# Normal Extension Equivalent to Same Degree Irreducible Factors (Hungerford, Exercise V.3.24)

I am doing exercises from Hungerford's Algebra for preparation of my exam.
I would appreciate some help in a part of the proof of the following question:

Exercise V.3.24. An algebraic extension $F$ of $K$ is normal over $K$ if and only if for every irreducible $f\in K[x],f$ factors in $F[x]$ as a product of irreducible factors all of which have the same degree.

The part I have problem with is showing $F$ is normal when all factorization is of the same degree.

Here is my progress so far:
Let $u\in F$ be a root of $f\in K[x]$.
We need to show that $f$ splits in $F$, so that $F$ is normal.
Let $g=f_K^u\in K[x]$ be the minimal polynomial of $u$.
Since $f(u)=0=g(u)$, we may conclude that $g\;|\;f$.

Since $f,g\in K[x]$ and $g\;|\;f$ this shows that $g=f$, since $f$ is irreducible.
Therefore for all irreducible $f\in K[x]$, if it has a root in $F$, then $f$ is necessarily the minimal polynomial of some $u\in F$.

Let deg $f=d$, such that $\lbrace u_1,\dots,u_r\rbrace$ are the roots in the Algebraic closure $\overline K$ of $K$.
(Such that $r\leq d$, assuming some roots may have multiplicity $>1$)
$u_1\in F$ by assumption, but why can we assume that $u_2,\dots,u_r\subset F$?

My guess is for some $\sigma\in Aut_K\overline K$, I can have $\sigma(u_1)=u_j$ that extends to an isomorphism $K(u_1)\cong K(u_j)$, since $\sigma(f)=f$ and $\sigma(u_1)=u_j$ are both roots.
Then using some restriction on $\sigma$, I should be able to show that $\sigma(u_1)\in F$.
But I am not sure how to do it.

Given an irreducible $f\in K[X]$, $f$ has a root in $F$ if and only if $f$ admits a linear (necessarily irreducible) factor in $F[X]$. So if $f$ has a root, the assumption implies that all of the irreducible factors of $f$ in $F[X]$ are linear, i.e., $f$ splits over $F$. This is the definition of normality.
Thanks for the answer. I kept thinking that it may be possible for $u\in F$ to be a root and $g(x)$ a degree 2 factor such that $g(u)=0$ and $g(x)\;|\;f(x)$. – Yong Hao Ng Dec 2 '12 at 9:13