3
$\begingroup$

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 $\{ 1, a_1, \dots a_n\}$?

2 Answers 2

5

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 $n$.

  • 2
    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 $\{a_1,\ldots,a_n\}$.2011-02-13
4

$1$ 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.

  • 2
    @Matt: Yes - for example, $\{1+\sqrt{2},\sqrt{2}\}$ is a basis for $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$.2011-02-13