Understanding Dummit & Foote p.528 Sec 13.2 Algebraic Extensions

Question

I can't understand why the authors conclude

Hence, the elements $\alpha_i\beta_j$ span the composite extension $K_1K_2$ over $F$.

I would like to understand what the authors mean and how they conclude this.

I tried to conclude in a quite different way:

Let $x\in F(a_1,\ldots,a_n,b_1,\ldots,b_m)=K_1K_2$, then, for some collection of index sets $\mathscr{A}=\{I_t\}_{t=1}^n$ and $\mathscr{B}=\{J_s\}_{s=1}^m$

\begin{align*} x=&\sum_{\displaystyle i_t\in I_t, j_s\in J_s} a_{i_1,\ldots,i_n,j_1,...,j_m}(a_1^{i_1}\cdots a_n^{i_n})(b_1^{j_1}\cdots b_m^{j_m})\\ =&\sum a k_1 k_2\quad\text{where $a\in F,\ k_j\in K_j$}\\ =&\sum a(\sum p_ia_i)\cdot (\sum h_jb_j)\quad\text{where $p_i,h_j\in F$}\\ \\ =&\sum f_{ij}a_ib_j\text{where $f_{ij}\in F$} \end{align*}

Where $I_t=\{0,1,2,\ldots, -1+[F(a_1,\ldots,a_t):F(a_1\ldots a_{t-1})]\}$ and $J_s=\{0,1,2,\ldots, -1+[F(a_1,\ldots,b_{s}):F(a_1\ldots b_{s-1})]\}$

Then $\{a_ib_j\}$ spans $K_1K_2$.

Am I right?

Answer 1

The author is not doing it in a much easier way, they just aren't writing the details, as you do in your answer. Here is why your answer is more or less the same as the book's.

First, you write:

$$x=\sum a_{i_1\ldots i_nj_1\ldots j_m}(\alpha_1^{i_1}\ldots\alpha_n^{i_n})(\beta_1^{j_1}\ldots\beta_m^{j_m})$$

Now, let $S=\operatorname{span}_F\{\alpha_i\beta_j\}_{i=1,\ldots,n}^{j=1,\ldots,m}$. The book makes three claims about $S$:

1. $\alpha_i^k$ and $\beta_j^k$ are members of $S$ for all $i,j$ and any positive integer $k$.
2. $S$ is closed under multiplication
3. $S$ is closed under addition

This is enough to immediately detect that $x$ is in $S$. By 1., we know that $\alpha_1^{i_1},\ldots,\alpha_n^{i_n},\beta_1^{j_1},\ldots,\beta_m^{j_m}$ are all in $S$. Now, by 2., it follows that the product $\alpha_1^{i_1}\ldots\alpha_n^{i_n}\beta_1^{j_1}\ldots\beta_m^{j_m}$ is also in $S$, as is the scalar multiple $a_{i_1\ldots i_nj_1\ldots j_m}(\alpha_1^{i_1}\ldots\alpha_n^{i_n})(\beta_1^{j_1}\ldots\beta_m^{j_m})$. This is true for any indices $i_1\ldots i_nj_1\ldots j_m$, so by 3., we can sum over all indices and remain in $S$. This shows, by our definition of $x$, that $x\in S$.

Notice, we never explicitly wrote out any of the intermediate linear combinations, we just know that they must exist.