Degree of normal closure

Question

how could one prove that if the degree of a separable field extension $\ell/k$ is at most $a$, then the degree of its normal closure $N/k$ is at most $a!$?

Answer 1

Since $l/k$ is separable, we can write $l=k(a)$ for some separable $a\in l$. Let $p(x)$ be the minimal polynomial $\in k[X]$ over $k$ of $a$. Also, $p(x)$ is (of course) separable.

Let $\operatorname{deg}(p)=m$ and let $a_1,a_2,\dots,a_m$ (where $a_1=a$) be all the roots of $p$.

We have $m\leq n$ ($n$ is the upper bound on the degree).

We consider the set of ascending fields: $$k\leq k(a_1)\leq k(a_1,a_2)\leq k(a_1,a_2,a_3)\leq \dots\leq k(a_1,a_2,a_3,\dots,a_m)$$

Clearly (by assumption) $[k(a_1):k]=[l:k]=m$.

And for every $1<j\leq m$, the polynomial $\frac{p(x)}{(x-a_1)(x-a_2)\dots (x-a_{j-1})}$ belongs to $k(a_1,a_2,\dots,a_{j-1})[X]$ (due to the division algorithm). But this polynomial has $a_j$ as a root. This means its degree is at least that of the minimal polynomial of $a_j$ over $k(a_1,a_2,\dots,a_{j-1})$.

So $$[k(a_1,a_2,\dots,a_{j-1},a_j):k(a_1,a_2,\dots,a_{j-1})]=[k(a_1,a_2,\dots,a_{j-1})(a_j):k(a_1,a_2,\dots,a_{j-1})]$$$$\leq \operatorname{deg}\left(\frac{p(x)}{(x-a_1)(x-a_2)\dots (x-a_{j-1})}\right)=(m-j+1)$$

Since this result holds for arbitrary $1<j\leq m$, we get, by the tower theorem: $$[k(a_1,a_2,\dots,a_m):k]=[k(a_1):k]\prod_{j=2}^m[k(a_1,a_2,\dots,a_{j-1},a_j):k(a_1,a_2,\dots,a_{j-1})]$$$$\leq m\prod_{j=2}^m (m-j+1)=m \prod_{r=1}^{m-1} r=m!$$

Then, since $m\leq n$, we get $[k(a_1,a_2,a_3,\dots,a_m):k]\leq n!$

Lastly, since $p$ is separable and $k(a_1,a_2,a_3,\dots,a_m)$ is the splitting field of $p$ over $k$, $k(a_1,a_2,a_3,\dots,a_m)/k$ is normal.

So $N\leq k(a_1,a_2,a_3,\dots,a_m)$ giving us $$[N:k]\leq [k(a_1,a_2,a_3,\dots,a_m):k]\leq n!$$