So, I am having troubles with the following proof from Dummit & Foote (3rd ed, 13.2, Prop. 21, or, rather, the considerations before it):
Suppose that $K_1$ and $K_2$ are finite extensions of $F$ in $K$. Let $\alpha_1, ..., \alpha_n$ be an F-basis for $K_1$ and let $\beta_1, ..., \beta_m$ be a basis of $K_2$. Then it is clear that these give generators for the composite $K_1 K_2$ over F:
\begin{equation} K_1 K_2 = F(\alpha_1, ..., \alpha_n, \beta_1, ..., \beta_m) \end{equation}
Since $\alpha_1, ... \alpha_n$ is an $F$-basis for $K_1$, any power of $\alpha_i^k$ of one of the $\alpha$'s is a linear combination with coefficients in $F$ of the $\alpha$'s and a similar statement holds for the $\beta$'s. It follows that the collection of linear combinations: \begin{equation} \sum_{i, j} a_{ij} \alpha_i \beta_j \end{equation} with coefficients in $F$ is closed under multiplication and addition since in a product of two such elements any higher powers of the $\alpha$'s and $\beta$'s can be replaced by linear expressions. Hence, the elements $\alpha_i$; $\beta_j$ for $i = 1, . . . , n$ and $j = 1, . . . , m$ span the composite extension $K_1 K_2$ over $F$.
I don't understand how the authors arrive at the final conclusion. Indeed, I see that the set of sums of $\alpha_i \beta_j$ (let's call it $A$) is closed under addition and multiplication, which is clearly a necessary condition for the span of $\alpha_i \beta_j$ to be a field (I see that if it was a field, it would be equal to $K_1 K_2$), but why would this be already sufficient? Of course, $A$ must be a ring, but how does the "closedness" under multiplication and addition show that every element has an inverse?
I tried to prove that every element of $A$ will have an inverse myself, but I don't see how I can do that. Of course, we know that there is an inverse for every element $\sum_i a_i \alpha_i$ and $\sum_j b_j \beta_j$, but we need to prove that there is an inverse for each element of the form $\sum_{i,j} a_{ij} \alpha_i \beta_j$. Could someone help me with that?