Working out the structure of a finitely generated field

Question

If I want to work out write the structure of the elements in the field $\mathbb{Q}(\omega,2^{1/3})$, where $\omega$ is the 3rd root of unity - what is the easiest way to do so?

Since $\omega$ and $2^{1/3}$ are algebraic does that mean $\mathbb{Q}(\omega,2^{1/3})=\mathbb{Q}[\omega,2^{1/3}]$ which means we can forget about the inverses of elements as they automatically be invertible - i.e. just focus on creating the smallest ring containing $\mathbb{Q},w$ and $2^{1/3}$ and we already get a field.

Is $\mathbb{Q}[\omega,2^{1/3}]$ easy to work out? For example as we know $\omega$ and $2^{1/3}$ can both range from powers of $0$ to $2$ before they belong to $\mathbb{Q}$ again so does this mean:

Let z=2^{1/3}

$\mathbb{Q}(\omega,2^{1/3})=\mathbb{Q}[\omega,2^{1/3}]=\{a +b{\omega} +c{\omega}^2 + dz + ez^2 + f{\omega}z + g{\omega}z^2 + h{\omega}^2z + l{\omega}^2z^2 :a,b,c,d,e,f,g,h,l \in \mathbb{Q}\}$

Is there easier ways of working out field structure?

score 2 · Answer 1 · answered Apr 25 '20 at 14:36

2

There are several possible answers to the question "Is there easier ways of working out field structure?"

Use the primitive element theorem. We obtain, for example, $$ \Bbb{Q}(2^{1/3},\omega)=\Bbb{Q}(2^{1/3}+\omega), $$ and this is easier to write down.

Reference: How to compute a primitive element for the splitting field of $x^3-2 \in \Bbb{Q}[x]$?

Describe it as the splitting field of $x^3-2$.

Reference: How should I find Splitting Field of $x^3-2$ over $\mathbb Q$.

answered Apr 25 '20 at 14:36

Dietrich Burde

130,978

Was my reasoning correct? – Partey5 Apr 25 '20 at 14:56
No, the basis must have $6$ elements, since the degree over $\Bbb Q$ is $6$. In fact, its Galois group is $S_3$ and has $6$ elements, see here. – Dietrich Burde Apr 25 '20 at 14:59
How does knowing there is a primitive element help us write the structure? – Partey5 Apr 25 '20 at 15:04
I’ve no idea how he worked out a basis for that group. Surely $\omega ^2 $ must be a basis vector. – Partey5 Apr 25 '20 at 15:08
If we have a primitive element, say $\Bbb Q(\sqrt{2})$, then it is easy to write a basis, i.e., just $1,\sqrt{2}$ here, or $\Bbb Q(\sqrt[3]{2})$ with basis $(1,\sqrt[3]{2},\sqrt[3]{2}^2)$. – Dietrich Burde Apr 25 '20 at 16:54
Ok thanks. How many powers would I know to put $2^{1/3} + \omega$ to? – Partey5 Apr 25 '20 at 20:36
Since $[\Bbb Q(\alpha):\Bbb Q]=(S_3:1)=6$, the basis is ${1,\alpha,\alpha^2,\ldots ,\alpha^5}$, with $\alpha=2^{1/3}+\omega$. – Dietrich Burde Apr 26 '20 at 08:26
Not sure about the $S_3$ part but the rest makes sense. – Partey5 Apr 26 '20 at 11:47
The degree of the Galois extension is the order of the Galois group, which is $S_3$ here. You know that $S_3$ has $6$ elements. – Dietrich Burde Apr 26 '20 at 14:05

Working out the structure of a finitely generated field

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