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If $K$ is an extension field of $F$ such that $[K:F]=2$. Then $K$ is normal?

I know that if $[K:F]=2$ then $K=F(u)$ where $u$ is the root of $f \in F[x]$. But how do you prove that dimension $2$ implies simple (i.e $K=F(u)$).

And $[K:F]$ is finite, and hence $K$ is algebraic over $F$. Then if an irreducible polynomial in $F[x]$ has one root in K, then it splits over $K$, thus it is normal?

Thanks for the help in advance!

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    $\begingroup$ What definition of normality are you using? $\endgroup$ – Alex Becker May 9 '14 at 21:17
  • $\begingroup$ Fixed the tags. The tag [finite-fields] is appropriate only when the fields have a finite number of elements. Not when the degree of the field extension is finite. $\endgroup$ – Jyrki Lahtonen May 11 '14 at 5:20
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Take $u\in K\setminus F$. The minimal polynomial $m_F(x)$ of $u$ over $F$ is necessarily of degree $2$. But it has $u$ as a root in $K$, and since it's quadratic, it splits over $K$. Also, $K$ is the smallest such field over which $m_F(x)$ splits, since any splitting field must contain $u$, but $K=F(u)$, since $K/F$ has no proper intermediate subfields as an extension of prime degree, and $K\supseteq F(u)\supsetneq F$.

So $K$ is a splitting field of $m_F(x)\in F[x]$, thus normal by one of the several equivalent definitions of a normal extension.

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