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Let $E/F$ be an extension of fields with $F \leq E_1, E_2 \leq E$. Prove that there exists an $F$-basis for $E_1E_2$ given as fololows. Let $\mathscr{B}_1$ be an $F$-basis for $E_1 \cap E_2$. Let $\mathscr{B}_2$ be an $(E_1 \cap E_2)$-basis for $E_1$, and let $\mathscr{B}_3$ be an $(E_1 \cap E_2)$-basis for $E_2$. Prove that $$\mathscr{B}= \{uv \text{ }:\text{ }u \in \mathscr{B}_1, v \in \mathscr{B}_2 \} \cup \{uw \text{ }:\text{ }u \in \mathscr{B}_1, w \in \mathscr{B}_3 \} $$ is an $F$-basis for $E_1E_2$.

Note that $E_1E_2$ is the composite of the two subfields $E_1$ and $E_2$ -- i.e., it is the smallest such subfield containing both $E_1$ and $E_2$.

We are given that for all $\lambda \in E_1E_2$, we have $$\lambda = \sum_{i=0}^n \alpha_i \beta_i$$ w where $\alpha_i \in E_1$ and $\beta_i \in E_2$.

I don't know where to go from here, however.

1 Answers 1

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Hint:

Note $$ [E_1E_2 : F] =[E_1E_2 : E_1] [E_1 : E_1\cap E_2] [ E_1\cap E_2 : F] $$ $$ [E_1E_2 : F] =[E_1E_2 : E_2] [E_2 : E_1\cap E_2] [ E_1\cap E_2 : F] $$ Prove that $[E_1E_2 : F]$ is the union of above two extensions. Note that bases of $[E_1E_2:F]$ are of $\alpha_i\beta_j, \:\alpha_i\in [E_1:F], \beta_j\in [E_2:F]$.

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    That makes so much more sense. Thank you again! – 2017-01-29