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)
 A: Your interpretation of "independent" coordinates is very accurate indeed.
It seems reasonable that the dimension of $\mathbb A^n$ is $n$, because you have $n$ "independent" coordinate functions on affine space. If you believe that, the Noether normalization lemma is a great way to intuitively understand dimension. 
Picture a curve $C\subset\mathbb A^2$. Choose some random line $L\subseteq\mathbb A^2$ and consider the projection $\pi: C\to L\cong \mathbb A^1$. This projection will be finite, i.e. the fibers $\pi^{-1}(t)$ will be finite sets. That's because $C$ has dimension one.
More generally, let $X\subseteq\mathbb A^n$ be a subvariety. Noether normalization says (when translating it to a geometric statement) that there is an affine subspace $\mathbb A^s\cong L\subseteq \mathbb A^n$ and a projection $\pi: X\to L$ which is finite. So in other words, $X$ is something that looks like an affine space of dimension $s$, but possibly folded, twisted and bent - however, only a finite number of times! After all, it's all polynomials.
More algebraically speaking, the function field $K(X)$ of $X$ is an algebraic extension of $\Bbbk(x_1,\ldots,x_s)$ via the morphism $\pi^\ast$. So the "coordinate functions" on $X$ are algebraic over the coordinate functions of an affine $s$-space. This is just another way of saying that $\pi$ is finite, by the way. In the most literal sense, $\Bbbk(x_1,\ldots,x_s)$ has $s$ independent elements, or more precisely speaking: $\Bbbk(x_1,\ldots,x_s)$ has transcendence degree $s$ over $\Bbbk$. Since $K(X)$ is algebraic over $\Bbbk(x_1,\ldots,x_s)$, it also has transcendence degree $s$ over $\Bbbk$. And indeed, the transcendence degree of $K(X)$ over $\Bbbk$ is the dimension of $X$. 
A: 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.
