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Let D be a domain and $a, b \in D^*$. Show that $a$ is a proper divisor of $b$ if and only if $b=ax$ for some nonzero nonunit $x$.

I'm just really not sure how to start this. Any advice would be great!

user2553807
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  • Suppose that a is a proper divisor of b. What does that mean? – user133458 Mar 14 '14 at 04:44
  • If $a$ divides $b$, then $b = ax$ for some $x$. Note that $x$ is unique because if also $b = ax'$, then $ax = ax'$ and hence $x = x'$. – D_S Mar 14 '14 at 05:02
  • a being a proper divisor of b means that a|b (a divides b) and there is no remainder so (b) $\subset$ (a). Is there a better way to phrase this? – user2553807 Mar 14 '14 at 12:39
  • @user2553807 "No remainder" doesn't make any sense because there may not be any division algorithm, and besides, $a|a$ with no remainder, and you wouldn't want to call that a proper divisor. The proper containment definition is probably the best way to go. – rschwieb Mar 14 '14 at 13:13

2 Answers2

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Unpacking the first hypothesis, you're showing this:

For $a,b\in D^\ast$, $(b)\subsetneq (a)$ if and only if $ax=b$ for a nonzero nonunit $x\in D$.

For $\Rightarrow$: from $(b)\subseteq (a)$ conclude that $b=ax$ for some $x$. Show that if $x$ is zero or a unit, you have a contradiction with $(b)\neq (a)$, therefore $x$ is a nonzero nonunit.

For $\Leftarrow$: conclude that $(b)\subseteq (a)$, and work to show that $(b)=(a)$ is not possible. You will be able to show that if $(a)\subseteq (b)$ and $b=ax$, then $x$ is actually a unit.

rschwieb
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It's not clear which definition of "proper divisor" you use. In any case, this should work:

Theorem $\ $ The following are equivalent for $\,a,b\in D^*$

$(1)\ \ \ ax = b,\quad\ x\nmid 1,\ $ for some $\, x\in D$

$(2)\qquad\ a\mid b,\quad b\nmid a$

$(3)\ \ \ (a)\supseteq (b),\,\ (b)\not\supseteq (a)$

Proof $\ $ (sketch) $\ \ \ \exists x\!:\, b=ax\iff a\mid b\iff (a)\supseteq (b)$

If so, by $\ \dfrac{a}b = \dfrac{1}x\, $ we infer $\, x\mid1 \!\iff\! b\mid a\!\iff\! (b)\supseteq(a)$

Remark $\ $ Above we used $ $ Divides $\!\iff\!$ Contains for principal ideals.

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Bill Dubuque
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