Let $A$ be a commutative unitary ring.
Can we see the localisation of a ring $S_f^{-1}A$, where $S_f=\{f^n,n\geq 0\}$, as the ring $A[f^{-1}]=A[x]/(xf-1)$?
I'm having problem showing that the obvious map $S_f^{-1}A\to A[f^{-1}]$ is injective.
More generally, is it it known the presentation of the localisation $S^{-1}R$ with $S$ any multiplicative system?
One way to see this is to use the (very useful) universal property of localization:
Universal property of localization:
Let $A$ be a ring, $S$ a multiplicative subset, and $i:A\to S^{-1}A$ the natural map. Then given any ring morphism $\varphi:A\to B$ such that $\varphi(S)\subseteq B^*$, there exists a unique morphism $u:S^{-1}A\to B$ such that $\varphi=u\circ i$, i.e. the following commutes:
$$\begin{array}{ccc} A & \rightarrow & B\\ \downarrow & \nearrow \\ S^{-1}A & &\end{array}$$
How to use this here? Well, suppose you have a map $\varphi:A\to B$ such that $\varphi(S_f)\subseteq B^*$. This is the same as saying $\varphi(f)$ is invertible, call this element $y$. Then extend $\varphi$ to a map $A[x]\to B$ by sending $x\mapsto y^{-1}$. Since clearly $(xf-1)\mapsto0$, this induces a map $u:A[x]/(xf-1)\to B$. It's not hard to see that this map satisfies the desired properties, so indeed $A[f^{-1}]$ satisfies the universal property of $S_f^{-1}A$ and the two must be isomorphic.
As for your second question, as far as I know there isn't a nice way to describe localizations in general using generators and relations. That's really why we use this messy construction anyways; if we always had a simple description like in the case of $S=\{1,f,f^2,\dots\}$, then we'd just use that instead.
Suppose $f$ is not nilpotent. Consider the category $C_f$ whose objects are morphisms $g:A\rightarrow B$ such that $g(f)$ is invertible. A morphism $h:(g:A\rightarrow B)\rightarrow (g':A'\rightarrow B')$ is a morphism of rings $h:B\rightarrow B'$ such that $h\circ f=f'$. $S^{-1}_fA$ is the initial object of this category. It is enough to show that $A[x]/(xf-1)$ is this initial object.
The canonical map $i:A\rightarrow A[x]$ composed with the quotient map $p_f:A[x]\rightarrow A[x]/(xf-1)$ defines a morphism of rings $i_f:A\rightarrow A[x]/(xf-1)$. We have $i_f(f)p_f(x)=1$, so $i_f$ is an element of $C_f$.
Let $g:A\rightarrow B$ be any element of $C_f$, we can define $g':A[x]\rightarrow B$, by $g'(a)=g(a)$ and $g'(x)=f^{-1}$, we have $g'(xf-1)=0$, so $g'$ factors by a morphism $g_i:A[x]/(xf-1)\rightarrow B$ wich is a morphism between $i_f$ and $g$. A morphism $h$ between $i_f$ and $g$ satisfies $h(i_f(a))=g(a), h(i_f(f))=g(f)$. This implies that $h(x)=h(f)^{-1}=g(f)^{-1}$ and $h=g_i$. This shows that $i_f$ is the initial element of $C_f$ so it is the localization of $A$ by $\{f,...,f^n,...\}$.
