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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.

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  • $\begingroup$ what can you say about $F = K[x] / (f(x))$ ? $\endgroup$ – reuns Jul 6 '16 at 19:34
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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. :)

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  • $\begingroup$ Would there be a constructive way of determining that there is a degree 7 polynomial whose Galois group is the metacyclic group of order 21? $\endgroup$ – user346096 Jul 8 '16 at 15:29
  • $\begingroup$ I guess if you exhibited a degree 7 polynomial with that as its Galois group. :) I don't know a concrete one myself. In general, cooking up examples to try to hit a target Galois group is really really difficult. The general problem "Is there a polynomial f(x) (over the rational numbers) such that its Galois group is G?" (i.e. the inverse Galois problem) is a open problem. Reverse engineering is notoriously difficult. :) $\endgroup$ – Bill Cook Jul 8 '16 at 17:15

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