# Compute the degree of K over Q

1. Feb 7, 2012

### 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(31/3)

[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(31/3)) = 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

2. Feb 7, 2012

### micromass

Staff Emeritus
Indeed, $X^3-1$ 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?

3. Feb 7, 2012

### Firepanda

Ah ok so X2+X+1

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

(a+b(31/3))2 + (a+b(31/3)) = -1

so it is irreducible over Q(31/3)

By the tower law [K:Q]=6

Are my basis elements

{1, (31/3), (91/3), w, (31/3)w, (91/3)w}?

Thanks

4. Feb 7, 2012

### micromass

Staff Emeritus
That seems correct!!