If R is a Principal Ideal Domain and let a and b be two non-zero elements of R. Prove that lcm(a,b)*gcd(a,b)=uab, where u is unity of R. 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. 
 A: $LCM(a,b)$ 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.  
