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Here is the question I am trying to solve:

Let $A = \bigoplus_{i \geq 1}A_i$ be a graded algebra such that the vector spaces $A_i$ are all finite-dimensional. Define the Poincare series of $A$ as the formal series $$P(A) = \sum_{i \geq 0} \operatorname{dim}(A_i)t^i.$$

Prove that $$P(k\{x_1, \dots , x_n \}) = \frac{1}{1 - nt} \text{ and } P(k [x_1, \dots , x_n ]) = \frac{1}{(1 - t)^n}$$

Definition 1.

Let $X$ be a set. Consider the vector space $k\{X\}$ with basis the set of all words $x_{i_{1}} \dots x_{i_{p}}$ in the alphabet $X,$ including the empty word $\emptyset.$ A word will be called a monomial. Define the degree of the monomial $x_{i_{1}} \dots x_{i_{p}}$ as its length $p.$ Concatenation of words defines a multiplication on $k\{X\}$ by $$(x_{i_{1}} \dots x_{i_{p}})(x_{i_{p + 1}} \dots x_{i_{n}}). \quad (2.1)$$

Formula $(2.1)$ equips $k\{X\}$ with an algebra structure, called the free algebra on the set $X.$ If $X = \{x_1 , \dots , x_n \}$ we denote $k\{X\}$ by $k \{x_1, \dots , x_n\}.$

Definition 2.

$k\{x_1, \dots , x_n \}/I \cong k[x_1, \dots , x_n]$ where $I = x_i x_j - x_j x_i$ where the later is a commutative algebra.

My Questions are:

1- I found on the internet that, for the commutative case, we know that

$$ \dim k^i [x_1, \cdots, x_n] = \sum_{e_1 + \cdots + e_n = i} 1. $$ but I am not sure why this is correct, could anyone show me a proof of this please?

2- Here is the proof of the statement for the commutative ones (I found here Closed formulas for two Poincaré series) but I am not sure what are the $e_i's$ and why their sum must be $i$? and why the second summation sign disappears in the statement before last? and how is the summation before last leads to the last line?

\begin{align*} P_{\Bbb{C}[x_1, \cdots, x_n]}(t) &= \sum_{i=0}^{\infty} \sum_{e_1 + \cdots + e_n = i} t^i \\ &= \sum_{i=0}^{\infty} \sum_{e_1 + \cdots + e_n = i} t^{e_1} \cdots t^{e_n} \\ &= \sum_{e_1, \cdots, e_n} t^{e_1} \cdots t^{e_n} \\ &= \frac{1}{(1-t)^n}. \end{align*}

Could someone help me in answering those questions please?

Intuition
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1 Answers1

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A basis of $\left(k[x_1,\dots,x_n]\right)_i$ is the set of monomials $x_1^{e_1}\dots x_n^{e_n}$ such that $e_1+\dots+e_n=i.$

Any double sum $\sum_{i\in\Bbb N}\sum_{e_1,\dots,e_n\in\Bbb N\atop e_1+\dots+e_n=i}f(e_1,\dots,e_n)$ is equal to $\sum_{e_1,\dots,e_n\in\Bbb N}f(e_1,\dots,e_n).$

The last equality may be detailed as follows: $\sum_{e_1,\dots,e_n\in\Bbb N}t^{e_1}\dots t^{e_n}=\left(\sum_{e_1\in\Bbb N}t^{e_1}\right)\dots\left(\sum_{e_n\in\Bbb N}t^{e_n}\right)=\left(\frac1{1-t}\right)^n.$ (This computation is purely formal, in formal series there is no question of convergence.)

Anne Bauval
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