Intuitive computation of localizations I am taking a class in algebraic geometry, where one of the needed basic skills is to be able to compute localizations. Since this computation is usually skipped in our lecture, I have the feeling that we should be able to do this "by intuition", but most often I am unable to do that.
In here, let's restrict ourselves to localizations of rings (commutative and with $1$) at some multiplicative set. So let $A$ be a ring and let $S$ be a multiplicative subset (containing $1$), and denote the localization of $A$ at $S$ by $S^{-1}A$. Then my first question is:
1. What is the strategy (in general) to determine the localization $S^{-1}A$, not by raw computation but rather by intuition or by expectations on "how it should look like"?
From my understanding, $S^{-1}A$ should be something like a smallest "ring extension" of $A$ such that all elements of $S$ become invertible, so basically one introduces something like fractions. In some cases, this is enough for me to intuitively see the result. One important case (for algebraic geometry) are polynomial rings. For instance, let $k$ be a field and $A=k[T]$ the polynomial ring in one variable, and let $S=\{T^n\mid n\in\mathbb{N}_0\}$, then we get
$$k[T]_T:=S^{-1}A=k[T,T^{-1}],$$
since the elements can be seen as fractions $f/(T^n)$ and one can see that this will lead to any rational function. In general, since we basically deal with fractions, my next question is:
2. If we localize $k[T_1,\dots,T_n]$ at some multiplicative subset, do we always get something which is isomorphic to a subset of $k[T_1,T_1^{-1},\dots,T_n,T_n^{-1}]$?
My last question will be something more specific from algebraic geometry.
3. Let $X=Spec(k[T_1,T_2])$ (as a scheme) and let $U=X\setminus \{(T_1,T_2)\}$. Show that $U$ is an open subscheme of $X$ which is not affine.
It is clear that $U=X\setminus V(\{T_1,T_2\})$ is open in $X$. Assume that $U=Spec(A)$ was affine. My idea is to somehow show how $A$ must look like in this case and lead this to a contradiction. We have
$$A=\Gamma(Spec(A),\mathcal{O}_{Spec(A)})\cong \Gamma(U,\mathcal{O}_X\mid_U)=\mathcal{O}_X\mid_U(U) = \varprojlim_{D(f)\subseteq U} k[T_1,T_2]_f=\varprojlim_{f\notin k} k[T_1,T_2]_f.$$
(I hope the notations are clear). Now here I was stuck, because I was unable to compute this. I was told that this should be simply $k$, but I do not see why, I am just lacking of that intuition I guess, which is why I opened this topic. If $A=k$ is true, then it follows that $U=Spec(k)=0$ contradicting $U=X\setminus\{(T_1,T_2)\}$. This proof by contradiction is the usual strategy to show that something is not affine, I believe.
I appreciate your help!
 A: Ok, a lot to say here.
1) I'm not sure how to answer this question. Just allow yourself to invert things in $S$. $S\newcommand{\inv}{^{-1}}\inv A$ is just $A$ where you allow yourself to divide by things in $S$, or fractions with numerators in $A$ and denominators in $S$. Not sure what you're looking for exactly here.
2) No. Note that $k(T)\ne k[T,T\inv]$. $\frac{1}{T+1}\not\in k[T,T\inv]$. 
3) Note that $\newcommand{\Spec}{\operatorname{Spec}}\Spec k[x,y]$ is an integral scheme. Hence it has a well-defined field of rational functions, $k(x,y)$. An element of $\newcommand{\calO}{\mathcal{O}}\calO(U)$ is a collection of pairs $(f_i/g_i,D(g_i))$ of rational functions that agree on their overlaps, i.e. such that $f_i/g_i=f_j/g_j$ on $D(g_i)\cap D(g_j)$.
I.e. as an element of the field of rational functions, $f_i/g_i$ is well defined, call it $F$. Now because $k[x,y]$ is a UFD, there is a unique 
smallest denominator $g$ such that $Fg = f\in k[x,y]$. Hence $D(g_i)\subset D(g)$ for all $g_i$. Then we must have that $U\subset D(g)$, or that the zeros of $g$ are at most the origin. However, over an
algebraically closed field, no polynomial in two variables has only one maximal ideal as a zero, in fact if $g\not\in k$, then $g$ will have a zero at $(p)$ for every prime factor of $g$ anyway. Hence $g$ must have no zeroes, so $g\in k$. But then $F\in k[x,y]$. Hence $k[x,y]\subset \calO(U)\subset k[x,y]$, so $\calO(U)=k[x,y]$. But $U$ is missing a point compared to $\Spec\calO(U)$, so $U$ is not affine.
