Let $L/K$ be a normal extension and a irreducible polynomial $f(X) \in K[X]$. Prove that, if $f$ is reducible over $L$ then $f$ is factored into product of irreducible factors with same degree. Furthermore, if $f$ has roots in $L$ then $f$ splits over $L$.

Help me a hint.

Thank for any insight.


1 Answer 1


First, note that the first part of the question implies the second part: i.e. suppose $f$ has roots in $L$. Then, $f$ factors into product of irreducible factors with same degree, one of which has degree 1 (since $f$ has a root in $L$). Therefore, $f$ splits over $L$.

For the first part, take $\bar{K}$ to be the algebraic closure of $K$. And consider $\sigma\in \textrm{Gal}(\bar{K},K)$ where $\sigma$ permutes the roots of $f$. Now, suppose $f$ factors as the product of $f_1\cdots f_n$ over $L$. Can you use the fact that $L/K$ is normal to show $\sigma|_L$ sends each $f_i$ to some $f_j$? (we are not ruling out the case $i=j$)

After that, you can choose appropriate $\sigma$ to show that you can actually send $f_1$ to $f_j$ for any $j$.

I hope this helps!

  • $\begingroup$ Thank you for your support $\endgroup$
    – user109584
    Nov 19, 2013 at 4:44
  • 1
    $\begingroup$ But why $\sigma|_L$ sends each $f_i$ to some $f_j$? $\endgroup$
    – user109584
    Nov 19, 2013 at 4:47
  • 2
    $\begingroup$ Can you explain more precisely to me? $\endgroup$
    – user109584
    Nov 19, 2013 at 4:48
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    $\begingroup$ Since $L/K$ is normal, $\sigma$ is an automorphism of $L$. Let's denote $\sigma(f_i)=f_\sigma$ and suppose $f_\sigma\ne f_j$ for all $j$. Note that the roots of $f_\sigma$ are also roots of $f$. But this means that $f_\sigma$ divides $f$ over $L$. Moreover, $f_\sigma$ and $f$ have the same degree. $\endgroup$ Nov 19, 2013 at 7:24

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