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There is a very known theorem that says: let $D$ be an integral domain, then $D$ is a field if only if $D[x]$ is a PID.

I found that the condition of $D[x]$ being a PID can be relaxed. More exactly, I found in this paper the following result (there it is theorem 16):

Theorem: Let $D$ be an integral domain. TFAE$\colon$

i) $D$ is a field.

ii) $D[x]$ is a PID.

iii) $D[x]$ is a Bézout domain.

iv) $D[x]$ is a Prüfer domain

All the implications are immediate, except for the implication iv)$\implies$ i). I'm stuck at that part of the proof. My idea is to show that every nonzero element of $D$ is an unit, but given that there are many characterizations of Prüfer domains, I don't know which one to use.

I tried to imitate the proof of the standard theorem ($D[x]$ is a PID, then $D$ is a field). More exactly, if $a\in D\setminus \{0\}$, then I considered the ideal $(a,x)$. Since is clearly a nonzero ideal, by the hypothesis $(a,x)$ is an invertible ideal, therefore there are $q_1(x), q_2(x)\in D(x)$ such that $q_1(x)(a,x)\subseteq D[x]$, $q_2(x)(a,x)\subseteq D[x]$ and $$q_1(x)a+q_2(x)x=1.$$

From here I was trying to write $q_i=\frac{f_i}{g_i}$, for some $f_i, g_i\in D[x]$ ($g_i\neq 0$) and then to use some degree arguments to show that $a$ is an unit, but I didn't success.

Any hints or solutions are appreciated.

Community
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Xam
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3 Answers3

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Since $xq_1(x)\in D[x]$, we have $q_1(x)=f(x)/x$ for some $f(x)\in D[x]$. We also have $aq_1(x)\in D[x]$, which means $af(x)$ is divisible by $x$ in $D[x]$. That is, the constant term of $af(x)$ is $0$, which implies the constant term of $f(x)$ is $0$ as well since $a\neq 0$. Thus $f(x)/x=q_1(x)$ is actually in $D[x]$. Similarly, $q_2(x)\in D[x]$. Evaluation of $q_1(x)a+q_2(x)x=1$ at $x=0$ now gives $q_1(0)a=1$, with $q_1(0)\in D$. Thus $a$ is a unit in $D$, as desired.

Eric Wofsey
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Hint $ $ Here $\ a,x\,$ are coprime, i.e. $\,a\mid xf\,\Rightarrow\, a\mid f,\, $ so $\, (a,x)\,$ invertible $\,\Rightarrow\, (a,x)=(1) = D[x],\,$ i.e. here (Euclid-)coprime $\,\Rightarrow\, $ comaximal, i.e. the converse of Euclid's Lemma (below) holds true

$${\rm if}\ \ (a,b)=(1)\ \ {\rm then}\ \ a\mid bc\,\Rightarrow\, a\mid c\quad\ \ \rm [Euclid's\ Lemma]$$

This can be proved in the same short simple way as the special case in Eric's answer.

Bill Dubuque
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  • Mr. Dubuque thanks for your answer. Just one thing, $a$ and $x$ are coprime because otherwise there would be a non-unit $d$ such that $d\mid a, x$, but then $d\mid 1$, contra, right? – Xam Jul 19 '17 at 15:58
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    @Xam Above I mean "coprime" in the stated sense (see "i.e. ..."), which is immediate: if $a$ divides all coefs of $xf$ then $a$ divides all coefs of $f$, since they have the same list of coefs. Note: the proof in Eric's answer has $a,x$ swapped, i.e. $,x\mid a f,\Rightarrow, x\mid f.\ \ $ – Bill Dubuque Jul 19 '17 at 16:10
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I found a way to solve my question, so although there are already two nice answers, for sake of completeness I've decided to post my own answer.

I'll use another characterization of Prüfer domains. Namely, this one (which is kind of funny because is another post of mine).

Let as usual $a\in D\setminus \{0\}$. By hypothesis $(a:x)+(x:a)=D[x]$. In particular, $1\in (a:x)+(x:a)$, so there are $f\in (a:x)$ and $g\in (x:a)$ such that $f+g=1$.

Now, as $g\in (x:a)$, then $ga\in xD[x]$, so $ga$ has constant term $0$, which implies that $0$ is the constant term of $g$ (since $a\neq 0$). This together with the above equality gives us that $f$ has constant term $1$, so the coefficient of the linear term of $fx$ is $1$. On the other hand, since $fx\in aD[x]$, then $fx=ah$ for some $h\in D[x]$. If $b$ is the coefficient of the linear term of $h$, then by equality of polynomials we deduce that $1=ab$. Hence, $a$ is an unit and we're done.

Xam
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