Biggest splitting field degree given a polynomial of degree n

Question

It's a well know fact that, given $f(x) \in \mathbb{K}[x]$ with $\deg(f) = n$, and being $\mathbb{L}$ its splitting field, we have that $[\mathbb{L}:\mathbb{K}] \leq n!$

What I'd like to know are some examples for which the equality holds. As far as I know:

• $n=2$ is trivial

• $n=3$ we can take $f(x) = x^3-2$ (am I correct?)

What about $n > 3$? Can someone give me some examples please? Is there a way to construct polynomials of any degree which satisfy what I'm asking for?

Thank you!

Answer 1

The splitting field of the generic polynomial of degree $n$: $\,x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n$ has degree $\,n!\,$ over the field $K(a_1,\dots,a_n)$. Its Galois group is $S_n$.

Using this fact, Galois proved the impossibility to solve the general equation of degree $n$ as soon as $n\ge 5$.

Numerical examples

• The splitting field of $x^3-2$ is $K=\mathbf Q(\sqrt[3]2,\omega)$, where $\omega=\mathrm e^{\frac{2\mathrm i\pi}3}$. As $[K:\mathbf Q]=6$, $\,\lvert\operatorname{Gal}(K/\mathbf Q)\rvert=6$.
• The polynomial $x^5-6x+3$ is irreducible, by Eisenstein criterion, hence its Galois group has order divisible by $5$. By Cauchy's theorem is contains an element of order $5$. As we're in $S_5$, it as actually a $5$-cycle. On another hand, elementary analysis shows it has three real toots and two complex roots. These are conjugates, so that the Galois group contains a transposition.

One shows that renumbering the roots adequately, one may suppose the Galois group contains the cycle $(1\,2\,3\,4\,5)$ and the transposition $(1\,2)$. But it is a classical result on symmetric groups that these two permutations generate $S_5$. Hence the Galois group is $S_5$.

• For quartic polynomials, a more technical example: $x^4-4x^2+x+1$ is irreducible, of discriminant $19\cdot103$. Its Ferrari resolvent is the polynomial $y^3+4y^2-4y-17$ and it is irreducible over $\mathbf Q$. A theorem states that is the discriminant is not a square and the resolvent is irreducible, the Galois group is $S_4$.

By definition the Ferrari resolvent of a quartic equation with (complex) roots $x_1, x_2, x_3,x_4$ is the polynomial in $y$: $$\bigl(y-(x_1x_2+x_3x_4)\bigr)\bigl(y-(x_1x_3+x_2x_4)\bigr)\bigl(y-(x_1x_4+x_2x_3)\bigr).$$

Answer 2

I guess, there are two separate ideas:

1. If you have $K\subset L$ fields and take $a\in L$ such that $\exists f\in K[X]$ such that $f(a)=0$. Then if you construct the minimal polynomial of $a$ which we will denote by $irr_a$, i.e. a polynomial that is irreducible in $K[X]$ and has $a$ a root, you have that $[K(a):K]=\deg irr_a$ where a basis in $K(a)$ as a vector space over $K$ is given by $\{1,a,\dots a^{\deg(irr_a)}\}$. All other polynomials $g$ in $K[X]$ that have $a$ as a root have $\deg(g)\geq \deg(irr_a)$. For example, you shall observe that: $[\mathbb Q(\sqrt[3] 2):\mathbb Q]=3$ where the minimal polynomial of $\sqrt[3] 2$ is $X^3-2$.
2. If you have $K\subset L$ fields and take $f\in K[X]$, then $[K(f):K]>\deg(f)$. If you take $f=x^3-2\in\mathbb Q(X)$, $\mathbb Q(f)=\mathbb Q(\sqrt [3]2, a, b)=\mathbb Q(\sqrt [3]2, a)$ where $a,b$ are the other two complex roots of $f$ and $[\mathbb Q(f):\mathbb Q]=6>3$. $[K(f):K]=\deg(f)$, if the extension is Galois, i.e. if it is normal (all the roots of $f$ are in $K$) and separable ($f$ has no multiple roots).