In a now deleted question on MathOverflow here, I asked the following question.
Let $A$ be a $k$-algebra equipped with a grading $A = \oplus_{i \in \mathbb{Z}}A_i$, as a $k$-vector space.
Question. Where can I find a reference to derivations of closed formulas for the formal power series$$P(A, t) := \sum_{i = 0}^\infty (\dim_k A_i) \cdot t^i$$in the cases where:
- $A = k[x_1, \ldots, x_n]$,
- $A = k\langle x_1, \ldots, x_n\rangle$,
and $n \ge 1$? Or would anybody be so kind as to supply a quick derivation themselves?
Qiaochu Yuan gave the following answer in the comments.
These are straightforward combinatorics exercises.
- Poincaré series are multiplicative under tensor product, so the problem reduces to the case $n = 1$, where the Poincaré series is ${1\over{1 - t}}$. So in general it's ${1\over{(1 - t)^n}}$.
- The number of words on $n$ letters of length $k$ is $n^k$, so the answer is ${1\over{1 - nt}}$.
The question was put on hold, presumably because it was at too low of a level for MathOverflow. However, I don't quite understand Qiaochu's answer, is it possible someone could add a bit more detail as to how he got his answers?
First, you should be sure you understand that $P(A\otimes B,t)=P(A,t)P(B,t)$. Now, $$K[x_1,\ldots,x_n]=K[x_1]\otimes K[x_2]\otimes\cdots\otimes K[x_n]$$ so the first problem reduces to computing $P(K[x],t)$. Well, the $k$th degree component of $K[x]$ is one-dimensional, spanned by $x^k$, so $$P(K[x],t)=1+t+t^2+t^3+\cdots=\frac{1}{1-t}.$$ It follows that $P(K[x_1,\ldots,x_n],t)=\frac{1}{(1-t)^n}$.
As for the free algebra $K\langle x_1,\ldots,x_n\rangle$, as Qiaochu noted, the number of words of length $k$ is $n^k$. As these words are linearly independent, the dimension of the $k$th degree component is $n^k$. So, $$P(K\langle x_1,\ldots,x_n\rangle,t)=1+nt+n^2t^2+n^3t^3+\cdots=\frac{1}{1-nt}.$$