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If $R$ is a commutative ring, define a relation $\equiv$ on $R$ by $a\equiv b$ if there is a unit $u\in R$ with $b=ua$.

Let $(a)=\{ra: r\in R\}$. Prove that if $R$ is an integral domain, then $(a)=(b)\implies a\equiv b$.

My attempt:

(I have proved that $\equiv$ is an equivalence relation. I don't know if that could help solve this problem.)

$(a)=(b)\implies(a)\subseteq(b)$ and $(b)\subseteq(a)$

Let $x\in(a)$.

So $x=r_1a$ for some $r_1\in R$.

Since $(a)\subseteq(b)$, $x\in(b)$.

So $x=r_2b$ for some $r_2\in R$.

$r_1a=r_2b$ --------------- (1)

Let $x\in(b)$.

So $x=r_3b$ for some $r_3\in R$.

Since $(b)\subseteq(a)$, $x\in(a)$.

So $x=r_4a$ for some $r_4\in R$.

$r_3b=r_4a$ --------------- (2)

Multiplying both sides of (1) by $r_4$, we get

$r_1r_4a=r_2r_4b$

$r_1r_3b=r_2r_4b$ [Using (2)]

If $b\neq 0$, I guess I can cancel $b$ from both sides of the equation because $R$ is an integral domain. But how would I proceed?

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    If $b=0$ were true then what can you say about $a$? Now remove that case and eliminate $b$, then what can you say about the $r_i$? Can you conclude now? – 2017-01-17
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    If $b=0$, $(b)=\{0\}\implies(a)=\{0\}\implies a=0$. So, $a=1b$. Since $1$ is a unit in $R$, $a\equiv b$. If $b\neq 0$, $r_1r_3=r_2r_4$. How do I proceed? – 2017-01-18
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    Can you say that some of these $r_i$ are units from here? – 2017-01-18
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    Sorry, I really don't see that. – 2017-01-18
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    Also, +1 for asking a model question. I like to encourage people who ask good questions. – 2017-01-18

2 Answers 2

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Since $b \in (a)$, there is $r$ such that $b=ra$. Similarly, $a \in (b)$, so there is $s$ such that $a = sb$. Hence, $b = ra = rsb$, so if we can cancel $b$, then $rs=1$, so each of $r,s$ is a unit, so $b=ra$, where $r$ is a unit. Otherwise, $b=0$, then $a=0$ and $b=a$, so we are done.

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If fraction fields of domains are known, then the nonzero case can be proved more intuitively:

$$ (b)\supset (a)\,\Rightarrow\, b\mid a\,\Rightarrow \color{#c00}{a/b\in R}$$

$$ (a)\supset (b)\,\Rightarrow\, a\mid b\,\Rightarrow \color{#c00}{b/a\in R}$$

Therefore we deduce that $\, u = \color{#c00}{b/a\ \text{ is a unit}}$, and $\,b = u a,\,$ as desired. The (common) proof in астон's answer is essentially the same, but the fractions have been eliminated to obtain a purely (integral) proof in $R$. Generally, as above, introducing fractions often serves to clarify proofs in number theory (and algebra).

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    Very late comment, but the last part is good to add. For example, going to fraction fields helps write negative powers of elements. If you have to prove that $a^m= a$ and $a^n = a$ implies that $a^{l} = a$ for $l = \gcd(m,n)$, then fraction fields would allow you to do $l = ms+nt \implies a^l = a^{ms}a^{nt}$ in the fraction field (otherwise you have to be careful, and do some transpositions). – 2018-10-04