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I am reading the following proof

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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 $L[x]$.

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    $\begingroup$ 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. $\endgroup$ Commented Jul 10 at 7:30

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$g$ 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.

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  • $\begingroup$ But why does this $gcd$ necessarily lies in $L[x]$ ? $\endgroup$ Commented Jul 10 at 17:55
  • $\begingroup$ 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. $\endgroup$ Commented Jul 11 at 6:32

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