# If $F/L$ is normal and $L/K$ is purely inseparable, then $F/K$ is normal

Let field extensions $K \subset L \subset F$ such that $F/L$ is normal extension, and $L/K$ is purely inseparable extension. Show that $F/K$ is normal extension.

My strategy: let $f(x)\in K[x]$ is irreducible polynomial, and $\alpha \in F$ is solution of $f(x)=0$. So if $\alpha \in L$ then $f(x)=0$ has only solution $\alpha$. Thus $f(x)$ split on $F$,then $F/K$ is normal.

PS: I don't what should I do when $\alpha \notin L$.

• You asked this question 2 days ago, I gave you a hint, and now you've deleted the question? If my hint wasn't sufficient, you could've commented showing what you'd tried, and where you got stuck... Again, let $f_i$ be a family of polynomials in $L[x]$ whose roots generate $K/L$. Since $L/F$ is purely inseparable, how can you bring the $f_i$ down to $F[x]$ without affecting their roots? You might find the Frobenius endomorphism ($x\mapsto x^p$ in characteristic $p$) useful. – Warren Moore Apr 23 '13 at 8:14
• @Muniain: It is extremely impolite to delete a question once someone has put effort into helping you - effectively you are just destroying their work. I have undeleted this original question, and merged your duplicate question here. Please don't do this again. – Zev Chonoles Apr 23 '13 at 8:30

## 2 Answers

Hint. Take a set of polynomials $f_i\in L[x]$ such that $F$ is generated by the roots of these polynomials. Since $L/K$ is purely inseparable, how can you make the $f_i$ into polynomials in $K[x]$ without affecting their roots?

You might find the fact that if $\text{char}(K)=p>0$, then $x\mapsto x^p$ is an endomorphism of $K$ (called the Frobenius endomorphism) useful.

Take a $$K$$ - homomorphism $$T$$ from the $$F$$ to the algebraic closure $$M$$ of $$F$$, we want to show that $$T(F)=F$$,(check this equivalence of normality in morandi's book) take an element "a" in $$L$$, so take then an irreducible polynomial in $$K$$ that solves "a" we say $$f(x)$$, then $$f(T("a))=O.$$ and since $$L/K$$ is Purele Ins. T("a")="a", in conclusion T(b)=b for every b in L, so T is really an L - homomorphism, since F/L is normal T(F)=F.