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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.

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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?

share|improve this question
    
@Zev thanks for your edit. –  Gastón Burrull Aug 5 '13 at 4:42
    
I see your answer @Gaston but it looks to me practically the same as the one in the book...but I think it is correct. –  DonAntonio Aug 5 '13 at 9:59
    
@DonAntonio I'm not using the fact that such expressions in $\alpha_i\beta_i$ are closed under sums and products. I just pick an element and I proved has an expression in $\alpha_i\beta_i$. I really want what Dummit (or Foote) means, probably autor are doing it in a much easier way. –  Gastón Burrull Aug 5 '13 at 20:17

1 Answer 1

up vote 1 down vote accepted
+50

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.

share|improve this answer
    
Thank you, now I understand what the author meant. The author forgot to explain closure under multiplication when we have $a_ia_j\in K_1$ with $i\neq j$ but is the same idea as the idea of $a_i^k$. (I must wait one day to award this bounty) –  Gastón Burrull Aug 7 '13 at 19:19

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