Let $E_1$ and $E_2$ be two extension of a field $F$ and $[E_2:F]=2$ such that $E_1\cap E_2=F$.Let $E_2=F(\alpha)$ then $[E_1(\alpha):E_1]?$
MY TRY:I think it will $2$ ,because $E_1\cap E_2=F\subset E_1 and \,E_2$ and $E_2=F(\alpha)$.
Let $E_1$ and $E_2$ be two extension of a field $F$ and $[E_2:F]=2$ such that $E_1\cap E_2=F$.Let $E_2=F(\alpha)$ then $[E_1(\alpha):E_1]?$
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field-theory
finite-fields
extension-field
1 Answers
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Yes, it is $ 2 $, because irreducibility is equivalent to having no roots for quadratic polynomials. Since the minimal polynomial of $ \alpha $ has no root in $ E_1 $, it follows that $ E_1(\alpha) \neq E_1 $, and thus $ [E_1(\alpha) : E_1] = 2 $.