Suppose $H$ is a finite $p$-group, $k$ a field of characteristic $p$, and consider the map $k[H] \rightarrow k$ via $h \rightarrow 1, \; \forall h \in H$, and extending $k$-linearly. Let $J_H$ be the kernel of this map. Then $J_H$ is nilpotent. (Presumably $k[H]$ is the $k$-algebra on $H$).
I am given the following proof of this result, which I am confused by:
We prove the result by induction.
We know that if $|H| > 1$, then $H$ has non-trivial centre, so picking a non-trivial element $z$ in it's centre, we note that $H/\langle z\rangle$ is a $p$-group of smaller order, and thus $J_{H/\langle z\rangle}$ is nilpotent, i.e. $J_{H/\langle z \rangle}^N = 0$ for some $N \in \mathbb N$. We note also that $J_{H/\langle z \rangle} \triangleright k[H/\langle z \rangle]$
It follows easily that $J_H ^N \subset \ker(k[H] \rightarrow k[H/\langle z \rangle]) = (z-1)k[H]$
Now if $\alpha_1, \dots, \alpha_p \in k[H]$, then $((z-1)(\alpha_1 - 1))\dots((z-1)(\alpha_p - 1)) = (z-1)^p(\dots) = 0 \in k[H]$
i.e. $J_H ^{pN} = 0$
Here are a list of things that are currently confusing me about this proof:
Why do we know that $J_H ^N \subset \ker(k[H] \rightarrow k[H/\langle z \rangle)$? (Presumably this map is the extension of the quotient map). I can't completely see where this is coming from. I might just be struggling with my visualisation of the problem and therefore not seeing the connection, but it is still confusing me.
Why is $\ker(k[H] \rightarrow k[H/\langle z \rangle]) = (z-1)k[H]$, and not $zk[H]$?
In the line where we note $((z-1)(\alpha_1 - 1))\dots((z-1)(\alpha_p - 1)) = (z-1)^p(\dots)$, this seems to rely on the fact that the group algebra $k[H]$ is commutative, which I believe is equivalent to saying that $H$ is abelian, but that is not necessarily true here, since all we know is that $H$ is some $p$-group. How then do we have this commutativity?
I'm sorry if these questions seem very basic, I may just be overlooking or forgetting some simple piece of theory that would make all of this clear, but if you could offer any help regarding this I would really appreciate it, thank you.
$J_H$ is sent to $J_{H/\langle z\rangle}$ by the quotient map, as is easily checked; so $J_H^N$ is sent to $J_{H/\langle z\rangle }^N=0$, thus is in the kernel.
Morally, the quotient map identifies $z$ with $1$, not with $0$, so $zk[H]$ wouldn't make sense, and $(z-1)k[H]$ does. Formally, if $\displaystyle\sum_h a_h h $ is in the kernel, it means for every $h\in H, \displaystyle\sum_{r\in \langle z\rangle} a_{hr} = 0$, thus $\displaystyle\sum_r a_{hr}r \in J_{\langle z\rangle}$, which is generated by $(z-1)$. The rest follows.
No it is not equivalent to $k[H]$ being commutative. It uses the fact that $z$ is in the center of $H$, therefore $z$ is in the center of $k[H]$, and thus so is $(z-1)$ : so you can freely move the $(z-1)$ factors in the expression to get the desired equality.
Note that in the proof you're asking for $z$ to be a nontrivial element of order $p$, otherwise there's no reason for $(z-1)^p$ to be $0$ (if you choose an arbitrary $z$ you can do the same reasoning but you have to take everything to the power of the order of $z$)