Quartic polynomial whose coefficients are product of symmetrical quantities

34 Views Asked by At

The following is taken from Problem 6.2 on p.86 of The Theory of Algebraic Numbers by Harry Pollard and Harold G. Diamond (Dover edition).

Consider the quartic polynomial $$(x-am)\,(x-an)\,(x-bm)\,(x-bn)\ \equiv\ x^4-c_1x^3+c_2x^2-c_3x+c_4$$

It is not stated what the $a,b,m,n$ are – they can be integers, rational numbers, etc. The question simply asks to verify that the coefficients $c_i$ can be expressed as the following product: $$c_i(a,b,m,n)\ =\ p_i(a,b)\cdot q_i(m,n)\quad(i=1,2,3,4)$$

where $p_i(a,b)$ is an expression in $a,b$ only and $q_i(m,n)$ is an expression in $m,n$ only. This is easily done for $c_1,c_3,c_4$ but I have a problem with $c_2$ (the coefficient of $x^2$). How can I write $c_2=p_2q_2$ where $p_2$ involves $a,b$ only and $q_2$ involves $m,n$ only?

Note that $p_2$ and $q_2$ need not be integers even if $a,b,m,n$ themselves are. If $a,b,m,n$ are integers, then $c_2$ is necessarily an integer, but $p_2,q_2$ themselves need not be – in fact, they need not even be rational.

I would appreciate some help here. Thanks.

1

There are 1 best solutions below

0
On

\begin{align*} c_{1} &= \alpha+\beta+\gamma+\delta \\ &= am+an+bm+bn \\ &= (a+b)(m+n) \\ c_{2} &= \alpha \beta+\alpha \gamma+\alpha \delta+ \beta \gamma+\beta \delta+\gamma \delta \\ &= (am)(an)+(am)(bm)+(am)(bn)+(an)(bm)+(an)(bn)+(bm)(bn) \\ &= (a^2+b^2)mn+ab(m^2+n^2)+2abmn \\ c_{3} &= \alpha \beta \gamma+\alpha \beta \delta+ \alpha \gamma \delta+\beta \gamma \delta \\ &= (am)(an)(bm)+(am)(an)(bn)+(am)(bm)(bn)+(an)(bm)(bn) \\ &= ab(a+b)mn(m+n) \\ c_{4} &= \alpha \beta \gamma \delta \\ &= a^2 b^2 m^2 n^2 \end{align*}

Note that $c_{2}$ is not the case.