the degree of the product of the minimal polynomials is same as $[E:K]$

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Let $E = K(a,b). $Suppose gcd of degree of minimal polynomial of $a$ and $b$ over $K$ is $1$, then does it imply that the degree of the product of the minimal polynomials is same as $[E:K]$?

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Yes. Let $m=[K(a):K]$ and $n=[K(b):K]$ be the degrees of the minimal polynomials. Then $[E:K]$ is divisible by $m$ and $n$ by the tower law. If $m$ and $n$ are co-prime, we get that $[E:K]$ is divisible $mn$, hence $mn \leq [E:K]$. The converse inequality holds trivially without any further assumptions.