I came across something related to the degree of a splitting field for a polynomial over a field $K$. Let's suppose $p \in K[x]$ with degree $n$, and $p$ has irreducible factors $f_{1}, \ldots, f_{c}$ with respective degrees $d_{1}, \ldots, d_{c}$.

Ok, I know we can construct the splitting field as a tower of extensions. BUT here is the question: According to Wikipedia, the degree of the splitting field is $\leq n!$, but why $n!$? Is there any way to achieve this bound? I would guess that the degree would be bound by $\prod_{i} d_{i}$, which could never be this big no matter the values of the $d_{i}$. Where is the error in my thinking?

  • 2
    $\begingroup$ I think you can see a construction of an extension of $\mathbb Q$ having Galois group $S_n$ in Milne's notes. $\endgroup$
    – Hoot
    Aug 5, 2015 at 6:34

2 Answers 2


It is certainly achievable. For a concrete example:

Let $f(x) \in \mathbb{Q}[x]$ be an irreducible quintic polynomial, and let $K$ be its splitting field. Furthermore, suppose $f$ has exactly $2$ complex roots. It will be convenient to think about $\operatorname{Gal}(K/\mathbb{Q})$ as a permutation group acting on the roots of $f$. Well, $[K:\mathbb{Q}] = | \operatorname{Gal}(K/\mathbb{Q})|$, and since $f$ is irreducible, then $5$ divides $|\operatorname{Gal}(K/\mathbb{Q})|$. Therefore, Cauchy's theorem tells us the Galois group contains an element of order $5$, which is necessarily a $5$-cycle. Next, complex conjugation is also a permutation in the Galois group, and it is a $2$-cycle. It is a theorem that any $2$-cycle together with any $p$-cycle will generate $S_p$ (for $p$ prime). Hence, $\operatorname{Gal}(K/\mathbb{Q}) \cong S_5$, which has order $5!$, and so finally, $[K:\mathbb{Q}]=5!$.

For an easier example, try computing the order of the Galois group of any irreducible cubic polynomial in $\mathbb{Q}[x]$ with two complex roots. You'll find that the Galois group is isomorphic to $S_3$ with order $3!$, and hence the degree of the splitting field over $\mathbb{Q}$ is $3!$.

Now one might ask whether we can always find extensions of $\mathbb{Q}$ with Galois group $S_n$. Hilbert showed that this is indeed true.

And these are all smaller questions of the more general Inverse Galois Problem: "Does every finite group appear as the Galois group of some Galois extension of $\mathbb{Q}$?", which is not yet known.

  • $\begingroup$ The order of the Galois group is certainly the same as the degree of the extension. See here: en.wikipedia.org/wiki/… $\endgroup$
    – Kaj Hansen
    Aug 5, 2015 at 7:03
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    $\begingroup$ To address your example though, the roots of $x^3 -2$ are $\{\sqrt[3]{2}, w\sqrt[3]{2}, w^2\sqrt[3]{2}\}$, and these can be viewed as the vertices of an equilateral triangle in the complex plane. You get reflections via complex conjugation, and rotations via $\sqrt[3]{2} \mapsto w \sqrt[3]{2}$. These two automorphisms generate all possible positions of the vertices, and so the order of the Galois group is $S_3$. Likewise, the splitting field is arrived at by $\mathbb{Q} \subset \mathbb{Q}[\sqrt[3]{2}] \subset \mathbb{Q}[w, \sqrt[3]{2}]$, and that is a degree $6$ extension. $\endgroup$
    – Kaj Hansen
    Aug 5, 2015 at 7:10
  • $\begingroup$ Yep, that was it. $\mathbb{Q}[\sqrt[p]{2}]$ isn't the splitting field. We're on the same page so far. $\endgroup$
    – Kaj Hansen
    Aug 5, 2015 at 7:12
  • $\begingroup$ Yep! that's right. $\endgroup$
    – Kaj Hansen
    Aug 5, 2015 at 7:19

Let $L_i$ be a splitting field of $f_i$ over $L_{i-1}$, whence $L_0$ is defined to be $K$ and $L_c=L$. Then $[L:K]=[L_c:L_{c-1}]\cdots [L_2:L_1][L_1:K]$. For each term in the product, $[L_j:L_{j-1}]\le [L_j:K]$, the degree of any splitting field of $f_j$ over $K$.

If $f_j$ has degree $1$, the result is obvious, so assume $f_j$ has degree $\gt 1$. By induction, consider $K[x]/(f_j(x))$ which is a field, then we have an extension $K_1/K$ such that $f_j$ has a root $\alpha$ in $K_1$ and thus $f_j(x)=(x-\alpha)g_j(x)$ in $K_1[x]$, which may be taken as $K(\alpha)[x]$. The degree of $g_j$ is $d_j-1$ and so the degree of any splitting field of $g_j$ over $K(\alpha)$ is $\le(d_j-1)!$ by the induction hypothesis. $f_j$ and the minimal polynomial of $\alpha$ over $K$ differ only by the leading coefficient, and hence have the same degree which is $[K(\alpha):K]=d_j$. Hence, $[L_j:K]=[L_j:K(\alpha)][K(\alpha):K]\le d_j!$

Therefore, $[L:K]\le \prod _{i=1}^c d_i!\le (\sum _{i=1}^c d_i)!=n!$. Note that the equality in place of the second inequality may occur when $p$ is itself irreducible, in which case we only have $d_1$ which is exactly $n$.


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