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If I understand correctly, the definition of the degree of a field extension $L/K$ is the dimension of $L$ over $K$ interpreted as a vector space. Now if the degree is $n < \infty$, the basis looks like $\{ 1, a_1, \dots , a_n\}$ where $a_i \in L - K$.

My question is:

Is it right to write $\{ 1, a, a^2, \dots , a^n\}$ for some $a \in L - K$ because of the primitive element theorem? If not, why exactly is the basis usually written $\{ 1, a, a^2, \dots , a^n\}$ instead of $\{ 1, a_1, \dots a_n\}$?

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up vote 5 down vote accepted

You are correct - there is a basis of the form $\{1,a,a^2,\ldots,a^{n-1}\}$ for some $a\in L-K$ precisely when the primitive element theorem does in fact hold, which is when $L/K$ is a finite, separable extension. Note that the exponents only go up to $n-1$, since the dimension, and hence the number of elements in a basis, is $n$.

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Thank you. I don't know what separable means (yet) but I looked up the conditions for the primitive element theorem: $char(K) = 0$ and $a_i$ algebraic over $K$. I assume this is equivalent. So when $char(K)=p$ you'd write $\{ 1, a_1, \dots , a_n\}$? – Rudy the Reindeer Feb 13 '11 at 22:06
@Matt: Those are not the most general conditions for the primitive element theorem. Any algebraic extension of fields that are of characteristic 0 is separable, which is why it is glossed over in most introductory treatments of the primitive element theorem. Given an algebraic extension $L/K$, we say that $a\in L$ is separable when its minimal polynomial over $K$ has no repeated roots. We then call the extension $L/K$ separable if every $a\in L$ is separable. – Zev Chonoles Feb 13 '11 at 22:11
Also, regardless of characteristic: there will always be $n=[L:K]$ elements in a basis, and 1 is not necessarily in a basis (as Qiaochu points out). So if you wanted to refer to an arbitrary basis for $L/K$ when $\text{char}(K)=p$, all you would have is just $\{a_1,\ldots,a_n\}$. – Zev Chonoles Feb 13 '11 at 22:17

$1$ doesn't necessarily have to be in the basis; we are ignoring the multiplication and just thinking about the vector space structure, so it should just look like $\{ a_1, ... a_n \}$ for some linearly independent $a_i \in L$ (and it doesn't matter whether they're in $K$ or not).

When people say that there exists $a$ such that $\{ 1, a, a^2, ... a^{n-1} \}$ is a basis, they are appealing to the primitive element theorem.

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Your answer is food for thought. Is there an example where none of the basis elements of $L/K$ are in $K$? I'm thinking about it. – Rudy the Reindeer Feb 13 '11 at 22:21
@Matt: Yes - for example, $\{1+\sqrt{2},\sqrt{2}\}$ is a basis for $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$. – Zev Chonoles Feb 13 '11 at 22:23

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