# 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.

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

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. :)

• 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? – user346096 Jul 8 '16 at 15:29
• 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. :) – Bill Cook Jul 8 '16 at 17:15