Why the Steinberg idempotent is idempotent? Consider the group $GL_n(\mathbb{F}_p)$. We have the following subgroups :
-$\Sigma_n$ the symmetric group (permutation matrices)
-$B_n$ the Borel subgroup (upper triangular matrices)
-$U_n$ the unipotent subgroup (upper triangular matrices with all diagonal entries equal to $1$).  
Consider the ring $R=\mathbb{F}_p[GL_n(\mathbb{F}_p)]$. If $H$ is a subgroup of $GL_n(\mathbb{F}_p)$, we denote by $\overline{H}:=\sum_{h\in H}h\in R$ and if $H\subseteq \Sigma_n$, we denote by $\tilde{H}:=\sum_{h\in H} sign(h) h\in R$, where $sign:\Sigma_n\rightarrow\{\pm 1\}$ is the usual map.
One defines the Steinberg idempotent as : $e_n=\overline{B}_n\tilde{\Sigma}_n/[GL_n(\mathbb{F}_p):U_n]\in R$. Here are my questions :  
(1) Why $e_n$ is idempotent in $R$ ? Is it possible to give a "bear hand" proof of this in this case? In the original paper of Steinberg, it isn't done explicitly. If an explanation is given by using Chevalley's group (which I have no knowledge of), can someone please state the properties which are used?   
(2) [Solved, it was an easy question] Why if $p=2$ we have $e_n=\overline{B}_n \overline{\Sigma}_n$ in $R$ ?
 A: $\def\ZZ{\mathbb{Z}}\def\FF{\mathbb{F}}\def\GL{\mathrm{GL}}\def\Id{\mathrm{Id}}$Here is a bare hands proof. This was fun to work out! One note: The OP asks to prove this result in $\FF_p[\GL_n]$, but I get a better result: We have $ \overline{B_n} \widetilde{\Sigma_n}  \overline{B_n} \widetilde{\Sigma_n}  = [\GL_n : U_n]  \overline{B_n} \widetilde{\Sigma_n}$ in $\ZZ[\GL_n]$, so the Steinberg element is idempotent with coefficients in $\ZZ[1/[\GL_n : U_n]]$ (the smallest ring where it makes sense). 
Notational convention: The OP defines $\overline{H}$ for $H$ a subgroup of $\GL_n$, but I will allow myself to use the notation for any subset $H$ of $\GL_n$.
Lemma 1: For any $w \in \Sigma_n$, we have
$$\overline{B_n} w \overline{B_n} = (p-1)^n p^{\binom{n}{2} - \ell(w)} \overline{B_n w B_n}$$
where $\ell(w) = \# \{ (i,j) : i<j,\ w(i) > w(j) \}$.
Proof: Fix a matrix $b_1 w b_2$ for $b_1$, $b_2 \in B_n$. We must determine in how many ways $b_1 w b_2$ can be written as $b'_1 w b'_2$ for $b'_1$, $b'_2$ in $B_n$.
Put $x = (b_1)^{-1} b'_1$, so $b'_1 = b_1 x$ and $b'_2 = w^{-1} x^{-1} w b_2$. We will have $b'_1$, $b'_2 \in B_n$ if and only if $x \in B_n$ and $w^{-1} x^{-1} w \in B_n$ or, in other words, $x \in B_n \cap w B_n w^{-1}$. So
$$\overline{B_n} w \overline{B_n} = \# ( B_n \cap w B_n w^{-1} )\  \overline{B_n w B_n}.$$
It remains to compute $\#(B_n \cap w B_n w^{-1})$. The diagonal entries of a matrix in $B_n \cap w B_n w^{-1}$ can be any nonzero elements of $\FF_p$; the only off-diagonal elements must be in positions $(i,j)$ where $i<j$ and $w(i) < w(j)$, and those elements may be anything. That is $n$ diagonal elements and $\binom{n}{2}- \ell(w)$ off diagonal elements, giving  $(p-1)^n p^{\binom{n}{2} - \ell(w)}$. $\square$
Summing on $w$, we obtain
$$ \overline{B_n} \widetilde{\Sigma_n} \overline{B_n} = (p-1)^n \sum_{w \in \Sigma_n} (-1)^{\ell(w)} p^{\binom{n}{2} - \ell(w)}  \overline{B_n w B_n}. \qquad (1)$$
We now recall the Bruhat decomposition -- every element of $\GL_n(\FF_p)$ is in precisely one of the double cosets $B_n w B_n$. 
For $g \in \GL_n(\FF_p)$, let $\sigma(g)$ be the unique permutation such that $g \in B_n \sigma(g) B_n$. 
So (1) is
$$ \overline{B_n} \widetilde{\Sigma_n} \overline{B_n} = (p-1)^n \sum_{g \in \GL_n(\FF_p)} (-1)^{\ell(\sigma(g))} p^{\binom{n}{2} - \ell(\sigma(g))} g .$$
Right multiplying both sides by $\widetilde{\Sigma_n}$, we have
$$ \overline{B_n} \widetilde{\Sigma_n}  \overline{B_n} \widetilde{\Sigma_n}  = (p-1)^n \sum_{g \in \GL_n(\FF_p)} \sum_{v \in \Sigma_n} (-1)^{\ell(\sigma(g))+ \ell(v)} p^{\binom{n}{2} - \ell(\sigma(g))} gv \qquad (2)$$
The idea of the rest of the proof is to find pairs $(g_1, v_1)$ and $(g_2, v_2)$ such that $\sigma(g_1) = \sigma(g_2)$, $(-1)^{\ell(v_1)} = - (-1)^{\ell(v_2)}$ and $g_1 v_1 = g_2 v_2$. Such pairs will cancel in the sum (2). After removing many such cancellations, the remaining terms will give $[\GL_n : U_n]  \overline{B_n} \widetilde{\Sigma_n}$. 
