Hello fellow mathematicians or students. I need some conceptual help on a proof, and I think it is not that hard, but I am just starting to study Ideals and Rings (Abstract Algebra in general to be accurate), that must be my difficulty.
Sometimes, I dare say almost always, the most simple things can be the most difficult ones to prove. And some of them are so simple that they are impossible to prove (axioms). With no further ado I'll introduce the problem.
Consider $m = q_{1}^{a_1} \cdots q_{t}^{a_t}$ and $n = q_{1}^{b_1} \cdots q_{t}^{b_t}$, where $q_i$ are prime integers and $a_i,b_i$ are non-negative integers, with $i \in \{1,\dots, t\}$.
Prove that $\gcd\{m,n\} = q_{1}^{c_1} \cdots q_{t}^{c_t}$ where $c_i = \min \{a_i,b_i\}$.
Okay, so I started with:
Let's call $d = \gcd\{m,n\}$. By definition follows that $d \mid m$. Indeed
$$
\begin{array}{ccc}
m &=& q_{1}^{a_1} \cdots q_{t}^{a_t} \\[0.3cm]
&=& q_{1}^{a_1 + c_1 – c_1} \cdots q_{t}^{a_t + c_t – c_t}\\[0.3cm]
&=& q_{1}^{c_1 + a_1 – c_1} \cdots q_{t}^{c_t + a_t – c_t} \\[0.3cm]
&=& q_{1}^{c_1}q_{1}^{a_1 – c_1} \cdots q_{t}^{c_t}q_{t}^{a_t – c_t}\\[0.3cm]
&=& (q_{1}^{c_1} \cdots q_{t}^{c_t}) \cdot (q_{1}^{a_1 – c_1} \cdots q_{t}^{a_t – c_t}) \\[0.3cm]
&=& d \cdot r,
\end{array}
$$
where $r = q_{1}^{a_1 – c_1} \cdots q_{t}^{a_t – c_t} \in \mathbb{Z}$. Also, $d \mid n$ can be shown analogously.
Now is where my problems begin. I need to show that
If $d' \in \mathbb{Z}$ is such that $d' \mid m$ and $d' \mid n$, then $d' \mid d$.
I'm just getting $m = d' \cdot r' = d \cdot r$ and $n = d' \cdot s' = d \cdot s$ and I don't know what to do next.
So, my path is correct? Is there another way to show this result?
Thanks in advance. Have a great day!
Best Answer
I finally did find a way to prove this statement, and it goes as it follows:
Let there be a $d' \in \mathbb{Z}$ such that $d' \mid m$ and $d' \mid m$. It's factorization is given by $$d' = {q_1}^{e_1} \cdots {q_t}^{e_t}$$ and it is possible to conclude that there is no other prime $q_{t+1}$ in it's factorization, otherwise $d' \nmid m$ and $d' \nmid n$.
Suppose by absurd that there exists $j \in \{1, \dots, t\}$ such that $e_j > c_j$, where $c_j = \min\{a_j,b_j\}$. If $c_j = a_j$, then $d' \nmid m$. If $c_j = b_j$, then $d' \nmid n$. Which are both contradictions, ergo $e_i \leq c_i$ for all $j \in \{1, \dots, t\}$, id est, $d' \nmid d$.
Therefore, $\gcd\{m,n\} = {q_1}^{c_1} \cdots {q_t}^{c_t}$ where $c_i = \min\{a_i,b_i\}$. cqd
I deeply appreciate everyone that tried to help me, thank you! Have a great day!