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Let $E/K$ be a field extension and let $\alpha \in E$ be algebraic over $K$. Let $f \in K[x]$.

My question is, if I have the following:

  • $f(\alpha)=0$
  • $f$ is monic
  • $\deg(f) = [E : K]$

Can I then conclude that

  • $f$ is the minimal polynomial for $\alpha$ over $K$?

Then I don't have to prove that $f$ is irreducible over $K$.

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1 Answer 1

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Take any element $a$ in $K$ and $f(x)=(x-a)^n, n=[E:F]$ $f$ is not the minimal polynomial of $a$.

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  • $\begingroup$ Do you have a counter example if $a \in E$ is algebraic over $K$? $\endgroup$
    – Johannes Jensen
    Dec 12, 2019 at 0:41
  • $\begingroup$ Take any $a$ with minimal polynomial $P$ of degree $l<n=[E:K]$, and consider $PX^{n-l}$. $\endgroup$
    – Tsemo Aristide
    Dec 12, 2019 at 0:46

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