Regarding the proof of Primitve element theorem (Theorem 3.3.2) in Murty's Problems in Algebraic Number Theory

I am reading the following proof

I understand that we can choose such $$\lambda$$ so that $$\beta$$ is only common root of $$\phi$$ and $$g(x)$$ but my question is that why the $$gcd$$ ($$\phi$$(x),$$g(x)$$) needs to be in $$L[x]$$? It is not explicitly mentioned that where the $$gcd$$ is being taken and does selection of such a $$\lambda$$ not require to consider all possible conjugates of $$\alpha$$ and $$\beta$$ ? Assuming this one i can follow the remaining proof but could someone please explain why it is in $$L[x]$$.

• Please include all relevant information in the post itself, rather than force potential volunteers to go to some other site and figure out what's relevant. Commented Jul 10 at 7:30

$$g$$ is the minimal polynomial of $$\beta$$ in $$\mathbb{Q}[x]$$ so it is in $$L[x]$$. If we take $$h = gcd(g,\phi)$$ in $$L[x]$$, its roots will be the common roots of $$g$$ and $$\phi$$, say in $$\mathbb{C}$$. So gcd is linear if they have only one common root.
• But why does this $gcd$ necessarily lies in $L[x]$ ? Commented Jul 10 at 17:55
• gcd trivially lies in $L[x]$ since we are taking gcd of two elements in $L[x]$, which is a PID, in there. Here the importance is that an irreducible polynomial cannot have multiple roots like $(x-\beta)^2$ since we are working over rational numbers, a perfect field. This demonstrates why primitive element theorem may fails in non-seperable extensions. Commented Jul 11 at 6:32