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Problem: Let $L|_k$ be an algebraic field extension . $f(x)\in L[x]$ a polynomial then $\exists g(x)\in L[x]$ such that $f(x)g(x)\in k[x]$

My approach: It is enough to prove for $f(x)$ irreducible in $L[x]$. If $f(x)$ is irreducible in $L[x]$ there is an extension $L(\alpha)$ over $L$ such that minimal polynomial of $\alpha$ over $L $ is $f(x)$.

Again $\alpha$ being algebraic over $k$ there is $h(x)\in k[x]\subseteq L[x]$ such that $h(\alpha)=0$ hence $f(x)$ divides $h(x)$ in $L[x]$. Hence it is done.

Now I am not sure if this argument is correct. Help me if I am wrong. And other approaches are also welcome

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    $\begingroup$ This looks good to me! $\endgroup$
    – Stahl
    Jul 28, 2020 at 8:46
  • $\begingroup$ Thanks @Stahl. Other approaches are also welcome $\endgroup$ Jul 28, 2020 at 8:50

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