I need some help with the following task:
Let $L/K$ be a finite field extension.
a) $[L:K]=41$, show that there exists an $a\in L$ such that $L=K[a]$.
b) $[L:K]=14$ and $f(x)=0$ with $deg(f(x))=3$. Assume that $f(x)=0$ has a solution in $L$ and show that $f(x)=0$ has a solution in $K$.
I'm not sure what to do,do I have to find a polynomial with degree 41? Thanks in advance for any help.
Edit: I think a) is supposed to be solved like this
$K \subset K[a] \subset L$, $a\in L$. Since $a\notin K$ it's $K \neq K[a]$, so $[K[a]:K]>1$ and since $[K[a]: K] | [L:K]=41$, $[K[a]: K] \neq 1$ implies that $[K[a]:K]=41$, so $[L:K[a]]=1$ has to hold and $L=K[a]$.