Finding minimal polynomials $f \in \mathbb{Z}[x]$ with additional restrictions

Question

Find every $f \in \mathbb{Z}[x]$, $f$ minimal and monic, such that

(i) one of its roots is a cubic root of unity,

(ii) at least one of its roots is imaginary pure,

(iii) $\sqrt{2}$ is root,

(iv) its constant coefficient is $-8$.

We know that $G_3 = \{1, \frac{1\pm\sqrt{3}i}{2} \}$ and that $G_3 \setminus \{1 \}$ is a primitive set of roots.

We can pick $(X+\frac{1-\sqrt{3}i}{2})(X+\frac{1+\sqrt{3}i}{2})$ as factors, so that its product is in $\mathbb{Z}[x]$. Combining these with factors $$(X-\sqrt{2})(X+\sqrt{2})(X-2i)(X+2i)$$

gives us a polynomial that verifies the restrictions:

$$f(X)=X^6+X^5+3X^4+2X^3-6X^2-8X-8$$

But is this $f$ minimal? If it is, how can I properly justify it? If it isn't, how can I find another one of lesser degree?

Answer 1

$f$ must be a multiple of the minimal polynomial of the primitive third root of unity, $X^2+X+1$. It also must be a multiple of the minimal polynomial of $\sqrt 2$, $X^2-2$. As these are of course co-prime, $f$ must be a multiple of their product. As we do not already "accidentally" has a purely imaginary root, we need to have the minimal polynomial of such a root, $X^2+c$, as yet another factor. Thus degree $6$ cannot be beaten.

Unless trick answers are allowed: Is $1$ a valid cubic root of unity (though not a primitive one) because $1^3=1$? Is $0$ purely imaginary because $0=0i$? (Okay, while each of these tricks could reduce the degree by $1$, both would conflict with the required constant part)

Answer 2

Yes, it is minimal. If $P(x)$ is such that all those condition holds, then:

• Since $\sqrt2$ is a root of $P(x)$, then so is $-\sqrt2$, and therefore $x^2-2\mid P(x)$.
• Since it has a root of the form $\lambda i$, with $\lambda\in\mathbb R\setminus\{0\}$ (if $\lambda=0$, the constant term of the polynonal would be $0$), $-\lambda i$ is also a root, and therefore $x^2+\lambda^2\mid P(x)$.
• So, $x^4+(\lambda^2-2)x^2-2\lambda^2\mid P(x)$.
• If $\deg P(x)=5$, then $1$ (which is a cubic root of unity) would have to be the fifth root of $P(x)$. But then $P(x)$ would not be monic and have constant term equal to $-8$.