Show that the ring of all rational numbers $m/n$ with $n$ an odd integer is a principal ideal domain.
We haven't really discussed principal ideal domains. I've heard that this is easy, but I just lack the basic knowledge of what a principal ideal domain is.
On
Hint $\ $ An ideal $\rm\,I\neq 0\,$ in a PID is generated by any $\rm\,0\neq b\in I\,$ with least number of prime factors, since such minimality implies that $\rm\,b\,$ divides every $\rm\,c\in I\ $ (else $\rm\ d = gcd(b,c)\mid b\ $ properly, $ $ therefore $\,\rm d\,$ has fewer primes factors than $\rm b,\,$ and $\rm\, (d) = (b,c) \subset I,\,$ contra minimality of $\rm b).$
In your PID $\rm\,D,\,$ all odd primes $\rm\,p\,$ are units by $\rm\,1/p \in D.\,$ So the only prime that survives in $\rm\,D\,$ is $\rm\,p=2.\,$ Thus, by above, an ideal of $\rm\,D\,$ is generated by any one of its elements having the least number of factors of $\,2.\,$ Thus every ideal has form $\rm\,(2^n),\,$ hence every ideal is principal.
Remark $\,$ Implicit in the above is the following pretty generalization of the Euclidean algorithm to arbitrary PIDs. The Dedekind-Hasse criterion states that a domain $\rm\,D\,$ is a PID iff given any $\rm\,0\ne b,c \in D,\,$ either $\rm\,b\mid c\,$ or there exists a $\rm D$-linear combination of $\rm\,b,c\,$ smaller than $\rm b,\,$ where size is measured by naturals (or any ordinal), so that induction (or descent) works.
It is clear that such a domain must be a PID, since the smallest element in an ideal must divide all others. Conversely, since a PID is UFD, an adequate metric is the number of prime factors (since if $\rm\,b\nmid c\,$ then their gcd $\rm\,d\,$ must have fewer prime factors; for if $\rm\,(b,c) = (d)\,$ then $\rm\,d\,|\,b\,$ properly, else $\rm\,b\,|\,d\,|\,c\,$ contra hypothesis). Notice Euclidean descent by the Division Algorithm is just a special case, hence Euclidean $\Rightarrow$ PID ($\Rightarrow$ {UFD, Bezout} $\Rightarrow$ GCD).
On
Perhaps we can attack your problem from the following point of view using some commutative algebra:
Let $R$ be your ring of all rationals which in lowest terms has an odd denominator. Then clearly this contains $\Bbb{Z}$ since any integer can be viewed as sitting inside $R$ via the map that sends $x \mapsto \frac{x}{1}$; clearly $1$ is an odd number and this fraction is always in its lowest terms. Now we can view $R$ as sitting inside of $\Bbb{Q}$ too, which is the fraction field of the integers. It follows that since
$$\Bbb{Z} \subset R \subset \Bbb{Q}$$
and $\Bbb{Z}$ is a PID (Principal Ideal Domain) that $R$ is also a PID. This result that I just used is proved here:
A subring of the field of fractions of a PID is a PID as well.
An integral domain is a commutative ring with unity that has no zero divisors. A Principal Ideal Domain is an integral domain in which every ideal is principal; that is, $R$ is a Principal Ideal Domain if and only if for every ideal $I$ of $R$ there exists $a\in R$ such that $I = (a) = aR = \{ax\mid x\in R\}$.
Examples of PIDs are: the integers (if $I$ is an ideal, then either $I=(0)$, or $I=(a)$ where $a$ is the smallest positive integer in $I$; a consequence of the division algorithm); polynomials with coefficients in $\mathbb{Q}$, in $\mathbb{R}$, in $\mathbb{C}$, or more generally in any field (again, a consequence of the division algorithm for polynomials); any field (the only ideals are $(0)$ and $(1)$) and others.
Let $R$ be the ring of all rational numbers which, when written in lowest terms, have odd denominator. Note that this is the same as the ring of all rationals that can be written with an odd denominator, since if $\frac{a}{b}$ is not in least terms and $b$ is odd, then the reduced fraction $\frac{m}{n}$ with $\frac{a}{b}=\frac{m}{n}$ has $n|b$, and so $n$ is also odd.
Let $I$ be an ideal of $R$, and assume that $I\neq (0)$. If $0\neq\frac{a}{b}\in I$ with $b$ odd, let $n$ be the largest nonnegative integer such that $2^n|a$. I claim that $2^n=\frac{2^n}{1}\in I$: indeed, write $a = 2^nc$ with $c$ an odd integer. Since $\frac{a}{b}\in I$ and $\frac{b}{c}\in R$, then $\frac{b}{c}\frac{a}{b} = \frac{a}{c} = \frac{2^n}{1}\in I$.
Now let $S$ be the collection of all positive integers $n$ such that $2^n\in I$; since $I\neq (0)$, then $S$ is nonempty. Let $m$ be the smallest positive integer in $S$. I claim that $I = (2^m) = (\frac{2^m}{1})$.
Clearly, $2^m\in I$ by construction, so $(2^m)\subseteq I$. Let $\frac{a}{b}\in I$ with $b$ odd; write $a=2^kc$ with $c$ odd. Then $k\geq m$ because from what we saw above, $\frac{2^kc}{b}\in I$ with $b$ and $c$ odd implies $2^k\in I$, hence $k\in S$, hence $k\geq m$. Therefore, $\frac{2^{k-m}c}{b} \in R$, and $\frac{a}{b} = \frac{2^{k-m}c}{b}\frac{2^m}{1}\in (2^m)$; that is, $I\subseteq (2^m)$.
Thus, $I$ is principal; this proves that all ideals of $R$ are principal. $\Box$
What's behind the argument is that every odd integer is a unit, so multiplication by units does not affect ideals in $R$; the only thing that matters are the primes of $\mathbb{Z}$ that are not units when considered in $R$, and the only such is $2$.
This is a consequence of the fact that $R$ is the localization of $\mathbb{Z}$ at the multiplicative set of odd integers; every ideal of $R$ is the extension of an ideal of $\mathbb{Z}$, and since every ideal of $\mathbb{Z}$ is principal, so is every extension of an ideal of $\mathbb{Z}$, hence so is every ideal of $R$.