Meaning of "extension field is a vector space"

Question

For example it is clear that the case $\mathbb{Q}\big(\sqrt 2\big) = \{ a+b\sqrt2 \vert a,b \in \mathbb{Q}\}$ (Here the $a,b$ is a scalar)

Then if we more generalize this thought, we can get a well known fact that:

(*) Given the field extension L / K, the larger field L is a K-vector space. (Source : https://en.wikipedia.org/wiki/Field_extension)

Hence, I've understood the statement, (*) means $\forall l(\in L) = \sum k_i l_i $ for some $k_i \in K$, $l_i \in L$.

(i.e. Considering the $L$ as a vector space, all the element of the $L $ can be expressed as a linear combination with scalars $k_i(\in K)$ ).

Does it still hold the $L$ like not a finite extension or not a algebraic extension etc like a $\mathbb{Q}(\pi)$ ? In my guess $\mathbb{Q}(\pi)$ is a vector space but not expressed as a finite linear combination.

So, is my thought right? I want to check my idea that I've known.

Any comment or advice would be appreciated. Thanks.

Answer 1

While your "understanding" statement is technically true, I would say that your interpretation misses most of the useful properties of thinking of $L$ as a $K$-vector space. It makes more sense to think of it in terms of the definition of a vector space.

How I would interpret it: $L$ satisfies all of the axioms of a vector space over $K$. I.e. you can consider the elements of $L$ to be "vectors", with addition and scalar multiplication (multiplication by elements of $K$) satisfying all of your familiar vector space properties. This lets you define the dimension of $L$ over $K$, if the dimension is finite you can define a basis, and you can represent $K$-linear transformations with matrices (over $K$).

Under this interpretation, $\mathbb{Q}(\pi)$ is an infinite dimensional vector space over $\mathbb{Q}$.

Answer 2

For any field extension $L/K$, $L$ is always a vector space over $K$; this follows directly from the field axioms, and we needn't consider any bases or further description of $L$ (such as stipulating $L$ is transcendental over $K$ as $\Bbb Q(\pi)$ is over $\Bbb Q$) to prove it; we mainly need show that for

$k_1, k_2 \in K, \tag 1$

and

$l_1, l_2 \in L, \tag 2$

we have

$k_1l_1 + k_2 l_2 \in L. \tag 3$

Since

$K \subset L, \tag 4$

$k_1, k_2 \in L, \tag 5$

and thus since $L$ is a field,

$k_1l_1, k_2l_2 \in L; \tag 6$

thus (3) binds as well. The other vector space properties such as

$k_1(l_1 + l_2) = k_1l_1 + k_1l_2 \tag 7$

etc., also follow directly from the field axioms.