Determine possible degree of monic irreducible polynomial given degree of Galois extension. 
Suppose $L/K$ is a separable normal field extension with $[L : K] = 21$.
(i) What integers n are possibly equal to the degree of a monic irreducible polynomial $f(x) ∈ K[x]$ for which $L/K$ is a splitting field of $f(x)$?
(ii) If the Galois group of $L/K$ is known to be an abelian group, what integers $n$ are possibly equal to the degree of a monic irreducible polynomial $f(x) \in K[x]$ for which $L/K$ is a splitting field of $f(x)$? Justify your answer.

Let $G$ be the Galois group of $L/K$, then $|G|=21$ implies that either $G\cong\mathbb{Z}_{21}$ or $G\cong\mathbb{Z}_7\times_\tau\mathbb{Z}_3,$ where the latter denotes the metacyclic group of order 21. But I don't know how this relates to possible orders of monic irreducible polynomials.
 A: Let $L$ split $f(x) \in K[x]$ where $f(x)$ is monic irreducible over $K$. Take any root of $f(x)$ in $L$, say $f(\alpha)=0$ where $\alpha \in L$. This means that $K \subsetneq K[\alpha] \subseteq L$.
$[K[\alpha]:K]=\deg(f(x))$. So $\deg(f(x))$ must divide $[L:K]=21$. This leaves us the options: $3$, $7$, and $21$.
However, $3$ is impossible. Splitting fields of irreducible cubics have degree either $3$ or $|S_3|=3!=6$ (can't get to 21). 
$7$ is possible. Check out the Wikipedia page on the Septic equation. There are degree 7 polynomials whose Galois group is the metacyclic group of order 21.
$21$ is also possible. For example, let $K$ be the rationals with all 21th roots of unity attached. Then split $x^{21}-2$ (irreducible by Eisenstein and splits after adjoining a single root so the degree of its splitting field is 21).
What if the group is abelian? Then $7$ is ruled out. Here's why: If the group is abelian it's $\mathbb{Z}_{21}$ so it has an element of order 21. 
Now the Galois group of a polynomial of degree $n$ can be embedded in $S_n$. Notice that $S_{10}$ is the smallest symmetric group having an element of order 21 (use disjoint 3- and 7-cycles). Thus this Galois group cannot be embedded in $S_7$ so $7$'s out. :)
