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In book by Roman 'Field Theory' it is written that it is straightforward that every extension of a finite field is normal. However I just cannot see it. Can you help me with this problem?

Thank

michael
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3 Answers3

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If the field $F$ is a subfield of $K$ with $K$ finite of order $q$, then the elements of $K$ are all roots of $x^q-x \in F[x]$, so this equation has $q$ (distinct) roots in $K$, and so $K$ is its splitting field over $F$. So $K$ is a Galois extension of $F$.

Derek Holt
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  • I think this is the best argument. – Hoot May 11 '14 at 20:04
  • But in this argument, we assume that extension of a finite field is also finite. Why is that? – michael May 11 '14 at 20:40
  • @michael Can we assume that they're still algebraic extensions? If so, then see Jakob's answer for the explanation. – Hoot May 11 '14 at 21:41
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    It depends on the definition of normal extension you are using. The one I am used to is that $K:F$ is normal if, whenever $K$ contains one root of $f \in F[x]$, then $f$ factorizes completely in $K[x]$. If $\alpha$ is the root in question, then $|F(\alpha):F|$ is finite, so $F(\alpha):F$ is normal by the above argument, and hence $f$ factorizes completely in $F(\alpha)[x]$. So $K:F$ is normal. – Derek Holt May 11 '14 at 22:22
  • Would that mean that every normal extension is finite? – michael May 12 '14 at 07:14
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    No. For example the field of all algebraic numbers is a normal extension of ${\mathbb Q}$ but is not a finite extension. It is less clear whether every normal extension needs to be algebraic: that seems to vary from boom to book. But since it is only the algebraic elements of the extension that determine whether or not it is normal, that is not so important. – Derek Holt May 12 '14 at 07:48
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An (algebraic) field extension is normal if and only if it is the splitting field of a family of polynomials, i.e. if the extension $K/k$ contains an element $\alpha$, it also contains all conjugates of $\alpha$ in an algebraic closure. Thus, we can prove the extension is normal by proving this property for every element! Now, consider an element $\alpha\in K$. Since $K$ is algebraic over $k$, there is a finite extension $k\subset L \subset K$ such that $\alpha \in L$. Now we'd like to prove that every finite extension of a finite field is normal. Why is that straightforward? For one, we can list all the finite extensions of finite fields, since we know all the finite fields...

Can you continue from here?

Jakob Oesinghaus
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I haven't thought about what this would mean for infinite and possibly non-algebraic "extensions", so I'll pretend they don't exist.

From a finite extension of fields $E/F$ we get inequalities $$ |\operatorname{Aut}(E/F)| \leq |\operatorname{Hom}_{F\text{-alg}}(E, \overline{F})| \leq [E:F]. $$ Normality means that the first inequality is actually an equality. The trick is that in this case it's easy to write $[E:F]$ elements of the automorphism group down.

Hoot
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