One element is given seperable and another purely inseperable then show that ...

Question

Kindly help with the following question asked in my mid term exam which has been concluded.

If u $\in F$ is separable over K and c $\in F$ is purely inseparable over K, then K(u,c) = K( u + c).

I am unable to get any idea on how this question will be solved. The irreducibility polynomial of u will be given by $x-u \in K[x]$ and irreducibility polynomial of c will be given by $(x-c)^m \in F[x]$ for some m >1 , $m\in \mathbb{Z}$.

But I am unable to use it to prove K(u,c). Let I choose any polynomial in K(u+v) , so it will be given by $ \frac{a_0 + a_1 (u+c) + a_2 (u+c)^2 +....}{ b_0 + b_1 (u+v) + b_2 (u+v)^2 ...}$ ,Now It is triavial to see that the polynomial also lies in K(u,v) .

If I choose a polynomial in K(u,v) then I struggle to prove that it lies as K(u,v). I think here we have to use seperability but unable to use it.

So, Do u mind helping me? Also , I Hope my rest of argument is fine.

Answer 1

We may assume $\operatorname{char}K=p>0$ (otherwise $c\in K$ and the statement is trivial). You already proved the inclusion $K(u+v)\subseteq K(u,c)$, for the other inclusion we only need to prove $u,c\in K(u+v)$. There is some $r\in\Bbb N$ with $c^{p^r}=d\in K$. Hence we have $(u+c)^{p^r}=u^{p^r}+d$, so $u^{p^r}\in K(u+c)$. Show

1. $K(u)$ is both separable and purely inseparable over $K(u^{p^r})$
2. $K(u)=K(u^{p^r})\subseteq K(u+c)$.
3. $u\in K(u+c)$
4. $c\in K(u+c)$

Conclude $K(u,c)\subseteq K(u+v)$ and then equality.