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Is it true that any irreducible polynomial of degree $5$ in $F_2[x]$ has distinct roots in any algebraic closure of $F_2$?
$F_2$ : field of characteristic $2$.

ts375_zk26
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Rhoades Maibam
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1 Answers1

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Finite fields are perfect, i.e. every finite extension is separable.

Bernard
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    @lomberlego: Do you know what a separable extension is? – Bernard Dec 05 '17 at 12:59
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    Yes, it is more or less that. More precisely, the algebraic closure of $\mathbf F_2$ is the direct limit of the inductive system of its finite extensions, which all have the form $\mathbf F_{2^d}$ (the order on the the set of indices is divisibility \ indeed, there exists a (transition) morphism from $\mathbf F_{2^d}$ into \mathbf F_{2^e}$ if and only if $d\mid e$. – Bernard Dec 05 '17 at 13:59
  • thanks a lot @ Bernard – jasmine Dec 05 '17 at 14:09