Find minimal polynomial over $\mathbb{Q}[x]$.

We have a root $9^{1/105}$ of some polynomial over $\mathbb{Q}[x]$. Find minimal irreducible polynomial with this property.

Actually it's easy to see that $x^{105}-9$ is good one. But will it be a minimal(non-zero polynomial with minimal degree) ?

I thought to find some polynomial which divide this one. But have no idea how to construct it. Also I thought about so : if we have $\mu_{min}(x)$ with this root , then in some basis we can represent it coefficients as a diagonal matrix. Also for $x^{105}-9$ and find common basis for diagonal view. So if we have that diagonal elements of the first matrix will be better (there will be less elements) then it will be minimal. But have no idea how to represent such matrix $M_{105\cdot105}$

Any hints?

• What is minimal? Minimal degree? Because then the null polynomial is a trivial solution – Francisco José Letterio Dec 16 '17 at 0:55
• Non-zero polynomial with minimal degree. – openspace Dec 16 '17 at 1:00
• $x^{105} - 9$ should be irreducible in the rationals since it is also irreducible in the reals. Note that the derivative, $105x^4$ is always positive or zero, thus the polynomial is strictly increasing and has only one real root at$9^{1/105}$ so it should be irreducible – Francisco José Letterio Dec 16 '17 at 1:15
• @FranciscoJoséLetterio Irreducible in the reals? $x-9^{1/105}$ is not a factor in $\Bbb{R}[X]$? You're sure that there are no other polynomials of degree less than 104 reducing it to something smaller? – GNUSupporter 8964民主女神 地下教會 Dec 16 '17 at 1:21
• Every polynomial in$\mathbb{R}[x]$ can be factored into the product of degree 2 and 1 polynomials, so any 104-degree polynomial is definitely reducible over $\mathbb{R}$. – Ashwin Trisal Dec 16 '17 at 5:08

It is clear $9^{\frac{1}{105}}$ is a root of the polynomial $x^{105} - 9$. We are going to show this polynomial is irreducible over $\mathbb{Z}$. By Gauss's lemma, this polynomial will be irreducible over $\mathbb{Q}$ and hence $x^{105} - 9$ is the desired minimal polynomial.

Given any prime number $p$ and polynomial $f(x) = \sum_{k=0}^n A_k x^k \in \mathbb{Z}[x]$ such that $A_0A_n \ne 0$.
We can construct something called "Newton polygon of $f$ with respect to $p$" as follows:

For each non-zero $A_k$, write it as $A_k = a_k p^{e_k}$ where $p$ doesn't divide $a_k$. Plot the point $(k, e_k)$ on the cartesian plane. The Newton polygon is the open polygon which forms the lower boundary of the convex hull of these points. Let $Q_0 = (0,e_0), Q_1, \ldots Q_m = (n,e_n) \in \mathbb{N}^2$ be the list of points on lower boundary with integer coordinates, ordered with increasing $x$ coordinates. Let $\mathcal{V}_p(f)$ be the list of vectors $( Q_1 - Q_0, Q_2 - Q_1, \ldots, Q_m - Q_{m-1})$.

There is a theorem by Dumas:

If $f = gh$ is a factorization of polynomial $f$ as a product of polynomials of lower degree, the list $\mathcal{V}_p(f)$ can be obtained by merging the two lists $\mathcal{V}_p(g)$, $\mathcal{V}_p(h)$ and reordering the entries.

As a corollary, if for some $p$, $\mathcal{V}_p(f)$ is a singleton,then $f$ is irreducible over $\mathbb{Z}$. I believe this is referred as Eisenstein-Dumas criterion by some author.

For the polynomial $f(x) = x^{105} - 9$ at hand. If we pick $p = 3$, the lower boundary of the Newton polygon consists of a single segment from $(0,2)$ to $(105,0)$. Since $\gcd(105,2) = 1$, asides from the end points, this segment doesn't intersect $\mathbb{N}^2$. This mean $\mathcal{V}_3(f)$ is a singleton. As a result, $f$ is irreducible over $\mathbb{Z}$.

For more accurate presentation and details, I will recommend Chapter 2, Irreducible polynomials, of the book Polynomials by Victor V. Prasolov. You can find other irreducible criterion there too.

As achille hui points out, by Gauss's "content lemma" irreducibility of the monic integer polynomial $x^{105} - 9$ over the rationals $\mathbb{Q}$ is equivalent to its irreducibility over the integers $\mathbb{Z}$.

A simple condition that often suffices is Eisenstein's criterion:

Suppose we have an integer polynomial $$f(x) \equiv a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0$$ and a prime integer $p$ such that:

• $p$ does not divide $a_n$,
• $p$ does divide each $a_i$ for $i \lt n$, and
• $p^2$ does not divide $a_0$.

Then the polynomial $f(x)$ is irreducible over the integers.

Unfortunately here the only prime $p$ that divides all the coefficients other than the leading coefficient of $x^{105} - 9$ is $3$, and the last part of Eisenstein's criterion fails since $3^2 = 9$ does divide the constant term.

If only we were instead dealing with $x^{105} - 3$, all the parts would work! Thus $x^{105} - 3$ is irreducible. We can use this observation to prove, with a little extra work, that $x^{105} - 9$ is also irreducible.

Let $r = 3^{1/105}$ be the positive real root of $x^{105} - 3 = 0$, and note that $s = 9^{1/105}$ (which we said is the root of $x^{105} - 9 = 0$) satisfies $s = r^2 = 3^{2/105}$.

Now the degree of the irreducible polynomial $x^{105} - 3$ is the same as the dimension of the field extension $\mathbb{Q}(r)$ as a vector space over $\mathbb{Q}$. Readers not familiar with this observation should reflect on the spanning set $\{1,r,r^2,\ldots,r^{104}\}$, and consider that irreducibility of the polynomial $x^{105} - 3$ implies the linear independence of these powers of $r$ less than $105$.

Since $s = r^2$, it is obvious that the field extension $\mathbb{Q}(s)$ is contained in $\mathbb{Q}(r)$. However it is also true that:

$$r = \frac{1}{3} 3^{106/105} = \frac{1}{3} s^{53}$$

This proves $r\in \mathbb{Q}(s)$, showing $\mathbb{Q}(r) = \mathbb{Q}(s)$.

Because these are equal as field extensions, their dimensions as vector spaces over $\mathbb{Q}$ are equal. We deduce that the minimal polynomial of $s = 9^{1/105}$ over the rationals will have degree $105$. Now the monic integer polynomial $x^{105} - 9$ is irreducible, and thus it is the minimal polynomial for $s$ over the rationals.