Why does Lang describe a field as a union and compositum of its subfields in this manner?

Question

I am reading Serge Lang's Algebra, revised 3rd edition, and on page 226, the author makes the following definitions and observations:

1. If $E,F \subset L$, the compositum of $E$ and $F$ in $L$ is defined to be the smallest subfield of $L$ containing both $E$ and $F$. It is denoted $EF$.
2. If $k \subset E$ and $\alpha_1,\dots,\alpha_n \in E$, then $k(\alpha_1,\dots,\alpha_n)$ is defined to be the smallest subfield of $E$ containing $k$ and $\alpha_1,\dots,\alpha_n$.
3. Observe that $$ E = \bigcup k(\alpha_1,\dots,\alpha_n), $$ where the union is taken over all finite families $\{ \alpha_1,\dots,\alpha_n \}$ of elements of $E$.
4. The compositum of an arbitrary family of subfields of a field $L$ is defined as the smallest subfield of $L$ containing all fields in the family.
5. $E$ is finitely generated over $k$ if there is a finite family $\{ \alpha_1,\dots,\alpha_n \}$ of elements of $E$ such that $E = k(\alpha_1,\dots,\alpha_n)$.
6. Observe that $E$ is the compositum of all its finitely generated subfields over $k$.

My question is regarding the two observations (point nos. 3 and 6). It appears to me that it is enough to take union (and compositum) over all subfields generated by a single element. For example for point no. 3, I can write $$ E = \bigcup_{\alpha \in E} k(\alpha) $$ because $E$ is clearly contained in the RHS, and each $k(\alpha)$ is contained in $E$ and hence so is the union, implying that $E$ contains the RHS. Similarly for point no. 6.

So, why does Lang emphasise to take the union and compositum over all finitely generated subfields? Is there some perspective that he wishes to emphasise that I am missing? Any help in understanding this will be appreciated.

Answer 1

I try to summarize some of the comments: The problem with considering simple subextensions of $E/k$, so of the form $k(a)$ for some $a\in E$, is that they do not form an filtered set: it is not true that given two simple extensions, say $k(a)$ and $k(b)$, there exists a $c\in E$ such that $k(a),k(b)\subset k(c)$.

If the extension $E/k$ is finite but it contains infinitely many subextensions (so it is not separable), then it is not simple (by the primitive element theorem).

For example, if $E=L(t,s)$ and $k=L(t^p,s^p)$, and $L$ is of characteristic $p$. In this case, although $E=\bigcup_{a\in E} k(a)$, you need infinitely many elements in the union to get $E$. But $E$ is finitely generated, so with only one element you are done.

The same is true for transcendental extensions: if $E=k(t,s)$, and $t$ and $s$ are algebraically independent, then $E/k$ is not a simple extension.

Why do you need the set to be filtered? The reason is that then you can construct the direct limit abstractly. And then prove that this direct limit is the initial $E$. So it is possible that the result he had in mind was the following.

Result: For any extension $E/k$, consider the direct system of finitely generated subextensions of $E/k$. Then $E$ is isomorphic to the direct limit of this system.

So, even it is hard to tell why some people does something, and specially if this people is Serge Lang (who did some weird things), it seems natural to me that he liked the system of extensions to consider to be a filtered.