Let $n\in\mathbb Z$ and let $A$ be the set of integers co-prime to $n$. Denote $A^{-1}\mathbb Z$ by $\mathbb Z_{(n)}$.
1) Find the invertible elements of $\mathbb Z_{(6)}$
My attempt: let $m$ be in $A$ and $n$ be in $\mathbb Z$. Then $\frac{n}{m}$ is invertible iff there exists $\frac{n'}{m'}$ with $\frac{nn'}{mm'}=1.$ So does that mean $n$ has to be invertible?
2) Show that $\mathbb Z_{(6)}$ has precisely two maximal ideals, namely $m_{1}=(3)\mathbb Z_{(6)}$ and $m_{2}=(2)\mathbb Z_{(6)}$.
I really don't know how to approach this one.
Thanks
Careful: the only invertible elements of $\Bbb Z$ are $\pm 1$, so certainly if $n\neq \pm 1$, it cannot be invertible. I think you mean to write that $A$ is the set of integers coprime to $n$. Note that $A$ is a submonoid of $(\Bbb Z,\,\cdot\,)$. If $m/k\in\Bbb Z_{(6)}$ is invertible we can find $m'/k'$ for which $$\frac mk\frac{m'}{k'}=1$$
This means $kk'=mm'$ in $\Bbb Z$. Since $kk'\in A$, we have $(6,mm')=(6,kk')=1$, whence $(6,m)=1$ and $(6,m')=1$. The invertible elements of $\Bbb Z_{(6)}$ are those with denominator coprime to $6$: one direction is the above, the other is trivial.
Since $\Bbb Z$ is a PID, so is $\Bbb Z_{(6)}$. We know that the ideals of $A^{-1}\Bbb Z$ arise from the prime ideals in $\Bbb Z$ with $\mathfrak a\cap A=\varnothing$. The only such ideals are $(2)$ and $(3)$. Suppose now that $(2)\subseteq \mathfrak a$ for some other proper ideal of $\Bbb Z_{(6)}$. The set $\mathfrak a'=\{x\in\Bbb Z:x/s\in \mathfrak a,{\rm for\; some}\; s\in A\}$ is seen to be an ideal of $\Bbb Z$, containing $(2)'$ properly. But $(2)'$ is a maximal ideal of $\Bbb Z$, so $\mathfrak a'=\Bbb Z$; so $\mathfrak a=\Bbb Z_{(6)}$. The same argument works with $(3)$.