Let a,b,c, be nonzero integers such that $\mathbb Za+\mathbb Zb+\mathbb Zc=\mathbb Z$. 
Let $a,b,c$ be non-zero integers such that $\mathbb Za + \mathbb Zb + \mathbb Zc = \mathbb Z$. Show that $\mathbb Za^{-1}\cap\mathbb Zb^{-1}\cap\mathbb Zc^{-1}=\mathbb Z$.

We have written out a bunch of definitions and implications that we got from this, as well as $\gcd$ and $\DeclareMathOperator{lcm}{lcm}\lcm$ implications. 

Hint: the intersection of $\mathbb Za^{-1}\cap\mathbb Zb^{-1}\cap\mathbb Zc^{-1}=\mathbb Zm$.  It is enough to show that $m$ exists in $\mathbb Z$, why?

 A: In general, for an integer $t$, if $\alpha\in\mathbf{Z}[t^{-1}]$,
then $\alpha=n\frac{1}{t^m}$ where $n, m \in \mathbf{Z}$. Hence, $t^m\alpha\in \mathbf{Z}$.
Now let $\alpha\in\mathbf{Z}[a^{-1}]\cap\mathbf{Z}[b^{-1}]\cap\mathbf{Z}[c^{-1}]$. Then there exist $s_1,s_2,s_3\in \mathbf{Z}$ such that
$a^{s_1}\alpha,b^{s_2}\alpha,c^{s_3}\alpha\in \mathbf{Z}$. Since
$\mathbf{Z}a+\mathbf{Z}b+\mathbf{Z}c=\mathbf{Z}$,
$x_1a+x_2b+x_3c=1$ for some $x_1,x_2,x_3 \in\mathbf{Z}$.
Hence 
$(x_1a+x_2b+x_3c)^{s_1+s_2+s_3}=x_1'a^{s_1}+x_2'b^{s_2}+x_3'c^{s_3}=1$,
and so $\alpha=1\alpha=(x_1'a^{s_1}+x_2'b^{s_2}+x_3'c^{s_3})\alpha\in
\mathbf{Z}$. Therefore, $\mathbf{Z}[a^{-1}]\cap\mathbf{Z}[b^{-1}]\cap\mathbf{Z}[c^{-1}]\subseteq\mathbf{Z}$.
Clearly
$\mathbf{Z}[a^{-1}]\cap\mathbf{Z}[b^{-1}]\cap\mathbf{Z}[c^{-1}]\supseteq\mathbf{Z}$,
and so
$\mathbf{Z}[a^{-1}]\cap\mathbf{Z}[b^{-1}]\cap\mathbf{Z}[c^{-1}]=\mathbf{Z}$.
A: We will break this up into three steps:


*

*Show $a, b, c$ are coprime.


We have that $\Bbb{Z}a+\Bbb{Z}b+\Bbb{Z}c=\Bbb{Z}$. This means that given any integer, there is some sum of an integer times $a$, another integer times $b$, and another integer times $c$ that equals the original integer. Therefore, we can apply this to the integer $1$ to find that there exist $z_1, z_2, z_3$ such that $z_1a+z_2a+z_3c=1$, meaning $a, b, c$ are coprime.


*Show $\Bbb{Z} \subseteq \Bbb{Z}a^{-1} \cap \Bbb{Z}b^{-1} \cap \Bbb{Z}c^{-1}$.


It is trivial to show that $\Bbb{Z} \subseteq \Bbb{Z}a^{-1}$. For any integer $z$, we simply choose $az$ to find that $az(a^{-1})=z \in \Bbb{Z}a^{-1}$. The same goes for $\Bbb{Z}b^{-1}$ and $\Bbb{Z}c^{-1}$, so $\Bbb{Z} \subseteq \Bbb{Z}a^{-1} \cap \Bbb{Z}b^{-1} \cap \Bbb{Z}c^{-1}$.


*$\Bbb{Z}a^{-1} \cap \Bbb{Z}b^{-1} \cap \Bbb{Z}c^{-1} \subseteq \Bbb{Z}$


Now, let's say $\frac p q \in \Bbb{Z}a^{-1} \cap \Bbb{Z}b^{-1} \cap \Bbb{Z}c^{-1}$, where $p$ and $q$ are coprime integers. Since $\frac p q \in \Bbb{Z}a^{-1}$, there exists an integer $z$ such that $\frac p q=\frac z a$. Cross products show that $ap=qz$, meaning $q \mid ap$. However, since $p$ and $q$ are coprime, we can deduce that $q \mid a$. By a similar argument, $q \mid b$ and $q \mid c$. However, $a, b, c$ are coprime, so their only common divisor is $1$, meaning $q=1$. Thus $\frac p q=p \in \Bbb{Z}$. We started with a generic rational number in the set of $\Bbb{Z}a^{-1} \cap \Bbb{Z}b^{-1} \cap \Bbb{Z}c^{-1}$ and shown that it is in $\Bbb{Z}$, which implies $\Bbb{Z}a^{-1} \cap \Bbb{Z}b^{-1} \cap \Bbb{Z}c^{-1} \subseteq \Bbb{Z}$.
From steps 2 and 3, we can deduce $\Bbb{Z}a^{-1} \cap \Bbb{Z}b^{-1} \cap \Bbb{Z}c^{-1}=\Bbb{Z}$.
