Some doubts in algebraic geometry Let be $X=\mathbb{C}^{2}\setminus \{(0,0)\}$
For me it's clear to see that $X$ is not algebraic if I think in $\mathbb{C}[x,y]$. 
Being an algebraic set depends on the ring of polynomials that one chooses? Could you give me an example of set that is not algebraic in any ring of polynomials? Some books define affine variety as an irreducible algebraic set and others define affine variety as an space with functions isomorphic with an algebraic set. Are these definitions equivalents?
I am beginner in algebraic geometry and I want to understand very good these basic concepts. I would appreciate your answers. 
 A: It is not precisely as you say, $xyz = 1$ does not contain $(x,y) = (0,1)$ while $X$ does contain it. Rather, you need to further cover it by complement of $x$-axis and complement of $y$-axis.
But you are correct that the conception of what is an algebraic variety is multi-step. First you consider subsets of $\mathbb{C}^n$ which are zeros of polynomials, I think this is what you call algebraic subsets.
But then you also consider open subsets of $\mathbb{C}^n$, with respect to the Zariski topology. So each such open subset will be union of subsets of the form $\{ x\in \mathbb{C}^n | \ f(x) \neq 0\}$. Such a subset is in bijection with the closed subset of $\mathbb{C}^{n+1}$ given by $\{ (x,c) \in \mathbb{C}^{n+1}| \ f(x)c - 1 = 0 \}$.
Then you also need to figure out what functions you allow on these sets. On the set of zeros of polynomials you allow the restrictions of polynomials. On the open subsets you allow also to divide by polynomials which are non-vanishing on the subset. Then you need to see that the arsenals of allowable functions agree w.r.t. the bijection above. And so on...
In the end, you would like to formulate the concept of an algebraic variety, which is a set equipped with a sheaf of functions, which locally looks like an algebraic set with the algebraic functions on it (i.e. restrictions of polynomials). Then the algebraic sets will correspond to closed subvarieties of $\mathbb{A}^n$, while your $X$ will be an example of an open subvariety of $\mathbb{A}^2$ which is not affine. 
A: Recall that an algebraic set is a set of the form $V(I)$ for some ideal $I\subset K[x_1,\dots,x_n]$ for some (alg. closed) field $k$ and some $n$. 
Therefore of course algebraic sets depend on the choice of the ring of polynomials: you might have different fields but also on a different number of variables. Take for example $V(x-1)$ in $\mathbb{C}[x]$ and $V(x-1)$ in $\mathbb{C}[x,y]$. In the first case the algebraic set is a point while in the second it is not (you may picture it as a "line"). 
To answer your second question, you can have sets that are not algebraic: take for example $\mathbb{Q}\subset\mathbb{C}$.  
As for your third question: different authors use different definitions and that is frustrating sometimes- especially in the beginning. My suggestion is that you use the definition of the notes you are reading until you are familiar with the concepts. In algebraic geometry it is more common for affine varieties to be irreducible. 
