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?
 A: 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.
A: 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 $n$.
