I recently found the primitive element theorem a little bit unnatural, and hence I was trying to replace the proofs of some theorems using the primitive element theorem, and here is one.
In the book by Jurgen Neukirch, the theorem that states (1) and (2) are equivalent, and uses the primitive element theorem:
\begin{array}{rcl} &L'&\\ &/\backslash\\ L &&K'\\ &\backslash/ \\ &K& \end{array} In this diagram, L' is just the compositum of $L$ and K'.
The following are equivalent:
(1) $L|K$ is unramified.
(2) L'|K' is unramified.
And he uses the primitive element to show one implication, while the other is a direct consequence of the first.
When I was trying to deal with this, the concept of value groups came across me, and I am wondering if the unit group of $L|K$ is contained in that of L'|K'; if this is the case, then since $L|K$ is unramified, the unit group of $L|K$ is just $\{1\}$, and hence L'|K' is unramified as well.
Nevertheless, this is how far I am now, thanks for any clarification or elaboration.
Edit: Here is a sketch of how the primitive element theorem works:
Fisrtly, since $L|K$ is unramified, the residue class field extension is separable, hence there exists a primitive element of the residue class field extension.
Then we lift it to an element of $L|K$, and use the minimal polynomial of that element to deduce that [L':K'] is in fact equal to the degree of the residue class field extension of L'/K', therefore we have proved what was to be proved.
This only serves as a supplement, if anyone needs a full proof, in my opinion, this is contained in every book on the valuation and ramification theory.
Thanks very much.