Thread: Solvability by radicals

I understand what it means for a polynomial to be solved by radicals, but came across this statement in Stewart's "Galois Theory" book; I know it should make things clearer, but it's just confusing me:

"We emphasise that we do not require the splitting field extension to be radical. There is good reason for this. We want everything in the splitting field to be expressible by radicals, but it is pointless to expect everything expressible by radicals to be inside the splitting field."

What exactly is this getting at? If someone could provide an example I think that would clear everything up. I think it may be related to the question of showing an extension that is radical but where is an intermediate field with NOT radical - I've tried to come up with my own example but must be looking at this in the wrong way...

Originally Posted by Boysilver

I understand what it means for a polynomial to be solved by radicals, but came across this statement in Stewart's "Galois Theory" book; I know it should make things clearer, but it's just confusing me:

"We emphasise that we do not require the splitting field extension to be radical. There is good reason for this. We want everything in the splitting field to be expressible by radicals, but it is pointless to expect everything expressible by radicals to be inside the splitting field."

What exactly is this getting at? If someone could provide an example I think that would clear everything up. I think it may be related to the question of showing an extension that is radical but where is an intermediate field with NOT radical - I've tried to come up with my own example but must be looking at this in the wrong way...

See here in Example 7.4.

In summary, choose and . You can see that M/K is a radical extension. If you choose L as , where , then . Verify that L/K is not a radical extension. For instance, cannot be expressed by real radicals.