Assume $f(x) \in F[x]$ has degree $n$ and there exists a field extension that splits $f(x)$, that is

$$f(x)=c_0(x-c_1)\dots (x-c_n) $$

Prove that the degree of the field extension of $F$ is at most $n!$

Assume $f(x) \in F[x]$ has degree n.

Step 1

Worst case scenario $f(x)$ is irreducible in $F[x]$. We generate $F[x]/(f(x))=K_1$ where $c_1=[x]$ is a root. So in $K_1[x]$ $$f(x)=(x-c_1)g_1(x) $$ Step 2 Worst case scenario again $g_1(x)$ is Irreducible. We generate $K_2=K_1/(g_1(x))$ so their is a root $c_2$ in K_2 where $$f(x)=(x-c_1)*(x-c_2)g_2(x) $$ $\vdots$

After a maximum of $n$ steps

$\exists K_n[x]$ where $$f(x)=c_0(x-c_1)\dots (x-c_n) $$

Note that the important thing is that $$F[x] \subset K_1=F[x]/(f(x)) \subset K_2= (F[x]/(f(x)))/(g_1(x)) \subset \dots \subset K_n $$ We want to find the maximum degree of $K_n$ over $F$, that is, $[K_n:F]$. Using the tower law, $$[K_n:F]=[K_n:K_{n-1}]\dots [K_i:K_{i-1}]\dots [K_1:F]=1*2*3* \dots* (n-1)*n $$ Since $[K_i:K_{i-1}]=deg (g_i(x))$ that is the degree of irreducible that was used to generate the new extension field. Each extension field is $1$ degree lower so that is why it is a factorial $n!$ at the most

Questions I assumed it is irreducible I believe it should say doesn't have a root but the field generated would not be a field. Did not stated what happens when it was reducible. Also is a chunk of it a proof or close to a prove that $f(x)$ can be split in some extension field. I know induction is strongly preferred.


1 Answer 1


If $f$ is reducible, say $f(x)=h(x)g(x)$ where $h(x)$ is irreducible, then you can use a similar argument when adding a root of $h$ to the field. Note that it is just fine even if $f(x)$ has a root (in this case, $h(x)=x-a$ and adding $a$ simply wont extend the field).

If you could show how to add a root to a field, without assuming a containing splitting field, then your argument could extend to show the existence of a splitting field (it is not unique by the way!). Hint: consider a quotient of $K[x]$.


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