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.
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',\, $ so
$$\ \ \begin{align}\rm c\mid\color{#0a0}{\gcd(a,b)}\!\iff&\rm\ c\mid a,b\\[3px] \color{#c00}\iff&\ \rm b',a'\mid c'\\[3px] \iff &\ \rm lcm(b',a')\mid c'\\[3px] \color{#c00}\iff &\ \rm c\mid \color{#90f}{lcm(b',a')'} \\ {\rm thus}\rm\ \ \ \color{#0a0}{\gcd(a,b)}\, \ \approx\ \,&\rm \, \color{#90f}{lcm(b',a')'}\,=\ \dfrac{ab}{lcm(a,b)} \end{align} $$
i.e. having the same set of divisors $\,c,\,$ they divide each other, i.e. they are associate $(\approx)$.
Above the red arrows are $\rm\color{#c00}{cofactor\ reflections}$ and the black arrows are the definition (or universal property) of gcd and lcm.
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.
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 $$L= Gh_1h_2 = \frac{P_1P_2}{G}.$$
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.
