The definition of a normal extension in the book "Abstract algebra" is :

If $K$ is an algebric extension of $F$ which is the splitting field over $F$ for a collection of polynomials $f(x)\in F[x]$ then $K$ is called a normal extension

I think that there is something here I don't understand: If $K$ is an algebraic extension of $F$ then by definition each element of $K$ is a root of a polynimial with coefficients in $F$.

So each element of $K$ corresponds to a polynomial in $F[x]$ (s.t the element is a root of this polynomial).

So I deduced that $K$ is the splitiing field of the collections of polynomials in $F[x]$ that corresponds to the elements in $K$. Hence every algebric extension is also a normal one.

What part of my argument is wrong ?


Just because the extension contains a root of a polynomial doesn't mean that it contains all of the roots of the polynomial, which is a requirement for it to be the splitting field for the polynomial.

For example, $\mathbb{Q}(\sqrt[3]{2})$ is not normal because it is not the splitting field of the minimal polynomial of $\sqrt[3]{2}$, namely $x^3-2$. This can be seen through Theorem 13 on p. 572 in Dummit and Foote:

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If $\mathbb{Q}(\sqrt[3]{2})$ were the splitting field for some polynomial (which doesn't split in $\mathbb{Q}$) then it would be the splitting field for any irreducible factor of that polynomial. In a field of characteristic $0$ (e.g. $\mathbb{Q}$), every irreducible polynomial is separable. So, by the theorem, since $x^3-2$ has a root in $\mathbb{Q}(\sqrt[3]{2})$, it would split in $\mathbb{Q}(\sqrt[3]{2})$ if the extension were normal.

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    $\begingroup$ $\mathbb{Q}(\sqrt[3]{2})$ is not normal because it is not the splitting field of any collection of polynomials... $\endgroup$ – lhf May 3 '12 at 11:32
  • $\begingroup$ @lhf thanks, I have clarified the answer. $\endgroup$ – Antonio Vargas May 3 '12 at 16:21
  • $\begingroup$ It's been a long time since the question was answered but I don't understand the proof of the Theorem that @AntonioVargas posted from that book. I don't understand the step where they apply the induction hypothesis. Could you explain that proof to me, please? $\endgroup$ – synack Dec 29 '12 at 17:01
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    $\begingroup$ @Kits89 it would probably be best to open a new question about that. $\endgroup$ – Antonio Vargas Dec 29 '12 at 20:00

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