Kindly help me up with this proof. I have been to this statement with the many of books which prove using summation of power series method. I have been through many books they all pose this as the proposition but no proof is being discussed.
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PID implies UFD and from there you can compare factorizations of $a$ and $b$ to obtain the lcm and gcd. – green frog Sep 18 '18 at 05:38
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I am very curious how this could be approached "using summation of power series method" – Badam Baplan Sep 18 '18 at 12:49
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$LCM(a,b)$ is defined as an element $m$ such that $a,b \mid m$ and if $a,b \mid n$ then $m \mid n$. Note that $LCM(a,b) \mid ab$ since $a,b \mid ab$.
Similarly $GCD(a,b)$ is an element $d$ such that $d \mid a,b$ and if $e \mid a,b$ then $e \mid d$. Both $LCM$ and $GCD$ only make sense to define up to multiplies by a unit.
Note that these definitions make sense only up to unit multiples.
Let's prove that the element $\frac{ab}{LCM(a,b)}$ satisfies the definition of $GCD(a,b)$.
Firstly it is clear that $\frac{ab}{LCM(a,b)} \mid a$ since $\frac{ab}{LCM(a,b)} \frac{LCM(a,b)}{b} = a$, and likewise for $b$.
Secondly, if $d$ is an element such that $d \mid a,b$ then $a,b \mid \frac{ab}{d}$ and thus $LCM(a,b) \mid \frac{ab}{d}$. It follows that $d \mid \frac{ab}{LCM(a,b)}$
As mentioned in the comments, this fact is often approached in $PID$s and $UFD$s by examining prime factorizations of $a$ and $b$, but the above proof works in rings (allowing zero divisors) in which not all pairs of elements are even required to have $GCD$s and $LCM$s. The point is that regardless of the structure of the ring, if $a$ and $b$ possess an $LCM$ and $GCD$ which are not zero divisors, then the identity in question holds.
Badam Baplan
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