I recently asked a question about a type of tensor product of field extensions I didn't know how to handle. I was given a link to an excellent set of notes that I found really helpful. The example I was after was Example 6.9 in those notes. In particular, I am given a field $k$ and would like to describe the tensor product $$ k(s) \otimes_{k} k(t). $$ I am happy with the proof given in the notes, but I really like to frame things in terms of universal properties where possible. What I have here follows the idea of the notes, but as someone still getting comfortable with using universal properties, would someone be able to see if the proof I have here seems valid?
My claim is, as in the notes, that $k(s) \otimes_{k} k(t)$ is given by the localization of $k[s, t]$ at the multiplicative set of polynomials of the form $f(s)g(t)$.
Let $S$ denote the multiplicative set of polynomials of the form $f(s)g(t)$ in $k[s, t]$. It is simple to check that this is indeed a multiplicative set. We first observe that any arbitrary element $\tau \in k(s) \otimes_{k} k(t)$ is given by a finite sum of the form $$ \sum_{i, j} \frac{\alpha_{i}(s)}{\beta_{i}(s)} \otimes \frac{\gamma_{j}(t)}{\delta_{j}(t)}, $$ with each of the $\alpha, \beta, \gamma, \delta$ terms being polynomials in the respective rings $k[s]$ and $k[t]$. Now observe that since this sum is finite, we can cross multiply to get common denominators in each of the tensor factors. More explicitly, we can write the above as $$ \sum_{i, j} \frac{\alpha_{i}(s)}{\beta_{i}(s)} \otimes \frac{\gamma_{j}(t)}{\delta_{j}(t)} = \left( \frac{1}{b(s)} \otimes \frac{1}{d(t)} \right) \sum_{i, j} a_{i}(s) \otimes h_{j}(t) $$ where $b(s)$ and $d(t)$ are given by the products of the $\beta_{i}(s)$ and $\delta_{j}(t)$ respectively. The important fact to glean from the above form is that elements of $k(s) \otimes_{k} k(t)$ of the form $f(s) \otimes g(t)$ are invertible. We will now argue from universal properties. We will identify the polynomial ring $k[s, t]$ with $k[s] \otimes_{k} k[t]$. Now construct we construct the map $$ \phi: k[s] \otimes_{k} k[t] \longrightarrow k(s) \otimes_{k} k(t) , $$ which is just given by $$ \sum_{i, j} f_{i}(s) \otimes g_{j}(t) \mapsto \sum_{i, j} f_{i}(s) \otimes g_{j}(t), $$ which is just the natural embedding. So we have now constructed a map $\phi: k[s, t] \longrightarrow k(s) \otimes_{k} k(t)$ under which elements of the multiplicative set $S$ (defined above) are sent to invertible elements. Now consider any $k$-algebra $F$ with a morphism $$ \psi: k[s, t] \longrightarrow F $$ which also sends elements of the multiplicative set $S$ to invertible elements of $S$. Then this factors uniquely through $S^{-1}k[s, t]$ by its universal property. Our only task now is to construct a map $$ k(s) \otimes_{k} k(t) \longrightarrow S^{-1}k[s, t] $$ which, by composition with the mediating morphism for $S^{-1}k[s, t]$ will give us our desired universal property. I claim that this map comes from the universal property for the tensor product. Indeed we have natural maps $$ k(t) \longrightarrow S^{-1}k[s, t] \\ k(s) \longrightarrow S^{-1}k[s, t] \\ $$ and so the universal property for the tensor product does indeed give us a unique map $$ k(s) \otimes_{k} k(t) \longrightarrow S^{-1}k[s, t] $$ as required. This (I think) completes the proof.