# Degree of a splitting field

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?

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

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.

• The order of the Galois group is certainly the same as the degree of the extension. See here: en.wikipedia.org/wiki/… Aug 5, 2015 at 7:03
• 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. Aug 5, 2015 at 7:10
• Yep, that was it. $\mathbb{Q}[\sqrt[p]{2}]$ isn't the splitting field. We're on the same page so far. Aug 5, 2015 at 7:12
• Yep! that's right. 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 $$n$$.