Easy question about tower of fields

Question

If $E/F/G$ is a tower of fields and $[E:G]<\infty$, then does $[F:G]<\infty$? I suspect the answer to be "yes", but somehow the fact that a basis of $E$ over $G$ might have vectors in $E\backslash F$ blocks me.

Answer 1

Yes, take a basis $\{f_\alpha\}_{\alpha\in I}$ for $F$ over $E$, and another $\{g _\beta\}_{\beta\in J}$ for $G$ over $F$. Then $\{f_{\alpha}g_\beta\}_{(\alpha,\beta)\in I\times J}$ is a basis for $G$ over $E$. Since $G$ is finite dimensional over $E$, the set $I\times J$ is finite, therefore $J$ is finite.

Answer 2

Yes. In general, $[E:G] = [E:F][F:G]$. Therefore, $[E:G] < \infty$ if and only if $[E:F], [F:G] < \infty$

Answer 3

If $E/F/G$ is a tower of fields, and $A$ is a basis of $F$ over $G$ and $B$ is a basis of $E$ over $F$ then $$ AB:=\{ab\mid a\in A,b\in B\} $$

is a basis of $E$ over $G$.

Although this is usually proven for finite $A,B$ the same method of proof works even for infinite cardinalites. i.e if $|A|=\kappa,|B|=\tau$ then $[E:G]=\kappa\tau$

Answer 4

If $W\subset V$ are vector spaces over a field $k$ then $\dim W \le \dim V$. For completeness we will prove this result.

First some definitions. We call a basis of a vector space a system of vectors that is linearly independent and moreover is maximal with this property, that is any larger system of vectors is not linearly independent. Define the dimension of a vector space over a field $k$ as the cardinality of maximal subsets of $V$ consisting of linearly independent vectors. It turns out that every linearly independent subset is contained in a linearly independent subset that is maximal (that is, in a basis)and moreover any two bases have the same cardinality.

Let's prove now that $\dim W \le \dim V$. Indeed, consider a maximal subset of $W$ consisting of linearly independent vectors, that is, a basis of $W$. This system is also a system of linearly independent vectors of $V$ and therefore can be enlarged to a basis of $V$. We showed that any basis of $W$ is contained in a basis of $V$. Therefore $\dim W \le \dim V$.

For the case of the extension of fields $E/F/G$ , take in the above $k= G$, $W= F$ and $V=E$ and since $G\subset F\subset E$ we have $W \subset V$ vector spaces over $k=G$.