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?

  • 1
    $\begingroup$ $R$ 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. $\endgroup$ – Angina Seng May 2 '17 at 4:34
  • $\begingroup$ Note that the condition on $k$ is that it is non-negative. Nowhere does it say that $k$ must be positive. $\endgroup$ – mweiss May 2 '17 at 4:36
  • $\begingroup$ If $x\in R$ and it has odd denominator then $1/x$ should have even denominator? Because it is a factor of $2^k$. $\endgroup$ – am_11235... May 2 '17 at 4:37
  • $\begingroup$ 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 $R$. $\endgroup$ – 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 $I_k$.


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