Least possible polynomial deegree of complex roots

Question

What is the least possible deegree of polynomial with real coefficients having roots $2\omega , 2+3\omega , 2+3\omega ^2 , 2-\omega -\omega ^2$

As there are four roots so the deegree should be four but the answer is given as five . how ?

Answer 1

Let's look at the given roots. We see that one of these roots, $2-\omega -\omega ^2$, is in fact real: $$ 2-\omega -\omega ^2 = 3, \quad\mbox{ while }\quad (x-(2+3\omega))(x-(2+3\omega^2)) = x^2-x+7. $$

A fourth degree polynomial with the four given roots would have complex coefficients, namely, it's the polynomial $$ P_4(x) = (x - 3) ( x - 2\omega ) (x^2 - x + 7). \tag{1} $$

There exists a $5$th-degree polynomial that has all four given roots and an additional fifth root: $$ P_5(x) = (x - 3) ( x - 2\omega )( x - 2\bar\omega ) (x^2 - x + 7) $$ $$ = x^5 - 2 x^4 + 6 x^3 - 17 x^2 - 2 x - 84. \tag{2} $$

So the sought-for minimal degree is $5$; the polynomial $(2)$ proves this.

The additional root, $2\bar\omega$, is obtained by complex conjugation of the root $2\omega$ of $P_4(x)$, eq. $(1)$.

Answer 2

The least possible degree of polynomial with real coefficients having these roots is the product of $(x-\alpha)(x-\bar\alpha)$, where $\alpha$ are the given roots, unless $\alpha$ is real, in which case the second factor is omitted.

Since $\bar\omega=\omega^2$ and $1+\omega+\omega^2=0$, the polynomial is $$ (x-2\omega)(x-2\bar\omega)(x-2-3\omega)(x-2-3\bar\omega)(x-3) \\= (x^2+2x+4)(x^2-x+7)(x-3) \\= x^5 - 2 x^4 + 6 x^3 - 17 x^2 - 2 x - 84 $$

Answer 3

Such a polynomial cannot have degree $4$, because we have, using $1+\omega+\omega^2=0$, \begin{align}f(x) & =(x-2\omega)(x-(2+3\omega))(x-(2+3\omega^2))(x-(2-\omega-\omega^2))\\ & =x^4 + 2x^3( - \omega - 2) + 2x^2(4\omega + 5) + x( - 20\omega - 21) + 42\omega, \end{align} which does not have real coefficients. Also, no scalar multiple of it has real coefficients.