I am trying to solve the following problem:
(a) Find the smallest $n$ such that $\mathbb{F}_{3^n}$ contains a primitive $11$th root of unity.
(b) Let $(m+n\sqrt{3})^{24} = a+b\sqrt{3}$, where $m,n,a,b \in \mathbb{Z}$. Show that $b$ is divisible by $7$.
(Texas University at Austin, 2022 Algebra Qual Problem)
My attempt:
For part a, I did the following: Let $w$ be a $11$th primitive root of unity and let $n$ be the smallest integer such that $w \in \mathbb{F}_{3^n}$. Firstly, multiplicative group of finite field is cyclic, and $w$ has order $11$ there. As a result, $w \in \mathbb{F}_{3^n} \iff 11 | (3^n-1)$. A small computation, shows that the smallest $n$ satisfying this is $5$.
For part b, I consider the polynomial $x^2-3$ over $\mathbb{F}_7$, it is irreducible over $\mathbb{F}_7$ since it has no roots in there. Let $w$ be a root of this polynomial and consider the field extension $\mathbb{F}_7(w) \cong \mathbb{F}_{7^2}$. Consider the element $(m+nw)^{24}$ over $\mathbb{F}_7(w)$. We know that $(a+b)^7 = a^7+b^7$ in this field. Thus, we can write $$(m+nw)^{24} = (m^7+n^7w)^3(m+nw)^3 = (a+bw).$$ Then a long computation(I am skipping it here) shows that $b$ is divisible by $7$.
My questions: Is there a relation between part a and part b?
Is there an easier way to solve part b?
Thanks in advance.
I don't see a connection, but there is no need to do long computaiton in Part(b). Consider the map $x\mapsto x^{24}$ from $\mathbb F_{7^2}^{\times}$ to itself. Since the group $\mathbb F_{7^2}^{\times}$ is cyclic of order $48$, we must have the kernel of the morphism consists of exactly $24$ elements (the odd numbers in $\mathbb Z/48\mathbb Z$ if you want). Hence the image of the map has exactly two elements. Since there is only one subgroup of any specified order in a cyclic group, we know the image must be $\{\pm 1\}$. Hence $b=0$, and $a=\pm 1 \mod 7$.
Or even simpler, if $x$ is not $0$ in $\mathbb F_{49}$ then $x^{48}=(x^{24})^2=y^2=1$ and $y=\pm1$.