I am a beginner trying to understand Dan Quillen's idempotent map $\xi:MU_{(p)} \to MU_{(p)}$ described in his classic paper 5-page paper On formal group laws...
Let $F$ be a formal group law over a ring $R$ and and $f(x)\in R[[x]$ with $f(0)=0$. He defines $V_nf$ and $F_nf$ to be elements of $R[[x]$ defined by $$(V_nf)(x)=f(x^n)$$ $$(F_nf)(x)={\sum_{i=1}^n}^F f(\zeta^iX^{1/n}) $$ where the sum above uses $F$. Now fix a prime $p$. Given a formal groups law $F$, he defines a change of variables $$c_F^{-1}(x)=\sum_{(n,p)=1} \frac{\mu(n)}{n}V_nF_n \gamma_0$$ where $\gamma_0(x)=x$ that makes $(c_{F*}F)(x,y):=c_FF(c_F^{-1}x, c_F^{-1}y)$ into a p-typical formal group law. My first question is can someone show the elementary algebra that establishes that $c_Fx$ is a p-typical group law.
Now let $F$ denote the formal group law for $MU_{(p)}$, and let $\xi=c_FF$ (Quillen writes $C_F\Omega$ and I think this is what he means... or maybe he means $\xi=c_FF_{MU_{(p)}})$. So $\xi$ is a p-typical group law over $MU^*_{(p)}$, and we know $MU_*$ is universal for formal group laws. So we have a map of rings $\xi:MU_* \to MU^*_{(p)}$ and hence $\xi:MU^*_{(p)} \to MU^*_{(p)}$. Okay, I understand that. Quillen then claims that for every $X$ we have a map $\hat{\xi}:MU^*_{(p)}(X) \to MU^*_{(p)}(X)$...and my question is why? Edit: that has been answered by Dylan Wilson in the comments as the map $\xi$ tensored with $MU^*(X)$, since we have $MU^*(X)_{(p)}=MU^*(X) \otimes_{MU_*} MU^*_{(p)}$.