constant. Say $f(X)=\lambda \in \mathbb Z.$ In this case equating the coefficient of $X$ from both side we get that $\lambda$ is actually a unit. (This part you got right.) Now let $f(X)=aX+b, a \neq 0.$ Since $X \in \langle f(X) \rangle,$ we must have $a$ is a unit and $b=0.$ (Why?) Thus $\langle f(X) \rangle = \langle X \rangle.$