Hilbert-Samuel multiplicity of a local ring of positive dimension is positive? Let $(R, \mathfrak m)$ be a Noetherian local  ring of dimension $d>0$. Then $e(R)=(d-1)!\lim_{n\to \infty} \mu (\mathfrak m^n)/n^{d-1}$. 
(Here $e(R)$ denotes Hilbert-Samuel multiplicity of $R$) . 
Is it true that we always have $e(R)>0$ ? 
I know this is true if $R$ is Cohen-Macaulay, but otherwise I'm not sure 
Please help. 
 A: Yes, this is true (assuming your definition of the quantity $e(R)$ is correct - I've never seen this particular definition before). $e(R)$ is the leading term of the Hilbert polynomial of $\mathfrak{m}$ as a module over $R$ multiplied by some (nonzero) factorial, so all we need to do is to show that the leading term of the Hilbert polynomial has positive coefficient. As the Hilbert polynomial is of degree $d$, the coefficient of the degree $d$ term is nonzero - that is, it's either positive or negative. We show that it's not negative, so it must be positive.
For large $n$, the Hilbert polynomial coincides with the Hilbert function, which is defined to be the length of $R/\mathfrak{m}^n$. But length is a non-negative invariant, and the behavior of a polynomial in $n$ for large $n$ is approximated by it's highest term. If the coefficient of the highest-degree term were negative, then the polynomial would eventually be negative for large $n$, leading to a negative length, an impossibility.
