Prove the splitting field of is a radical extension...

when it comes to ,I am a little bit confused. Since many theorem is based on ..

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- September 5th 2009, 06:01 AMynjRadical extension
Prove the splitting field of is a radical extension...

when it comes to ,I am a little bit confused. Since many theorem is based on .. - September 5th 2009, 07:31 AMThePerfectHacker
The theorem about determining if something is a radical extension (that is, the polynomial is solvable) is about charachteristic zero fields. However, the definition of radical extensions is more general then that.

Remember, is a radical extension (by definition) iff there exists and such that , .

So construct the splitting field of this and argue that you can write it in the above form. - September 5th 2009, 07:39 AMynj
- September 5th 2009, 10:50 AMThePerfectHacker
Notice that is irreducible over . Therefore, as you know, there exists an extension field which has that solves this polynomial. Therefore, . Now . You need to ask now whether has a zero in . Sadly, it does not, this can be confirmed by checking where . Thus, is not the splitting field over . However, we know there exists such that there is which solves . The extension field will therefore become the splitting field over . Now it remains to argue this satisfies the conditions of being a radical extension.

- September 5th 2009, 03:42 PMNonCommAlg
this is maybe an easier way: let see that if then thus the splitting field of is a radical extension of

- September 5th 2009, 04:09 PMThePerfectHacker
- September 5th 2009, 04:19 PMynj
- September 5th 2009, 04:22 PMNonCommAlg