Let $R$ be a commutative ring. 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 $(f)=(1)$.

Answer 2

Let $R$ be a commutative ring such that $R[x]$ is a PID. We want to prove that every nonzero element of $R$ has a multiplicative inverse, i.e. that $R$ is a field.

Let $a\in R\setminus\{0\}$, and denote $\iota(a)\in R[x]$ the corresponding constant polynomial and with $x\in R[x]$ the "identity polynomial". Being $R[x]$ a PID, there is some $f\in R[x]$ such that $\langle x,\iota(a)\rangle=\langle f\rangle$. More explicitly, this means that there are $\tilde a,\tilde x\in R[x]$ such that $$x=\tilde x f, \qquad \iota(a)=\iota(\tilde a) f.$$ Note that $\iota(a)=\iota(\tilde a) f$ implies that $\deg f=0$. But then, this means that $\tilde x=x x'$, and thus $x' f=1$ for some $x'\in R$. This means that $f$ is a unit (i.e. it is invertible), and therefore $\langle x,\iota(a)\rangle=R[x]$.

We conclude that $\langle x,\iota(a)\rangle = \iota(a)R[x] + x R[x]=R[x]$. This is only possible if $a$ is also invertible. To see it, notice that for any $b\in R$ there must be some $c\in R$ and $g\in R[x]$ such that $$\iota(a)\iota(c) + x g= \iota(b).$$ But then clearly $g=0$, and thus $ac=b$, meaning $a$ is a unit, and thus $R$ a field.

Answer 3

You can prove this proposition another way.

Assume R[x] is a Principal Ideal Domain. Since R is a subring of R[x] then R must be an integral domain (recall that R[x] has an identity if and only if R does).The ideal (x) is a nonzero prime ideal in R[x] because R[x]f(x) is isomorphic to the integral domain R. By Every nonzero prime ideal in a Principal Ideal Domain is a maximal ideal, (x) is a maximal ideal, hence the quotient R is a field.

From Dummit's Abstract Algebra.

Answer 4

f is invertible. fg=1 for some g. Every r in <1> is 1s = fgs = f(gs) $\in \langle f\rangle$. Every $t \in \langle f\rangle$ is fh = 1(fh) $\in \langle 1 \rangle$. Therefore $\langle 1\rangle \subset \langle f\rangle$ and $\langle f\rangle \subset \langle 1\rangle$ and thus $\langle 1\rangle = \langle f\rangle$.

Answer 5

Because of the fact that $〈f〉=\{fx|x\in R\}$. Since $f$ is a unit, $f$ has an inverse $f^{-1}$. Hence $1=f\cdot f^{-1}\in〈f〉$. Also all ideals are contained in $〈1〉=R$. Hence the claim.