Dimension in Hilbert function definition. And "for large s" statement confusion in Eisenbud commutative algebra On page 42 of Eisenbud's commutative algebra textbook I ran into some problems I would like some expert guidance with. We have the definition:

Let $M$ be a finitely generated graded module over $K[x_1, \dots, x_r]$, with grading by degree, then the numerical function:
  $$ H_M (S) := \dim_K M_s $$
  is called the Hilbert function of $M$.

I have trouble understanding what is meant by "dimension" in this context. I could find some ideas relating to "homogeneous elements". But I like some form of a precise definition, if possible.
I also had trouble with the corresponding theorem:

If $M$ is a finitely generated graded module over $K[x_1, \dots, x_r]$, then $H_M(s)$ agrees, for large $s$, with a polynomial $P_M$ of degree $ \leq r-1$.

What is meant here with "for large $s$ ", this does not feel like a precise statement at all. Is this like saying that $$ \|H_M(s)-P_M (s)\| \to  0$$ for some $ s\in \mathbb N$? If so, with respect to which norm?
 A: As the comments have indicated, $\dim_K M_s$ is referring to the dimension, as a $K$-vector space, of $M_s$, the $s$th graded piece of $M$, and agreeing for large $s$ literally means there is an $n$ such that, if $s \ge n$, then $H_M(s)=P_M(s)$. 
I think the best way to get your hands on these concepts is to look at a couple examples: 
Perhaps the simplest nontrivial example to look at is when $R=K[x,y]$ and $M=R$.  The grading on $R$ is the standard grading, i.e., $M_s$ is a $k$-vector space with a basis consisting of the monomials of degree $s$.  For example, $M_0$ has basis $\{1\}$, $M_1$ has $\{x,y\}$, $M_2$ has $\{x^2,xy,y^2\}$ etc. So, to determine $H_R(s)$, we need to compute the number of monomials of degree $s$. One way to do this is to note that we get exactly the monomials of degree $s$ if we take those of degree $s-1$ and multiply by $x$, except we miss $y^s$. Thus $H_R(s)=H_R(s-1)+1$.  Since $H_R(0)=1$, we get that $H_R(s)=s+1$ for every $s$. So $P_M(s)=s+1$ and we have agreement for every value of $s$ in this case, and not just for sufficiently large $s$. 
Now let's keep $R=K[x,y]$ but now take $M=K[x,y]/(xy)$.  The grading on $M$ is inherited from that of $R$. In other other words, a basis of $M_s$ consists of the monomials of degree $s$ in $M$, or, equivalently, the monomials of degree $s$ in $R$ which aren't divisible by $xy$. The monomials of $R$ in degree $s$ which aren't divisible by $xy$ are exactly $x^s$ and $y^s$.  Thus, $H_M(0)=1$, and $H_M(s)=2$ when $s \ge 1$.  So $P_M(s)$ is the constant polynomial $2$. In this case, however, we see that $H_M(0) \ne P_M(0)$, but for $s \ge 1$ they agree. 
