Find what $A/S$ is isomorphic to, where $A = \Bbb Z_2[X_1,\dots,X_n]$ and $S=\langle X_1^2-X_1,\dots,X_n^2-X_n\rangle$ 
Find what $A/S$ is isomorphic to, where $A = \Bbb Z_2[X_1,\dots,X_n]$ and $S=\langle X_1^2-X_1,\dots,X_n^2-X_n\rangle$.

I'm sorry but I don't have anything to add here. I've been trying it with $n=2$ and the first theorem of isomorphism, but I can't get anywhere.
The real question of the exam was to probe that $\forall\ \bar{a} \in A/S$, $\bar{a}^2=\bar{a}$. I'm trying to find a simpler ring to work with...but I'm stuck.
 A: Your ring is isomorphic to the tensor product over $\mathbb Z/2\mathbb Z$ of the rings $(\mathbb Z/2\mathbb Z)[X_i]/(X_i^2-X_i)\simeq \mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z$, so it is isomorphic to $(\mathbb Z/2\mathbb Z)^{2^n}$.
An elementary approach should start from $n=1$: $(\mathbb Z/2\mathbb Z)[X]/(X^2-X)\simeq \mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z$ by $f(X)\mapsto(f(0),f(1))$, and then generalize this to several variables.
A: Nothing wrong with user26857's answer (+1). I just cannot resist adding a different point of view :-)
Let $V$ be an $n$-dimensional vector space over the field $\Bbb{F}_2=\Bbb{Z}/2\Bbb{Z}$. I claim that the ring $A/S$ is isomorphic to the ring $\Gamma(V)$
$$
\Gamma(V)=\{f:V\to\Bbb{F}_2\}
$$
of all functions from $f$ to the field $\Bbb{F}_2$. The set $\Gamma(V)$ becomes a ring with operations defined pointwise. Because $x^2=x$ for all 
$x\in\Bbb{F}_2$ we have $f^2=f$ for all the functions $f\in\Gamma(V)$.
The isomorphism is seen as follows. Let us fix a basis $\mathcal{B}=\{e_1,e_2,\ldots,e_n\}$ for $V$. We define a homomorphism $\phi$ of rings from the polynomial algebra $A$ to $\Gamma(V)$ by sending the indeterminate $X_i$ to the function that maps the point $\sum_i a_ie_i$ to its $i$th coordinate $a_i$. By the above property it is then clear that $X_i^2-X_i\in\operatorname{Ker}\phi$ for all $i=1,2,\ldots,n$. Thus $S\subseteq \operatorname{Ker}\phi.$
Next I claim that $\phi$ is surjective. If we fix an arbitrary point
$P=\sum_ia_ie_i\in V$, we see that the function $\phi(\prod_i(X_i-a_i-1))$
takes value $1$ at $P$ but vanishes at all other points $P'\in V, P'\neq P$.
Such functions form a basis of $\Gamma(V)$, so surjectivity of $\phi$ follows.
From user26857's answer (or a direct calculation) it follows that $\dim A/S=2^n=\dim\Gamma(V)$. Therefore $S=\operatorname{Ker}\phi$ and $A/S\cong\Gamma(V)$.
The ring $\Gamma(V)$ shows up naturally in various contexts. In coding theory the so called Reed-Muller codes are defined as subspaces of $\Gamma(V)$. In theoretical computer science the ring $\Gamma(V)$ is known as the ring of boolean functions in $n$ variables.
A: Since others already answered the main question pretty thoroughly, I will show you the most direct way to show that every element in $A/S$ is idempotent.
First note that every element of $A/S$ is of the form
$$
\sum_{k_1,\dotsc,k_n} a_{k_1,\dotsc,k_n} x_1^{k_1} \dotsm x_n^{k_n} \label{eq:element} \tag{1}
$$
with $a_{k_1,\dotsc,k_n} \in \Bbb{Z}/2\Bbb{Z} =: \Bbb{F}_2$ (at most finitely many non-zero), $k_1,\dotsc,k_n \in \Bbb{Z}_{\geq 0}$, and where $x_j$ is the class of $X_j$ in $A/S$.
Now, we know that every element of $\Bbb{F}_2$ is idempotent and by definition of $S$ we know that $x_j^2 = x_j$ in $A/S$ for every $j \in \{1,\dotsc, n\}$.
Finally, note that for every $a,b \in A/S$ idempotent we have $(ab)^2 = a^2b^2 = ab$ and
$$
(a + b)^2 = a^2 + b^2 = a + b
$$
because $\Bbb{F}_2$ has characteristic $2$. By \eqref{eq:element} this is enough to conclude that every element of $A/S$ is idempotent.
