Properties of associates in an integral domain (Ring theory)

Question

I was asked to show the following three exercises for applications to understanding PID is UFD, however, I am struggling a little bit with the third one

Let $a$ and $b$ be elements of an integral domain $R$. Then

(1) $(a) \subseteq (b)$ if and only if $b|a$

The principal ideal $(b)$ consists of all multiples of $b$ thus $a \in (b) \iff b|a$

So if $(a) \subseteq (b)$ then $a$ is in the ideal $(b)$ so $b|a$

To show the converse, if $b|a$ then $a \in (b)$ which implies that every multiple of $a$ is also in $(b)$, therefore $(a) \subseteq (b) \ QED$

(2) $(a)=(b)$ if and only if $b|a$ and $a|b$

$(a)=(b)$ if and only if $(a) \subseteq (b)$ and $(b) \subseteq (a)$ so from 1) $(a) \subseteq (b)$ if and only if $b|a$ and $(b) \subseteq (a)$ if and only if $a|b \ QED$

(3) $(a) \subsetneq (b)$ if and only if $b|a$ and $b$ is not an associate of $a$

(this is where I am a little bit lost, but I have got the hint that if I can show that $a|b$ and $b|a$ if and only if $b$ is an associate of $a$, and then use 1) and 2) then it should be possible.

Answer 1

(1) Suppose $(a)\subseteq(b)$; in particular $a\in(b)$, so $b\mid a$. Conversely, suppose $b\mid a$, with $a=bx$; if $r=as\in(a)$ then $r=bxs$ and therefore $r\in(b)$. Hence $(a)\subseteq(b)$.

^{Your idea is good, but it should be written out in greater detail.}

(2) This follows from (1); indeed $(a)=(b)$ is the same as “$(a)\subseteq(b)$ and $(b)\subseteq(a)$”. Thus $b\mid a$ and $a\mid b$.

Note. The condition “$b\mid a$ and $a\mid b$” implies $a$ and $b$ are associate. Indeed, $b=ax$ and $a=by$ for some $x$ and $y$. Therefore $a=axy$ and so $a(xy-1)=0$. Since $R$ is a domain, either $a=0$ (and so also $b=0$) or $xy=1$, implying $x$ is invertible.

^{Conversely, if $a$ and $b$ are associate, then $(a)=(b)$ because of (1).}

(3) Follows from (2) and the note: if $(a)\subsetneq(b)$, then $b\mid a$; but $b$ cannot be associate to $a$, otherwise $(a)=(b)$. You should be able to do the converse.