To this end, for every $g \in \GL_n(\FF_p)$, define a function $\tau(g)$ from $\{ 1,2, \ldots, n \}$ to itself. Let $\tau(g)(j)$ be the largest $i$ such that $g_{ij} \neq 0$; in other words, in column $j$, look for the bottom nonzero element. (Since $g$ is invertible, it does not have any $0$-columns.) We break $(2)$ into the sum over $g$ for which $\tau(g)$ is or is not a bijection.
Case 1: $\tau(g)$ is a bijection. In that case, we can consider $\tau(g)$ as an element of $\Sigma_n$. For $u \in \Sigma_n$, we have $\tau(g)=u$ if and only if $g$ is of the form $bu$ for $b \in B_n$. Also, note that $\sigma(bu) = u$. The contribution to $(2)$ from such $g$ is
$$(p-1)^n \sum_{b \in B_n} \sum_{u \in \Sigma_n} \sum_{v \in \Sigma_n} (-1)^{\ell(u)+ \ell(v)} p^{\binom{n}{2} - \ell(u)} buv $$
$$= (p-1)^n \left( \sum_{b \in B_n} b \right) \left( \sum_{u \in \Sigma_n} \sum_{v \in \Sigma_n}   (-1)^{\ell(uv)} p^{\binom{n}{2} - \ell(u)} uv \right)$$
$$= (p-1)^n \left( \sum_{b \in B_n} b \right)  \left( \sum_{w \in \Sigma_n} (-1)^{\ell(w)} w \right) \left(\sum_{u \in \Sigma_n} p^{\binom{n}{2} - \ell(u)}\right)$$
$$=  (p-1)^n  \overline{B_n} \widetilde{\Sigma_n} \left(\sum_{u \in \Sigma_n} p^{\binom{n}{2} - \ell(u)}\right).$$
In the second equality, we have made the substitution $w=uv$. Using the well known equality $(p-1)^n \left(\sum_{u \in \Sigma_n} p^{\binom{n}{2} - \ell(u)}\right) = [\GL_n(\FF_p):U_n]$, we get $[\GL_n(\FF_p):U_n] \overline{B_n} \widetilde{\Sigma_n}$.
We are done if we can show that the remaining terms sum to $0$.
Case 2: $\tau(g)$ is not a bijection. In that case, some fiber of $\tau$ has $\geq 2$ elements. Let $i$ be the largest value (lowest row) for which $|\beta(g)^{-1}(i)| \geq 2$ and let $j_1$, $j_2$ be the two least (leftmost) elements of $\beta(g)^{-1}(i)$. Let $s$ be the permutation which switches $j_1$ and $j_2$ and fixes everything else. We claim that the terms $(g,v)$ and $(gs, sv)$ in $(2)$ cancel. Concretely, $gs$ is $g$ with columns $j_1$ and $j_2$ switched.
We start with some preliminary observations. First of all, $\tau(gs) = \tau(g)$, since the two columns which are switched both have their bottom nonzero element in row $i$. So our recipe of pairing $(g,v)$ to $(gs, sv)$ also pairs $(gs, sv)$ to $(g,v)$. Secondly, $(gs)(sv) = gv$ and $(-1)^{\ell(sv)} = - (-1)^{\ell(v)}$. So the corresponding terms in $(2)$ will cancel if we can show that $\sigma(g) = \sigma(gs)$.
Our task now is to show that $g$ and $gs$ are in the same double coset. I originally did this using a standard criterion: $g_1$ and $g_2$ are in the same double coset if and only if every bottom left submatrix of $g_1$ has the same rank as the corresponding submatrix of $g_2$. But it turned out to be only slightly longer to give a sequence of left and right multiplications by $B_n$ turning $g$ into $gs$. 
Note that left multiplying by $B_n$ corresponds to upward row operations and right multiplication corresponds to rightward column operations. At each reduction step, we will preserve the property that our two matrices are related by swapping the $j_1$ and $j_2$-columns.
Step One: For each row below that $i$-th row, we have assumed there is at most one column of $g$ whose bottom nonzero element is in that row. (Using that $g$ is invertible, there is exactly one, but we won't need that.) Using the same upward row operations on $g$ and $gs$, we may and do assume that, in each of these columns, the $i$-th row is $0$. We will never perform column operations using these columns again, and the future row operation (Step Four) will add row $i$ to other rows above it, so these columns will have no future impact.
Step Two: We have $g_{i j_1}$ and $g_{i j_2} \neq 0$. Rescaling those columns by a right multiplication by a diagonal matrix, we may assume that $g_{i j_1} = g_{i j_2} = 1$.
Step Three: Using rightward column operations, we may assume that $g_{ij} =0$ for all $j \neq j_1$, $j_2$. In step four, we will add row $i$ to rows $<i$; this step ensures that only columns $j_1$ and $j_2$ are affected.
Step Four: From now on, our operations will only affect rows $i$ and up, and only columns $j_1$ and $j_2$. Let that portion of $g$ be $\left[ \begin{smallmatrix} \vec{x} & \vec{y} \\ 1 & 1 \end{smallmatrix} \right]$. Then
$$\begin{bmatrix} \vec{y} & \vec{x} \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} - \Id_{i-1} & \vec{x}+\vec{y} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \vec{x} & \vec{y} \\ 1 & 1 \end{bmatrix}.$$
So $\left[ \begin{smallmatrix} \vec{x} & \vec{y} \\ 1 & 1 \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} \vec{y} & \vec{x} \\ 1 & 1 \end{smallmatrix} \right]$ are in the same $B_i$ right coset. We see that, having made our previous reductions, $g$ and $gs$ are in the same $B_n$ right coset, and our original matrices are in the same double coset. $\square$
