I need a check for this:
Let $n=2^p -1$, with $p$ prime and let $\alpha$ a primitive $n$-th root of unity. Show that $\mathbb{F}_{2^p}$ is the splitting field of the minimal poly $M_{\alpha^i}(x)$
I know that $$M_{\alpha^i}(x)=\prod_{j \in C_i}(x-\alpha^j)$$ where $C_i$ is the $i$-th cyclotomic coset. So it's a product of linear factors and since $\alpha$ is a primitive n-th root, it belongs to $\mathbb{F}_{2^t}$, where $t=ord_n(2)$. So I need to find the smallest $t$ s.t. $$2^p-1 | 2^t-1$$ and hence $t=p$. So $\alpha \in \mathbb{F}_{2^p}$. Hence $M_{\alpha^i}(x)$ is the product of linear factors like $(x - \alpha^j)$ and since they belong to $\mathbb{F}_{2^p}$, it is the splitting field, by definition
Is it okay, or am I missing something?
An easier way that I don't want to use would be to see that each cyclotomic coset has the same cardinality, which is exactly $p$. Hence, $M_{\alpha^i}(x)$ is an irreducible poly of degree $p$ over $\mathbb{F}_2$. Therefore, its splitting field is exactly $\mathbb{F}_{2^p}$.