# A question about ring theory

Consider the ring $$R=\bigg\{\frac{2^km}{n} \ \bigg| \ m,n \ \text{odd integers; k is a non-negative integer}\bigg\}$$.

$$(a)$$ Describe all the units (invertible elements) of $$R$$.

$$(b)$$ Exhibit one nonzero proper ideal $$I$$ of $$R$$.

$$(c)$$ Examine whether the ideal $$I$$ you have chosen is a prime ideal of $$R$$.

Since $$2^k$$ is always a power of $$2$$, it is mutually prime with every odd $$m,n$$. Again if $$b\in R$$ is a unit of $$R$$, then there exists nonzero $$a\in R$$ such that $$ab=ba=1$$. So how to "describe" exactly the units of $$R$$ here? Again using the definition of an ideal of a ring, left or right multiplying any element of $$I$$ with that of $$R$$ takes it into $$I$$. So how to show this is prime?

• $R$ consists of rationals with odd denominator. Which rationals $x$ have odd denominator and also have $1/x$ with odd denominator? If you are looking for ideals, remember that principal ideals are the easiest ones to work with. – Angina Seng May 2 '17 at 4:34
• Note that the condition on $k$ is that it is non-negative. Nowhere does it say that $k$ must be positive. – mweiss May 2 '17 at 4:36
• If $x\in R$ and it has odd denominator then $1/x$ should have even denominator? Because it is a factor of $2^k$. – am_11235... May 2 '17 at 4:37
• You're missing that $k$ can equal $0$. As an example, both $5/3$ and $3/5$ are in $R$. So is $10/3$, but $3/10$ isn't in $R$. – Mark May 2 '17 at 4:40

Since you need odd denominators, you will never cancel a positive power of $2$. So the only option is $k=0$: the invertible elements are precisely $$L=\left\{\frac mn:\ m,n\ \text{ odd }\right\}.$$
A nonzero ideal will consist of non-invertible elements. So this suggests fixing $k$ and taking $$I_k=\left\{\frac{2^hm}n:\ h\geq k, m,n\ \text{ odd }\right\}.$$ To check whether this is prime, if $ab\in I_k$, we have $a=2^hm/n$, $b=2^r s/t$, and $$ab=\frac{2^{h+j}ms}{nt}.$$ We see that we may obtain $k$ as the sum of two numbers strictly less than it.
In conclusion: $I_1$ is prime, but $I_k$, for $k\geq2$, is not. Because $2^k=2^{k-1}\times 2\in I_k$, while neither $2^{k-1}$ nor $2$ are in $I_k$.