Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

Thanks for any reading!

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.