# Proving $f\in K(\beta)[X]$ irreducible iff $g\in K(\alpha)[X]$ irreducible [duplicate]

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?

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

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 $\deg(f)\deg(g)$.
• 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. – quid Dec 6 '14 at 19:00
• @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? – user337254 Oct 17 '18 at 13:56
• @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. – quid Oct 17 '18 at 14:07