Field Extension Let $F$ be an extension field of $K$. let $L$ and $M$ be intermediate fields,  with both finite algebraic extensions of $K$. 
 Suppose {$a_1,...,a_n$} is a basis for $L$ over $K$ and {$b_1, ...,b_m$} is a basis for $M$ over $K$. Show that {$a_ib_j$} is a spanning set for the field $LM$ ($LM$ is the smallest field between $K$ and $F$ containing both $L$ and $M$) as a vector space over $K$.
What I have done so far is this:
Let $x$ $\in$ $L$. Then $x$ can be written as a linear combination of $a_i$. Also $y$ $\in$ $M$ implies that $y$ can be written as a linear combination of $b_j$. This is where I'm stuck.
 A: 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 $K$. 
A: 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?
