Prove that every monic polynomial $f$ irreducible in $\mathbb {F}_q[x]$ has $\deg(f)$ roots in a finite extension $ E $ such that $\deg(f) \mid s$ and $[E: \mathbb{F}_q] = s$.

The book I'm reading uses this fact without giving a proof; why is this true?


Since finite fields are perfect, the polynomial $f$ is separable, so it has $\deg(f)$ distinct roots in its splitting field. Actually, in this case the splitting field is $\mathbb{F}_q(a)$, where $a$ is a root of $f$, because of uniqueness of finite fields of a given cardinality.

We also know that $[\mathbb{F}_q(a):\mathbb{F}_q]=\deg(f)$.

  • $\begingroup$ because $[\mathbb{F}_q(a): \mathbb{F}_q] = def (f)$ ?? $\endgroup$ – Joew and ellie Jan 19 '18 at 23:06
  • $\begingroup$ @Joewandellie That's standard theory: $\mathbb{F}_q(a)\cong\mathbb{F}_q[x]/(f)$. $\endgroup$ – egreg Jan 19 '18 at 23:15

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