polynomial algebras and their coefficients in prime fields According to the definition of the polynomial algebras $A(n)$ and $A(n,m)$ for $ n \in \mathbb {N} $ and $ m \in \mathbb {N}^n$, if $\mathbb{F}$ be field $GF(2)$ and $ X_1,...,X_n$ be $n$ pairwise commuting indeterminates over GF(2), then for $a\in \mathbb{N}$ we write $ X^a$  for $ X_1 ^ {a_1}...X_n^{a_n}$. Now $A(n)$ denote the comutative algbera consisting all formal sums over GF(2) with the form $\sum_{a\in\mathbb{N} } \alpha_a X^a$ equipped with the usual addition and the multiplication $ (\sum _a \alpha _a X^a) ( \sum _b \beta _b X^b)= \sum_c (\sum_{a+b=c} \alpha _a \beta_b {c \choose a}) X^c$, where the multi-binomial coefficient is evaluated modulo $2$ and thus is considered as an element in the prime field. 
I need help of you to explain more about "multi-binomial coefficient which is evaluated modulo $2$" ? Can anyone please help me to understand this evaluation through examples? Thank you in advance.
 A: My educated guess as to the meaning of this product rule is the following. The product of
two monomials like
$$
P=X_1^{a_1}X_2^{a_2}\cdots X_n^{a_n}\qquad\text{and}\qquad 
Q=X_1^{b_1}X_2^{b_2}\cdots X_n^{b_n}
$$
is defined to be
$$
P\cdot Q=\left(\prod_{i=1}^n\binom{a_i+b_i}{a_i}\right)
X_1^{a_1+b_i}X_2^{a_2+b_2}\cdots X_n^{a_n+b_n},
$$
where the binomial coefficients are calculated modulo two. This rule is then extended linearly to all polynomials.
Lucas' theorem is the key to understanding the binomials modulo two. The binomial coefficient $\binom n k$ is an odd integer, if and only if in base two expansions $n$ has a $1$ whenever $k$ does. Otherwise the binomial coefficient is even, and thus zero modulo two.
So for example
$$
X_1^1\cdot X_1^1= 0,
$$
because $\binom{1+1}1=2\equiv0$, but
$$
X_1^2\cdot X_1^1=X_1^3,
$$
because $\binom{2+1}1=3\equiv1$.
Because $13=8+4+1=1101_2$ in base two, you get that
$$
X_1^aX_1^{13-a}=X_1^{13},
$$
if $a$ is one of $0=0000_2$, $1=0001_2$, $4=0100_2$, $5=0101_2$, $8=1000_2$, $9=1001_2$,
$12=1100_2$ or $13=1101_2$. Because the second bit from the right of $13$ is equal to $0$, the binomial coeffcient $\binom{13}a$ is, by Lucas, an even number, whenever that bit is $1$ in base-2 representation of $a$. That is the case for all $a\in\{2,3,6,7,10,11\}$.
Check this from Pascal's triangle. It is kinda neat!
The different variables then seem to play independently from each other. So for example
$$X_1^4X_2^3 \cdot X_1^3 X_2^8=X_1^7X_2^{11},$$
because both $\binom 73$ and $\binom {11}8$ are odd, but
$$
X_1^4X_2^3 \cdot X_1^3 X_2^2=0\cdot X_1^7X_2^5=0,
$$
because $\binom 52=10$ is even.

Educated as it may be this was still a guess.

This looks vaguely familiar though. In my distant youth I played with rings a bit like this (something like rings of differential operators reduced modulo a prime number, here $p=2$) - the "powers" of a generator were multiplied like these. The way I thought about them (naturally coming from the characteristic zero structures) was as follows. Think of $X_i^{a_i}$ as $T_i^{a_i}/(a_i!)$, where $T_i$ is a "normal" indeterminate, so $T_i^{a_i}$ was the usual product of $a_i$ copies of $T_i$. Given this, we get
$$
X_i^{a_i}\cdot X_i^{b_i}=\frac{T_i^{a_i}}{a_i!}\frac{T_i^{b_i}}{b_i!}=
\frac{T_i^{a_i+b_i}}{a_i! b_i!}=\binom{a_i+b_i}{a_i}X_i^{a_i+b_i}.
$$
For the algebra to work out correctly in characteristic $p$ I needed to play with combinations of $X_i^{a_i}$ instead of $T_i^{a_i}$.
