Compute the degree of K over Q

[Firepanda]

I'll skip out some tedious bits that I'm confident I have correct and get to the part I'm stuck on

Let L = Q(3^1/3)

[L:Q] = 3

Clearly [K:L] >1

w is a root of X³-1

So

[K:L] is less than or equal to 3, in particular it is either 2 or 3

Here is where I'm stuck

I tried to show that X³-1 is the min poly of K over L by supposing there exists an a,b in Q such that

(a+b(3^1/3)) = 1

with a hope of finding a contradiction, but when a=1 this is clearly a root and 1 is an element of L, so this canot be the minimal polynomial of K over L.

Can anyone help me find the minimal polynomial of K over L?

Thanks

Matt

 

[micromass]

Indeed, [itex]X^3-1[/itex] is not irreducible. Can you factor this polynomial over L?? I.e. can you factor out x-1? Maybe what's left WILL be the minimal polynomial?

 

[Firepanda]

Ah ok so X²+X+1

By the same method as before I showed thre is no element a,b satisfying

(a+b(3^1/3))² + (a+b(3^1/3)) = -1

so it is irreducible over Q(3^1/3)

By the tower law [K:Q]=6

Are my basis elements

{1, (3^1/3), (9^1/3), w, (3^1/3)w, (9^1/3)w}?

Thanks

 

[micromass]

That seems correct!