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I try to calculate the degree of splitting field of $(x^{15}-1)(x^{12}-1)$ over $\mathbb{F}_7$:

  • order of $7$ in $(\mathbb{Z}/15\mathbb{Z})^*$ is $8$;
  • order of $7$ in $(\mathbb{Z}/12\mathbb{Z})^*$ is $4$;
  • so degree is $\operatorname{lcm}(8,4)$; I'm not sure about the last deduction.
Shaun
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  • The degree of the splitting field is, indeed, the least common multiple of the degrees of splitting fields of the individual factors. This is because there is a unique extension field of $\Bbb{F}_7$ of a given degree (inside an algebraic closure or some other suitable ambient field). However, please check the calculation of order of $7$ in those two groups. Your data would imply that those two groups are cyclic, but neither of them is. – Jyrki Lahtonen Jun 20 '17 at 11:53
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    Also if $\zeta_n$ is a primitive $n$-th root of unity then $\mathbb{F}p(\zeta_n) = \mathbb{F}{p^k}$ where $k$ is the least integer such that $n | p^k-1$ ie. $k = ord(p \bmod n)$ – reuns Jun 20 '17 at 11:58

1 Answers1

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Assuming $p\nmid m$ the splitting field of the cyclotomic polynomial $\Phi_m(x)$ over $\mathbb{F}_p$ is given by $\mathbb{F}_{p^h}$ where $h$ is the least positive integer ensuring $p^h\equiv 1\pmod{m}$. This follows from the fact that $\mathbb{F}_{p^h}^*$ is a cyclic group with order $p^h-1$ and the roots of $\Phi_m(x)$ are exactly the primitive $m$-th roots of unity.

In your case you need an extension of $\mathbb{F}_7$ containing both the $12$-th and $15$-th primitive roots of unity, hence the answer if given by $\mathbb{F}_{p^h}$ with $h$ being the least positive integer ensuring $12\mid(7^h-1)$ and $15\mid(7^h-1)$, i.e. $(3\cdot 4\cdot 5)\mid(7^h-1)$. It follows that the splitting field of $(x^{12}-1)(x^{15}-1)$ over $\mathbb{F}_7$ is $\color{blue}{\large \mathbb{F}_{7^4}}$.

Here it is the full factorization over $\mathbb{F}_7$:

$$\scriptsize\begin{eqnarray*} (x^{12}-1)(x^{15}-1)&=& (x+1)(x+2)(x+3)^2 (x+4)(x+5)^2 (x+6)^2\\ &&\cdot(x^2+1)(x^2+2)(x^2+4)\\ &&\cdot\left(x^4+x^3+x^2+x+1\right) \left(x^4+2x^3+4x^2+x+2\right). \left(x^4+4x^3+2x^2+x+4\right)\end{eqnarray*} $$

Jack D'Aurizio
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