Show that the ring of all rational numbers, which when written in simplest form has an odd denominator, is a principal ideal domain. 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.

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 $R$. 
