Dimension of a Finite Field Extension is a Power of 2

Question

I have the following question:

Prove that if $E$ is a finite dimensional extension of $F$ which is generated over $F$ by a set $S$ of elements $a$ satisfying $a^2\in F$ for all $a\in S$, then $|E:F|=2^k$ for some $k$. Also, give an example that shows 2 cannot be replaced by 3.

Now, I can show the first part easily enough (or I think) by showing the basis of $E$ over $F$ by just constructing it manually, but I was wondering if there was a simpler way, such as constructing the extension field one element at a time, like $F(\alpha_1),F(\alpha_1,\alpha_2),\ldots$ etc. The part with 2 cannot be replaced by 3 I'm having trouble with as well.

score 2 · Accepted Answer · answered Jan 05 '13 at 04:49

2

Part 1: it's cleaner to do this by induction, building a tower of subextensions and then using the tower law. The inductive step is showing that an extension which is generated by a single element whose square lies in the base has dimension either $1$ or $2$, which is straightforward.

Part 2: $\mathbb{Q}(\sqrt[3]{2}, \omega \sqrt[3]{2})$ has dimension $6$ over $\mathbb{Q}$. Note that the argument from Part 1 fails because when you adjoin the second cube root, the corresponding extension has dimension $2$ over $\mathbb{Q}(\sqrt[3]{2})$.

answered Jan 05 '13 at 04:49

Qiaochu Yuan

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3

For part 2, an even simpler example would just be $\mathbb{Q}(\omega)$ over $\mathbb{Q}$. – Zev Chonoles Jan 05 '13 at 04:50
Yes, for part one, that's kind of what I was getting at. I was fumbling showing that $[F(\alpha,\beta),F(\alpha)]=2$, for example. – Frank White Jan 05 '13 at 04:57
Wait, thats not bad at all. Nevermind. – Frank White Jan 05 '13 at 05:02

Dimension of a Finite Field Extension is a Power of 2

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