Krdytkyu

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.

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.

This proof works in any gcd domain. We use the $,overbrace{{rm involution}, x' :=, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab,,$ which exposes $ $ cofactor reflection $,rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus

$$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[3px]
color{#c00}iff& rm b',a'mid c'\[3px]
iff & rm lcm(b',a')mid c'\[3px]
color{#c00}iff & rm cmid lcm(b',a')' \
{rm Thus}rmquad gcd(a,b), cong &rm , lcm(b',a')'= dfrac{ab}{lcm(a,b)}
end{align}quad $$

Above the black arrows are the definition of gcd and lcm, and the red arrows are cofactor reflections, and $,acong b,$ means $,a,b,$ are associates, i.e. $,amid bmid a$.

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
$$L= Gh_1h_2 = frac{P_1P_2}{G}.$$

Think of the irreducible factors of P$_1$ and P$_2$ as your prime factors. Suppose P$_1$ = gcd(P$_1$,P$_2$)($q_1q_2ldots q_n$) and P$_2$=gcd(P$_1$,P$_2$)($r_1r_2ldots r_m$).
Thus
$$frac{P_1P_2}{gcd(P_1,P_2)}=gcd(P_1,P_2)(q_1ldots q_n)(r_1 ldots r_m).$$
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.