Lemma about composite fields in Weintraub's "Galois Theory" I'm working through Weintraub's Galois theory book and I'm having trouble understanding the following lemma (which as I understand will be used later on). So, given a base field $F$, its extension $E$, and subfields $B,D$ that are themselves extensions of $F$, i.e. $F\subset B,D\subset E$, Weintraub defines the composite field $BD$ as the smallest field containing both subfields and makes a passing comment about how it consists of all expressions of the form $\big(\sum_i b_id_i\big)\big(\sum_j b_jd_j\big)^{-1}$ (I suppose this is plausible). He goes on to state and prove the following lemma, about the case where the denominators $\big(\sum_j b_jd_j\big)^{-1}$ may be omitted:

[With $F,E,B,D$ defined as above] suppose $|D:F|$ is finite. Then $|BD:B|\leq|D:F|$ and $BD=\{\sum_i b_id_i\}$.


Proof: Let $K=\{\sum_i b_id_i\}$ and $\{\delta_j\}$ be a basis of $D$ over $F$, then $\{\delta_i\}$ spans $K$ over $B$ (I suppose this is true, just express the $d_i$s in $\sum_i b_id_i$ terms of the basis and you get a $B$-linear combination of the basis vectors), therefore $|K:B|\leq|D:F|$ and [by a previous lemma, about how a finite-dimensional integral domain is a field] $K$ is field. Therefore $BD=K$.

I do not understand the last step, why from the fact that $K$ is a field does it follow that it is equal to the composite $BD$? Aside from a technical explanation of why this works, some intuition as to why we can dispense with denominators in this case would be helpful as well. Thank you for reading.
 A: The fact that $K$ equals $BD$ follows because both $B$ and $D$ are definitely contained in $K$ (note that $K$ was defined as the set of all sums of products of elements of $B$ with elements of $D$; it contains $B$ by taking a sum with a single summand and expressing $b\in B$ as $b\cdot 1$, and symmetrically for $D$). Since $BD$ is, by definition, the smallest field that contains both $B$ and $D$, if $K$ is indeed a field then this shows that $BD\subseteq K$. But since you already noted that $BD$ is the set of all products of elements of the set $K$ and inverses of elements of the set $K$, then we have $K\subseteq BD$, giving the desired equality.
The intuition is essentially that in a finite field extension, the inverse of an element $\alpha$ can be written as a polynomial in $\alpha$: for if $F\leq E$ is an extension, and $\alpha\in E$, then $|F(\alpha):F|$ is finite, so there is an $n\gt 0$ such that $1$, $\alpha,\ldots,\alpha^n$ is linearly independent. Picking the smallest such $n$ we get an expression of the form
$$a_0 + a_1\alpha + \cdots +a_n\alpha^n=0$$
with $a_0\neq 0$, and from this we get $\alpha(a_1+a_2\alpha+\cdots+a_n\alpha^{n-1}) = -a_0$, so
$$\frac{1}{\alpha} = -\frac{1}{a_0}\left(a_1+a_2\alpha_\cdots+a_n\alpha^{n-1}\right).$$
Similarly, any element $\alpha=\sum_i b_id_i$ has an inverse, which can be expressed as a polynomial expression in $\alpha$, so you don't have to explicitly throw in the multiplicative inverse of $\alpha$, it comes for free once you throw  in $\alpha$ and close under ring operations.
Thus, the field $\mathbb{Q}(\sqrt{2},\sqrt{3})$ which includes $\sqrt{2}+\sqrt{3}$, does not require you to also throw in $1/(\sqrt{2}+\sqrt{3})$ explicitly, since it can be expressed as
$$10\left(\sqrt{2}+\sqrt{3}\right) - \left(\sqrt{2}+\sqrt{3}\right)^3$$
(you can verify this by multiplying this expression by $(\sqrt{2}+\sqrt{3})$ and seeing that you get $1$).
