I've tried to prove that the reduced symmetric algebra is graded as an algebra and a coalgebra. $V$ is a vector space on a field of characteristic $p$. $T(V)$ is the tensor algebra.
$s(V):=\frac{T(V)}{(xy-yx,x^{p})}$
My definition of graded coalgebra: $C=\oplus{C^{n}}$ where each $C^{n}$ is a subspace, and : $\Delta(C^{n})\subseteq \sum_{i=0}^{n}C^{i}\otimes C^{n-i}$ and $\epsilon(C^{n})\subseteq \delta_{n,0}k$
I know that $T(V)$ is a graded coalgebra with $C^{n}=V^{\otimes n}$.
My proof:
$I=(xy-yx,x^{p})$ is a homogeneous ideal and it is graded. In fact: $I^{n}:=I\cap V^{\otimes n}$. I want to prove that $I=\oplus I^{n}$:
$I=\sum_{x,y\in V}T[x,y]T +\sum_{x\in V} Tx^{p}T\subseteq \sum_{x,y}\sum_{a,b\in N} V^{\otimes a}[x,y]V^{\otimes b}+\sum_{x}\sum_{a,b\in N} V^{\otimes a}x^{p}V^{\otimes b}\subseteq \sum_{x,y}\sum_{a,b\in N} V^{\otimes a}I^{2}V^{\otimes b}+\sum_{x}\sum_{a,b\in N} V^{\otimes a}I^{p}V^{\otimes b}\subseteq \sum_{a,b\in N} I^{a+b+2}+\sum_{a,b\in N}I^{a+b+p}\subseteq \sum_{n\geq2}I^{n}\subseteq \sum_{n}I^{n}$.
Since the other inclusion is trivial, $I$ is graded. Then, put $s^{n}:=\pi({V^{\otimes n}})$. We clearly have that $s(V)=\sum_{n}s^{n}$. To prove that this sum is direct, pick $x\in s(V)$, then $x=\sum_{n=0}^{d} \pi(x_{n})$. If $x=0$, $\sum_{n=0}^{d}x_{n}\in \ker(\pi)=I=\oplus I^{n}$ . Since each $x_{n}$ has degree $n$, it must be $x^{n}\in I^{n}$ hence $\pi(x_{n})=0$ for all $n$.
So the sum is direct. The proof of the formula on comultiplication and counit is straightforward. Is this correct? I've seen this proof for the symmetric algebra and I'm wondering if the presence of $x^{p}$ in the ideal causes problems. Thank you very much!