Is the set of all rational numbers with odd denominators a subring of $\Bbb Q$?(When the fraction is completely reduced)
I have tried to apply the subring test on this, and this means I want to show that it is closed under subtraction and multiplication.
To show it is closed under multiplication, I have done the following:
$$\frac{a}{2b+1}\frac{c}{2d+1},a,b,c,d\in\Bbb Z$$
$$=\frac{ac}{4bd+2b+2d+1}$$ Which still has odd denominator.
With regard to subtraction we have:
$$\frac{a}{2b+1}-\frac{c}{2d+1}$$
$$=\frac{a(2d+1)-c(2b+1)}{(2b+1)(2d+1)}=\frac{a(2d+1)-c(2b+1)}{4bd+2b+2d+1}$$
and since the denominator is odd, even if it isn't reduced, it will reduce to another odd denominator.
So we have closure under both of these operations it would seem. Is this sufficient?
Also if it were to show the set of all rational numbers with even demoninator, all I would need is $\frac{1}{2}-\frac{3}{2}=-\frac{2}{2}=-1$ to get a counter example right?
Best Answer
Yes. Actually, it is the ring $\mathbf Z_{(2)}$ – the localisation of $\mathbf Z$ at the prime ideal $(2)$. It has only one maximal ideal, namely $\,2\mathbf Z_{(2)}$, and its residue field is isomorpphic to $\mathbf Z/2\mathbf Z$.