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Let $K\subset L$ a field extension and $\alpha,\beta\in L$ with minimal polynomials $f,g\in K[X]$.

How to show that $f\in K(\beta)[X]$ is irreducible iff $g\in K(\alpha)[X]$ is irreducible?

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marked as duplicate by Marc van Leeuwen, Mark Bennet, Lord_Farin, user7530, apnorton Jan 2 '15 at 15:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Consider the degree of the extension $K\subset K(\alpha, \beta)$.

Note that "$f\in K(\beta)[X]$ is irreducible" and "$g\in K(\alpha)[X]$ is irreducible" are both equivalent to that degree being $\deg(f)\deg(g)$.

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  • $\begingroup$ I see from math.stackexchange.com/a/900889/109299 that the degree of both is equal, but why is that equivalent to the irreducibility of the minimal polynomials? $\endgroup$ – sj134 Dec 6 '14 at 18:53
  • $\begingroup$ I am not sure what you mean by "the degree of both is equal." Note that the degree of the extension $K(\beta) \subset K(\alpha, \beta)$ is the degree of the minimal polynomial of $\alpha$ over $K(\beta)$ this is $f$ if and only if $f$ is irreducible over $K(\beta)$. If it is not it's degree would be smaller. $\endgroup$ – quid Dec 6 '14 at 19:00
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    $\begingroup$ @quid You say that '$[K(\alpha , \beta) : K]$ is the degree of the minimal polynomial of $\alpha$ over $K(\beta)$ this is if and only if f is irreducible over $K(\beta)$. If it is not its degree would be smaller.' Could you better explain what you mean by this? $\endgroup$ – user337254 Oct 17 '18 at 13:56
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    $\begingroup$ @MathWolf no. Let the degree of the fields extension be $d$. Then $d = deg(f)deg(g)$ is equivalent to $f$ being irred. Same for $g$. Maybe also consult the duplicate. $\endgroup$ – quid Oct 17 '18 at 14:07

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