Degree of splitting field of $(x^{15}-1)(x^{12}-1)$ over $\mathbb{F}_7$. 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.

 A: 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 $\mathbb{F}_7$:
$$\scriptsize\begin{eqnarray*} (x^{12}-1)(x^{15}-1)&=& (x+1)(x+2)(x+3)^2 (x+4)(x+5)^2 (x+6)^2\\ &&\cdot(x^2+1)(x^2+2)(x^2+4)\\
&&\cdot\left(x^4+x^3+x^2+x+1\right) \left(x^4+2x^3+4x^2+x+2\right). \left(x^4+4x^3+2x^2+x+4\right)\end{eqnarray*} $$
