Abstract Algebra – If $R[X]$ is a Principal Ideal Domain, Then $R$ is a Field

Question

I've just read a proof of the statement:

Let $R$ be a commutative ring. If $R[X]$ is a principal ideal domain, then $R$ is a field.

In one part of the proof there is a step which I don't understand. I'll copy the proof:

Let $u\in R$ be non-zero.

Then using the principal ideal property, for some $f\in R[X]$ we have $\langle u,X\rangle = \langle f \rangle \subseteq R[X]$. Therefore, for some $p,q∈R[X], u=fp$ and $X=fq$.

By properties of degree we conclude that $f=a$ and $q=b+cX$ for some $a,b,c\in R$.

Substituting into the equation $X=fq$ we obtain $X=ab+acX$ which implies that $ac=1$, i.e. $a\in R^\times$, the group of units of $R$.

Therefore, $\langle f\rangle =\langle 1\rangle=R[X]$.

Therefore, there exist $r,s\in R[X]$ such that $ru+sX=1$ and if $d$ is the constant term of $r$, then we have $ud=1$.

Therefore $u\in R^\times$.

Our choice of $u$ was arbitrary, so this shows that $R^{\times}⊇R \setminus \{0\}$, which says precisely that $R$ is a field.

I don't understand this:

"$X=ab+acX$ which implies that $ac=1$, i.e. $a\in R^\times$, the group of units of $R$. Therefore, $\langle f\rangle = \langle 1\rangle =R[X]$."

Why does the fact that $a=f$ is a unit implies that $\langle f\rangle = \langle 1\rangle$?

I would appreciate if someone could explain this to me. Thanks in advance.

Answer 1

If $f$ is a unit then there exists $g \in R[x]$ such that $1=gf \in (f)$. So $1 \in (f)$ implies that $(f)=(1)$.