LCM and GCD polynomial relationship I need some help with constructing a proof for the following statement,$ \frac{P_1 P_2}{hcf(m,n)} = lcm(P_1,P_2)$  where $P_1$ and $P_2$ are polynomials with real coefficients.
I know how to do the same for integers using prime factors and their exponents but not sure where to go with polynomials.
 A: Do it the exact same way.  Suppose that the hcf/gcd of $P_1$ and $P_2$ is $G$.  Because $G$ is a factor of $P_1$, there exists an $R_1$ such that $P_1$ is equal to $R_1G$, and likewise $P_2$ equals some $R_2G$.  
$P_1P_2 = R_1R_2GG$
${P_1P_2 \over G} = R_1R_2G$
$R_1$ and $R_2$ can have no factors in common as any factor $H$ could be multiplied by $G$ to obtain a new GCD.   
Because $R_1R_2G$ is a multiple of $R_1G$, it is a multiple of $P_1$, and likewise for $P_2$.  It is a multiple of both, and no factor can be removed which would preserve its multiplicity.  Therefore, it is the Least Common Multiple.
Q.E.D.
A: This proof works in any gcd domain. We use the $\,\overbrace{{\rm involution}\,\  x' :=\, ab/x}^{\rm\large cofactor\ duality\ \ }\ $ on the divisors of $\rm\:ab,\,$ which exposes  $\rm\color{#c00}{cofactor\  reflection}$ $\rm\ x\mid y\color{#c00}\iff y'\mid x',\ $  by ${\,\ \rm\dfrac{y}x = \dfrac{x'}{y'} \ }$ by $\rm\, \ yy'\! = ab = xx'.\, $ Thus
$$\begin{align}\rm c\mid\gcd(a,b)\!\iff&\rm\ c\mid a,b\\[3px]
 \color{#c00}\iff&\ \rm b',a'\mid c'\\[3px]
\iff &\ \rm lcm(b',a')\mid c'\\[3px]
\color{#c00}\iff &\ \rm c\mid lcm(b',a')' \\
{\rm Thus}\rm\quad \gcd(a,b)\, \ \cong\ \,&\rm \,  lcm(b',a')'\,=\ \dfrac{ab}{lcm(a,b)}
\end{align}\ $$
i.e. having the same set of divisors $\,c,\,$ they divide each other (i.e. they are associate $\cong\,)$
Above the red arrows are $\rm\color{#c00}{cofactor\ reflections}$ and the black arrows are the definition (or universal property) of gcd and lcm.
A: It works pretty much the same for integers if you modify the argument a little. Let $L = lcm(P_1, P_2)$ and $G=gcd(P_1, P_2)$. Then
$$P_1 = Gh_1, P_2 = Gh_2,$$
with $gcd(h_1, h_2) = 1$. It's easy to see that $P_1$ and $P_2$ both divides $Gh_1h_2$ so $L$ also divides $Gh_1h_2$. Assume that 
$$ Gh_1h_2 = Lh,$$
then $P_1 h_2 = L h$, or $h_2 = \frac{L}{P_1} h$. That is, $h$ divides $h_2$. Similarly $h$ divides $h_1$. Since $gcd(h_1,h_2)=1$, $h$ must be as scalar as well. In other words
$$L= Gh_1h_2 = \frac{P_1P_2}{G}.$$
A: Think of the irreducible factors of P$_1$ and P$_2$ as your prime factors. Suppose P$_1$ = gcd(P$_1$,P$_2$)($q_1q_2\ldots q_n$) and P$_2$=gcd(P$_1$,P$_2$)($r_1r_2\ldots r_m$). 
Thus 
$$\frac{P_1P_2}{gcd(P_1,P_2)}=gcd(P_1,P_2)(q_1\ldots q_n)(r_1 \ldots r_m).$$
Note that the numerator has [gcd(P$_1$,P$_2$)]$^2$ as a factor.
So the RHS is a common multiple of P$_1$ and P$_2$. You should be able to show that if there is a "smaller" lcm, then we can get a "larger" gcd.
