Generic Element of Compositum of Two Fields [duplicate]

You are correct. In general, if $S$ is a subset of $\Omega$ and $F$ is a subfield, then $F(S)$ is defined to be the intersection of all subfields of $\Omega$ which contain $F$ and $S$. You can show that $F(S)$ is the set of all quotients $\frac{f(s_1, ... , s_n)}{g(s_1, ... , s_n)}$, where $n \geq 1, s_i \in S$, and $f, g \in F[X_1, ... , X_n]$ with $g(s_1, ... , s_n)$.

By definition, $KL$ is the field $K(L)$. But you can show this is the same thing as $L(K)$. In fact, $$KL = K(L) = L(K) = F(K \cup L)$$ (you can replace $F$ by any subfield of $\Omega$ which is contained in $L \cap F$) and this last field $F(K \cup L)$ you can easily show to be the same as the thing you said. I am not sure if any of these descriptions is simpler than another.

Again if $F$ is a subfield of $\Omega$ and $S$ a subset, $F[S]$ is defined to be the intersection of all subrings of $\Omega$ containing $F$ and $S$. You can show that $F[S]$ is equal to the set of all $f(s_1, ... , s_n)$, where $n \geq 1, s_i \in S$, and $f \in F[X_1, ... , X_n]$. Again $$K[L] = L[K] = F[K \cup L]$$

Here is a useful proposition:

Proposition: The following are equivalent:

(i): $F(S) = F[S]$.

(ii): Every element of $S$ is algebraic over $F$.

Proof: Let me think about it. It should follow from the case when $|S| = 1$, which is a pretty standard result.

From the proposition, we can make the following observation, $K[L]$ is equal to the set of all finite sums $k_1l_1 + \cdots + k_tl_t, k_i \in K, l_i \in L$. So we see that $KL$ admits the nice description you suggested if and only if one of the following equivalent conditions are satisfied:

(i): $KL$ is algebraic over $K$.

(ii): $KL$ is algebraic over $L$.

(iii): $KL$ is algebraic over $K \cap L$.

(iv): Every element of $L$ which is not in $K$, is algebraic over $K$.

(v): Every element of $K$ which is not in $L$, is algebraic over $L$.