# Splitting field of $x^6+x^3+1$ over $\mathbb{Q}$

I am trying to find the splitting field of $x^6+x^3+1$ over $\mathbb{Q}$.

Finding the roots of the polynomial is easy (substituting $x^3=t$ , finding the two roots of the polynomial in $t$ and then taking a 3-rd root from each one). The roots can be seen here [if there is a more elegant way of finding the roots it will be nice to hear]

Is is true the that the splitting field is $\mathbb{Q}((-1)^\frac{1}{9})$ ? I think so from the way the roots look, but I am unsure.

Also, I am having trouble finding the minimal polynomial of $(-1)^\frac{1}{9}$, it seems that it would be a polynomial of degree 9, but of course the degree can't be more than 6...can someone please help with this ?

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It makes no sense to talk about "minimal polynomial" of a polynomial. "Minimal polynomials" are associated to algebraic elements, not to polynomials. Do you mean, "splitting field", as in the title? – Arturo Magidin Apr 17 '12 at 19:14
@ArturoMagidin yes, I will edit. thanks for pointing out the typo – Belgi Apr 17 '12 at 19:15
Note that your polynomial is the ninth cyclotomic polynomial. – user21725 Apr 17 '12 at 19:17

You've got something wrong: the roots of $t^2+t+1$ are the complex cubic roots of one, not of $-1$: $t^3-1 = (t-1)(t^2+t+1)$, so every root of $t^2+t+1$ satisfies $\alpha^3=1$). That means that you actually want the cubic roots of some of the cubic roots of $1$; that is, you want some ninth roots of $1$ (not of $-1$).

Note that $$(x^6+x^3+1)(x-1)(x^2+x+1) = x^9-1.$$ So the roots of $x^6+x^3+1$ are all ninth roots of $1$. Moreover, those ninth roots should not be equal to $1$, nor be cubic roots of $1$ (the roots of $x^2+x+1$ are the nonreal cubic roots of $1$): since $x^9-1$ is relatively prime to $(x^9-1)' = 9x^8$, the polynomial $x^9-1$ has no repeated roots. So any root of $x^9-1$ is either a root of $x^6+x^3+1$, or a root of $x^2+x+1$, or a root of $x-1$, but it cannot be a root of two of them.

If $\zeta$ is a primitive ninth root of $1$ (e.g., $\zeta = e^{i2\pi/9}$), then $\zeta^k$ is also a ninth root of $1$ for all $k$; it is a cubic root of $1$ if and only if $3|k$, and it is equal to $1$ if and only if $9|k$. So the roots of $x^6+x^3+1$ are precisely $\zeta$, $\zeta^2$, $\zeta^4$, $\zeta^5$, $\zeta^7$, and $\zeta^8$. They are all contained in $\mathbb{Q}(\zeta)$, which is necessarily contained in the splitting field. Thus, the splitting field is $\mathbb{Q}(\zeta)$, where $\zeta$ is any primitive ninth root of $1$.

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I still fail to understand how to see that if $(k,9)\not=1$ then it is not a root. – Belgi Apr 17 '12 at 19:32
@Belgi: The polynomial $x^9-1$ has exactly nine complex roots. Since it is relatively prime with its derivative, $9x^8$, none of the roots are multiple roots. So a power of $\zeta$ has to be a root of $x^9-1$, hence either a root of $x^6+x^3+1$, or of $x^3-1$. If $(k,9)\neq 1$, then it is a root of $x^3-1$; since $x^9-1$ has no repeated roots, it cannot be a root of both $x^3-1$ and $x^6+x^3+1$, so it cannot be a root of $x^6+x^3+1$. – Arturo Magidin Apr 17 '12 at 19:34
OK, can you also help me understand what is the minimal polynomial of the primitive ninth root of unity ? (I'm guessing that it's the one in the question, but why ?) ,+1 – Belgi Apr 17 '12 at 19:47
@Belgi: The minimal polynomial is $x^6+x^3+1$. The ninth roots of unity are the roots of $$x^9-1 = (x^3-1)(x^6+x^3+1)= (x-1)(x^2+x+1)(x^6+x^3+1).$$ By definition, a complex number $\zeta$ is a *primitive $n$th root of unity* if and only if $\zeta^n=1$, and $\zeta^k\neq 1$ for all $k$, $1\leq k\lt n$. So a primitive ninth root of unity must satisfy $\zeta^9=1$, but $\zeta^k\neq 1$ for $1\leq k\lt 9$. If $\zeta^9=1$, then either $\zeta$ is a root of $x^3-1$, or of $x^6+x^3+1$, but not both. So it is a root of $x^6+x^3+1$ if and only if $\zeta^3\neq 1$. (cont) – Arturo Magidin Apr 17 '12 at 19:50
@Belgi: (cont) So the primitive $9$th roots of unity are precisely the roots of $x^6+x^3+1$, and you just need to verify that $x^6+x^3+1$ is irreducible over $\mathbb{Q}$. You can check that $(t+1)^6+(t+1)^3+1$ is Eisenstein at $3$, so it is irreducible, hence $x^6+x^3+1$ is also irreducible (it's the same shift that is used to prove that $x^{p-1}+\cdots+x+1$ is irreducible when $p$ is a prime, so the idea should suggest itself as well). – Arturo Magidin Apr 17 '12 at 19:53

This polynomial is $\Phi_9(x)$, the ninth cyclotomic polynomial whose roots are precisely the primitive ninth roots of unity. A $\mathbb{Q}$-basis for this extension is $\{\zeta_1,\zeta_2,\zeta_4,\zeta_5,\zeta_7,\zeta_8\}$. So you have your splitting field.

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How do you know what powers to take ? – Belgi Apr 17 '12 at 19:20
Thanks Arturo, fixed. They are precisely the powers that relatively prime to 9. Check out Garrett's book for more info, particularly the chapters on cyclotomic polynomials and Galois theory: math.umn.edu/~garrett/m/algebra – user21725 Apr 17 '12 at 19:22
@Belgi: You take those powers that are relatively prime to 9, ie. units mod 9. – Brett Frankel Apr 17 '12 at 20:07