The original question that I set out to answer is as follows:
Let $\mathbb{Q}$ be the field of rational numbers, let $p \in \mathbb{Q}[x]$ be a monic, irreducible polynomial with $n$ distinct roots $r_1, r_2, ..., r_n \in \mathbb{\bar{Q}}$ in the algebraic closure of $\mathbb{Q}$. What are the possible values of $deg(r_j, \mathbb{Q}(r_i))$? Note: $deg(\alpha, F)$ is defined as the degree of the unique, monic, irreducible polynomial of minimal degree $f(x) \in F[x]$ which satisfies $f(\alpha)=0$.
In working on this problem, I've deduced several things:
First, we know that $deg(r_j, \mathbb{Q}(r_i)) \leq deg(r_j, \mathbb{Q}) - 1$. This is because when working in $\mathbb{Q}(r_i)$, we can reduce $p$ as follows $p' = \frac{p}{x-r_i}$ and $p' \in \mathbb{Q}(r_i)$. In particular, we have equality if and only if $r_i$ is the only root of $p$ which is in $\mathbb{Q}(r_i)$. Clearly, if $r_j \in \mathbb{Q}(r_i), deg(r_j, \mathbb{Q}(r_i)) = 1$ as $p' = x - r_j \in \mathbb{Q}(r_i)$ is a degree one polynomial with $r_j$ as a root. Now, if $r_j \notin \mathbb{Q}(r_i)$, we can go further and write $deg(r_j, \mathbb{Q}(r_i)) =deg(r_j, \mathbb{Q}) - m$ where $m$ is the number of roots of $p$ in $\mathbb{Q}(r_i)$. This uses the same trick of factorization as above. Additionally, $r_j \in \mathbb{Q}(r_i) \rightarrow \mathbb{Q}(r_j) = \mathbb{Q}(r_i)$
So the question becomes (with the same set up): What are the possible numbers of roots which have the same simple field extension? I.E. What values can $m$ take for a given $n$ in the above equation?
Now, let's look at a few examples:
If $n=2$, we have $deg(r_j, \mathbb{Q}(r_i)) \leq deg(r_j, \mathbb{Q}) - 1$ implies $deg(r_j, \mathbb{Q}(r_i)) = 1$. But this effectively means $r_j \in \mathbb{Q}(r_i)$ as shown above so in terms of our reworded question, we have $m=2$ – both of the roots are in either simple extension field.
If $n=3$, we can clearly get $m=1$ with a polynomial such as $p(x) = x^3-2$. The three zeros $\sqrt[3]{2}, \sqrt[3]{2}e^{2\pi i/3}, \sqrt[3]{2}e^{4\pi i/3}$ all form distinct simple extensions. Can we get $m=3$? I'm not sure. I've yet to come up with a constructive example nor a proof as to why it is impossible.
If $n=4$, we can get $m=2$ with $p(x)= x^4-2$. This is because $\mathbb{Q}(\sqrt[4]{2})=\mathbb{Q}(-\sqrt[4]{2})\neq\mathbb{Q}(\sqrt[4]{2}i)=\mathbb{Q}(-\sqrt[4]{2}i)$. We can also get $m=4$ with $p(x)=x^4+x^3+x^2+x+1$. Similarly, this is because $\mathbb{Q}(\sqrt[4]{2}e^{2\pi i /5})=\mathbb{Q}(\sqrt[4]{2}e^{4\pi i /5})=\mathbb{Q}(\sqrt[4]{2}e^{6\pi i /5})=\mathbb{Q}(\sqrt[4]{2}e^{8\pi i /5})$ as the four zeros of this polynomial are simply powers of each other. But can we get $m=1$. Again, I'm not sure. I've yet to construct it but have no proof as to why it shouldn't be possible.
In fact, if $n$ is even, we can always get $m=n$ at least by using $p(x)=x^n+x^{n-1}+...+x+1$. Then the $(n+1)^{th}$ roots of unity (excluding $-1$) all have the same simple extension.
From here, I tried defining an equivalence relation $\sim$ as follows. Suppose $a$ and $b$ are zeros of $p$. Then $a\sim b$ if and only if $\mathbb{Q}(a)=\mathbb{Q}(b)$. Since this is an equivalence relation, we can partition the roots of $p$ into equivalence classes notated by representatives. E.G. $[a] = \{x \in \mathbb{\bar{Q}} | p(x) = 0, a \sim x\}$. Now, I believe we can apply an automorphism from the galois group $\phi \in Gal(\mathbb{Q}(p)/\mathbb{Q})$ where $\mathbb{Q}(p)$ is the splitting field of $p$. $\phi (a) \sim \phi (b)$ if and only if $a \sim b$. This provides an isomorphism between $[r_i]$ and $[r_{i}']$ if $\phi (r_i) = r_i'$ because $\phi [r_i] = \phi [r_i']$. This should tell us that all of these equivalence classes have the same number of elements. I.E. $|[r_1]| = |[r_2]| = ... = |[r_n]|$. In particular, this allows us to deduce that $|[r_i]|$ divides $n$ and thus that $m|n$.
This is the major thing that I'm working on proving at the moment: $m|n$. I am neither sure if it is a necessary or sufficient condition for $m$ to be a possibility. If it was a sufficient condition, it would imply that we can have $n=4$ with $m=1$ and $n=3$ with $m=3$ which I have yet to find. If it was necessary, it would imply that we can only have $m=1,n$ if $n$ is a prime number which I don't really have any intuition on.
I've tried a few times to incorporate galois theory and automorphisms of the roots more than I did above but couldn't find a simple way of tying it back to the degree of polynomials.
Any help is appreciated, whether it be further ideas or comments on mistakes made so far. Thanks!