Let $R$ be a commutative ring with unity. Let $s,t \in R$ such that $Rs+Rt=R$ . Consider the localisation maps $p_1 :R \to R_s$ and $p_2: R \to R_t$ .
Then how to show that $R_{st}$ is the Push-out ( https://en.m.wikipedia.org/wiki/Pushout_(category_theory)) of $R_s$ and $R_t$ along $R$ under the localisation maps ?
Here $R_s$ is the localisation of $R$ at the multiplicative set $S:=\{s^n : n \ge 0\}$. So $R_s \cong R[X]/(sX-1)$ .
I'm stuck. Please help.
You want to prove that $R_s\otimes_R R_t$ is canonically isomorphic to $R_{st}$.
Now there is an obvious map from $R_s$ to $R_{st}$, sending any $\frac{x}{s^n}$ to $\frac{xt^n}{(st)^n}$. Similarly, there is a map from $R_t$ to $R_{st}$.
Both maps being morphisms of $R$-algebras, we get a map $\alpha$ from $R_s\otimes R_t$ to $R_{st}$, sending $\frac{x}{s^n}\otimes \frac{y}{t^m}$ to $\frac{xys^mt^n}{(st)^{m + n}}$.
Let's show that the map $\alpha$ is injective and surjective.
Surjective: clear, since any element in $R_{st}$ is of the form $\frac{x}{(st)^k}$, which is the image of $\frac{x}{s^k} \otimes \frac{1}{t^k}$.
Injective: consider an element $I$ in the kernel of $\alpha$.
We first show that $I$ can be written as a pure tensor.
Being an element of $R_s\otimes R_t$, it is a finite sum of pure tensors $\sum_i u_i \otimes\frac{y_i}{t^{m_i}}$, where $u_i\in R_s$ and $y_i\in R$.
By $R$-linearity, every pure tensor $u_i\otimes\frac{y_i}{t^{m_i}}$ is equal to $u_i y_i\otimes\frac{1}{t^{m_i}}$.
Taking $m = \max_i(m_i)$, we may rewrite the above pure tensor as $v_i \otimes \frac{1}{t^m}$, with $v_i = u_i t^{m - m_i} \in R_s$.
Summing up, we see that the original element $I$ can be written as a pure tensor $v \otimes \frac{1}{t^m}$, with $v = \sum_i v_i \in R_s$.
Now write $v = \frac{x}{s^n}$. Since $I$ is in the kernel of $\alpha$, we see that $\alpha(I) = \frac{xs^mt^n}{(st)^{m + n}}$ is zero. This means there exists $k$ such that $xs^mt^n(st)^k = 0$ in $R$.
Therefore we have, by $R$-linearity, $I = \frac{x}{s^n}\otimes\frac{1}{t^m} = \frac{xs^mt^n(st)^k}{s^{m + n + k}} \otimes \frac{1}{t^{m + n + k}} = 0$.