intuitive interpretation of dimension of an affine variety On the technical level i do understand what is the dimension of an affine variety and the connection of the definition with the Krull dimension or rings.However,  what is the intuitive interpretation of the dimension of an affine variety? What does it mean if a variety of $\mathbb{A}^n$ has dimension $s$? Does it mean that it can be described by $n-s$ "independent" equations? "Independent" in what sense? How is the dimension of an affine variety connected to the dimensions of its irreducible components? Is it equal to the maximum dimension among all dimensions of irreducible components? (here by variety i just mean an algebraic set of $k^n$, $k$ a field)

First of all, if $X$ is an irreducible affine variety, then the dimension of $X$ is defined to be the Krull dimension of $k[X]$, the affine coordinate ring of $X$. This means that $X$ is of dimension $n$ if we can find an increasing sequence of prime ideals $(0)=\frak{p}_0\subsetneq\frak{p}_1\subsetneq\cdots\subsetneq\frak{p}_n$, and no other chain is longer than this one.
Since prime ideals correspond to irreducible closed subvarieties of $X$, then the chain above means that there is a chain of varieties
$$\{\mbox{point}\}=X_n\subsetneq X_{n-1}\subsetneq\cdots\subsetneq X_0\subsetneq X.$$
So basically, by definition a point is 0 dimensional, a curve is something where the only strictly smaller subvarieties are points (dimension 0), a surface is something that only has points and curves, etc.
To answer your last question, if $X$ is not irreducible, then dimension means the maximum dimension of all irreducible components.
