This is from the Stacks project https://stacks.math.columbia.edu/tag/0C2X
Let $L/K$ be a separable algebraic field extension. Then, there is a multiplicative subset $S \subseteq L \otimes_K L$ such that $S^{-1}(L \otimes_K L)$ is isomorphic to $L$, and the isomorphism commutes with the multiplication map $L \otimes_K L \rightarrow L$ and the localization map $L \otimes_K L \rightarrow S^{-1}(L \otimes_K L)$
The proof given skips a lot of steps. First, if $L/K$ is finite, then by the primitive element theorem, $L = K[a]$ for some $a \in L$ with minimal polynomial $p$. Then, $L \otimes_K L = K \otimes_K K[x] / p = L[x] / p$. Since $a$ is a root and $p$ doesn't have repeated roots, $p = (x-a) q$ for some $q \in L[x]$ that doesn't have $a$ as a root. Then, we invert $q$ (that is : $S = \{1, q, q^2, ... \}$) in $L[x] / ((x-a) q)$.
Why is this isomorphic to $L$? I know that in $S^{-1} L[x]$, $((x-a)q) = (x-a)$, so $S^{-1}L[x] / ((x-a) q) = S^{-1}L[x] / (x-a) = S'^{-1}L$ ,where $S'$ is $\{1, q(a), q(a)^2 \}$, which is nonzero, so $S'^{-1}L = L$. But, we in the previous paragraph, we quotient at $((x-a) q)$ first. Question 1 : Why can we swap localization and quotient?
Then, in the general case, write $L$ as a union of its finite extensions $L = \cup_i L_i$. For each $L_i$, let $S_i$ be from the previous paragraph. The proof then claims that we can use $S = \cup S_i$. Question 2: Why does $S = \cup S_i$ work? Why is $S$ even multiplicatively closed? A union of multiplicatively closed sets is not necessarily multiplicatively closed.