Field Extension

Consider the field $L(b_1,\ldots,b_m)=L[b_1,\ldots,b_m]$ (the equality because we are dealing with finite algebraic extensions). Suppose you can prove that it is equal to $LM$.

Then every element of $LM$ can be written as an $L$-linear combination of $b_1,\ldots,b_m$ (there may be a bit of work to be done here if it is not clear; certainly, you can write it as an $L$-linear combination of products of powers of the $b_i$, but since each of those lies in $M$, you can write them as $K$-linear combinations of the $b_i$; take it from there).

And every coefficient in that linear combination can be written as a $K$-linear combination of $a_1,\ldots,a_n$. See where that leads you (assuming you can prove $L[b_1,\ldots,b_m]=LM$, of course).

Corrected. In comments you ask about showing that if $[LM:K]=[L:K][M:K]$, then $L\cap M=K$. I messed up my first attempt in a rather silly way (feel free to look at the edit history to see the screw-up!). Sorry about that.

Again, let $E=L\cap M$ for simplicity. Then: \begin{align*} [L:K][M:K] = [LM:K] &= [LM:E][E:K]\\ &\leq [L:E][M:E][E:K]\\ &= [L:E][M:K]. \end{align*} Can you take it from here?

Comme l'a signalé Arturo Magidin, $LM$ est en fait l'anneau engendré par $L$ et $b_1, \ldots, b_m$. Pour montrer que tout élément de cet anneau est combinaison linéaire des $b_i$, il suffit de voir que $A = \sum Lb_i$ est un anneau. On se ramène donc à prouver que chaque produit $b_i b_j$ (et 1) est dans $A$. Mais en fait, $b_i b_j \in \sum Kb_i$ (et 1 aussi) par l'hypothèse que les $b_i$ forment une base de $M$ sur $K$.