Is there a relationship between these polynomial concepts? I'm currently doing a bit of reading on abstract algebra (more specifically Polynomial Theory), and noticed something that may have some sort of significance perhaps?
The section I'm reading on at the moment is on cyclotomic polynomials and their primitives. The example used is $\zeta_{5}$. The minimal extension of the rationals in order to contain all the roots of $t^4+t^3+t^2+t+1=0$ (normal extension?) is $\mathbb{Q}(\zeta_5)$, which has basis being $\zeta_5^k$ where $k=1,2,3,4$.
Algebraically, these bases are indistinguishable since each element generates the entire basis. So in some sense if I 'swap' the elements around, we can't 'tell' which one is which because they all generate the same subfield (being the basis in itself). 
Now earlier in my reading, there was some time spent on Vieta's Formulas and how some expressions can be swapped around without 'detection' whereas others will detect any changes immediately.
For example $\alpha+\beta+\gamma$ is symmetric so by permuting the elements around, there is no difference. However, $\alpha + \beta^2 + \gamma^3$ is not symmetric so we can do no such thing.
These two concepts seem quite similar to each other. Is there a relationship between them somewhere down the track?
 A: It sounds like you're asking about the beginning of what's called Galois Theory. In your example, you've observed that, informally, you can replace $\zeta := \zeta_5$ with $\zeta^k$ for any $k \in \{1, 2, 3, 4\}$ (which induces other replacements, namely $\zeta^2$ with $\zeta^{2k}$, $\zeta^3$ with $\zeta^{3k}$, and $\zeta^4$ with $\zeta^{4k}$) and that the resulting algebra of the field $\Bbb Q(\zeta)$ works just as well.
Put a little more precisely: The ring homomorphism $\phi_k: \Bbb Q(\zeta) \to \Bbb Q(\zeta)$, $k \in \{1, 2, 3, 4\}$ characterized by $$\phi_k(\zeta) = \zeta^k$$ (1) is a field isomorphism and (2) restricts to the identity on $\Bbb Q$. Now, by construction the set of such isomorphisms is a group under composition; we call it the Galois group and denote it $\text{Gal}(\Bbb Q(\zeta) / \Bbb Q)$. In our case, the maps $\phi_k$ exhaust all such isomorphisms and so $\text{Gal}(\Bbb Q(\zeta) / \Bbb Q)$ has order $4$ (of course, the identity element is $\phi_1$); the element $\phi_2$ has order $4$ under composition, so
$$\text{Gal}(\Bbb Q(\zeta) / \Bbb Q) \cong \Bbb Z_4.$$
(This alone tells us something we already knew, namely that the given polynomial is special: In a way that can be made precise, a typical quartic polynomial over $\Bbb Q$ has Galois group isomorphic to the much larger group $S_4$.) Note that this implies that not all linear maps determined by permutations of $\zeta, \zeta^2, \zeta^3, \zeta^4$ are such field isomorphisms, only the four permutations described in the first paragraph above are.
On the other hand, we can play this game just as well for any polynomial $p$ over any base field $\Bbb F$ (in our example, that field is $\Bbb Q$): We can ask which permutations of roots determine isomorphisms of the splitting field of $p$ (informally, the smallest field containing $\Bbb F$ that contains all of the roots of $p$) fixing the base field $\Bbb F$.
