Prove $a\mathbb{Z}[x]+x\mathbb{Z}[x]$ is a principal ideal on $\mathbb{Z}[x] \iff a=0$ or $a=1$ or $-1$.

Question

I've tried several things, but I don't know how to properly show it.

Prove $a\mathbb{Z}[x]+x\mathbb{Z}[x]$ is a principal ideal on $\mathbb{Z}[x] \iff a=0$, $a=1$ or $a=-1$.

My try:

Proof:

$\Leftarrow\mid$

Let $a=0$. Then $0\mathbb{Z}[x]+x\mathbb{Z}[x]=x\mathbb{Z}[x]=(x)$ is a principal ideal.

Let $a=1$. Then $1\mathbb{Z}[x]+x\mathbb{Z}[x]=\mathbb{Z}[x]+x\mathbb{Z}[x]=\mathbb{Z}[x]=(1)$ is a principal ideal.

Let $a=-1$. Then $-1\mathbb{Z}[x]+x\mathbb{Z}[x]=-\mathbb{Z}[x]+x\mathbb{Z}[x]=\mathbb{Z}[x]=(-1)=(1)$ is a principal ideal.

I would appreciate a hint to solve the direction $\implies$, because I've tried some things, but without success.

Thanks.

Answer 1

Start by supposing that $a\mathbb Z[x] + x\mathbb Z[x] = (f)$ for some $f\in \mathbb Z[x]$.

On the one hand, we must have $a = gf$ for some $g\in \mathbb Z[x]$, and since $$\deg(fg) = \deg(f)+\deg(g) \ge deg (f),$$ we must have $\deg(f) = 0$ or $f=0$.

When can this happen?

Answer 2

Try a proof by contradiction. Suppose there exists $f\in\mathbb{Z}[x]$ with $a\mathbb{Z}[x]+x\mathbb{Z}[x]=(f)$ Then we must have $a\in(f)$. What does this tell us about $f$?

Answer 3

It is clear if $\,a = 0.\ $ Else if $\rm\ (a,x) = (f)\ $ then using evaluations we find:

$\rm\ f \in (a,x)\ \Rightarrow\ f\ =\ a G + x H.\: $ Eval at $\rm\, x = 0\, \Rightarrow\, f(0) = a G(0) = an\,$ for some $\rm\, n\in \mathbb Z$

$\smash[t]{\rm\quad a \in (f)\, \Rightarrow\, a = f\, g\, \overset{a\,\neq\, 0}\Rightarrow\, deg\ f = 0\, \Rightarrow\, f = f(0) = an}$

$\rm\quad x \in (f)\, \Rightarrow\, x = f\, h = anh.\ $ Eval at $\rm\, x = 1\, \Rightarrow\, 1 = an\,h(1)\ \Rightarrow\ a\mid 1$