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I can't follow a statement in my notes:

"Let $K$ be a finite field, with $f \in K[X]$ an irreducible polynomial of degree $d$. Then any finite extension $L/K$ is normal, and so if $L$ contains one root of $f$ then it contains all the roots of $f$. Therefore, the splitting field $L$ of $f$ is of the form $K(\alpha)$, where $f$ is the minimal polynomial for $\alpha$."

I can see why $L$ must be normal (any finite extension of a finite field is Galois), and so by definition if $L$ contains one root of $f$ then it contains all the roots of $f$. I don't follow the next sentence at all:

i) Why must $L$ be the splitting field of $f$? EDIT: Is this $L$ now a 'new' $L$?

ii) If $L$ is the splitting field of $f$, why must it be of the form $K(\alpha)$?

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    For ii), I can see why it would be true if $f$ were separable (primitive element theorem), but not why it's true for inseparable $f$2011-12-13
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    Another way of explaining the problem in your comment would be to observe that over a finite field all the irreducible polynomials are separable. An irreducible inseparable polynomial is always a $p$th power of a polynomial with coefficients in $K^{1/p}$. But when $K$ is finite, the Frobenius automorphism is onto, and hence $K^{1/p}=K$. Thus the polynomial would be a $p$th power of a polynomial in $K[x]$, and hence couldn't be irreducible after all.2011-12-17

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