1. ## algebraic over K

Perhaps this is not the proper place but, nonetheless, there it goes: E, K are fields, K included in E. A is a subset of E and all elements of A are algebraic over K. And the author says: "Clearly K(A) is the union of all subfields of the form K(M), where M ranges over finite subsets of A".

It is this statement I cannot understand. I have proved that the union of the K(M) is included in K(A) but I cannot prove the inverse inclusion. Thank you for you time.

Enrique.

2. Originally Posted by ENRIQUESTEFANINI
Perhaps this is not the proper place but, nonetheless, there it goes: E, K are fields, K included in E. A is a subset of E and all elements of A are algebraic over K. And the author says: "Clearly K(A) is the union of all subfields of the form K(M), where M ranges over finite subsets of A".

It is this statement I cannot understand. I have proved that the union of the K(M) is included in K(A) but I cannot prove the inverse inclusion. Thank you for you time.

Enrique.

Well, take any element x in K(A): since this is a vector space over K generated by A (perhaps not finite dimensional, though) , there exist elements y_1,...,y_n in K(A) s.t. x is K-linear combination of y_1,...,y_n.
Now, each element y_i is a K-lin. comb. of a finite number of elements from A since, as pointed out above, K(A) is generated as K-vec. space by A ==> in y_1,...,y_n intervene only a finite number of elements of A ==>
the element x is a K-lin. com. of a finite number of elements of A, say a set M ==> x belongs to K(M), for M a finites subset of A.

Tonio

3. I see. But why not say directly: take any element x in K(A). Since this
is a vector space over K generated by A, x is a K-lin. comb. of a
finite number of elements from A, say a set M. Therefor, x belongs to
K(M)?

After all, afterwards you take y_i belonging to K(A) and say that y_i
is a lin.comb. of a finite number of elems. of A. But x was also in
K(A). Regards,

Enrique.

4. Hi

Take care, there can be more elements in $\displaystyle K(A)$ than just $\displaystyle K$-linear combinations of elements of $\displaystyle A$:

for instance, with $\displaystyle A=\{\sqrt[3]{2}\}$ and $\displaystyle K=\mathbb{Q}$,
$\displaystyle \sqrt[3]{2}^2\in\mathbb{Q}(\sqrt[3]{2})$ but $\displaystyle \sqrt[3]{2}^2$ can't be written as a $\displaystyle \mathbb{Q}$-lin. comb with $\displaystyle \sqrt[3]{2},$ it would implie that $\displaystyle \sqrt[3]{2}$ minimal polynomial has degree at most 2
(while $\displaystyle X^3-2$ is irreducible and has $\displaystyle \sqrt[3]{2}$ as a root, therefore it is its minimal polynomial)

But there is a similar argument as what tonio proposed, using that:

$\displaystyle K(A)=\{\frac{P(a_1,...,a_n)}{Q(b_1,...,b_m)}\ ;\ n,m\in\mathbb{N},$ $\displaystyle (P,Q)\in K[X_1,...,X_n]\times(K[X_1,...,X_m]-\{0\}),\ a_1,...,a_n,b_1,...,b_m\in A\}$

5. Alright. Then let x belong to K(A). Then x = P(a_1,...,a_m)/Q(b_1,...,b_n) for some m, n, P, Q and a_1,...,a_m,b_1,...,b_n belonging to A. Now, if I put M= {a_1,...,a_m,b_1,...,b_n} then M is a subset of A and, using the same definition for K(M) as that used for K(A), x belongs to K(M) and all the more so to the union of all the K(M) s.t. M is a finite subset of A.

I also see that the notion of a vector space has not been used here. Well,
thank you very much.
Enrique.

6. Originally Posted by ENRIQUESTEFANINI
I see. But why not say directly: take any element x in K(A). Since this
is a vector space over K generated by A, x is a K-lin. comb. of a
finite number of elements from A, say a set M. Therefor, x belongs to
K(M)?

After all, afterwards you take y_i belonging to K(A) and say that y_i
is a lin.comb. of a finite number of elems. of A. But x was also in
K(A). Regards,

Enrique.
.