minimal primes of a homogeneous ideal are homogeneous I am trying to study the proof of this result. It appears as part 3 of the proposition on page 2 of the following document
http://math.mit.edu/classes/18.721/projgeom6.pdf
I understand everything but the last line. Here the author says $\psi_{\lambda}(P_i)=P_j$ for all nonzero $\lambda$ implies $\psi_{\lambda}(P_i)=P_i$. I would appreciate if anyone can throw some insight on this.
Thanks
 A: Remember that $\psi_{\lambda}$ is the homomorphism on $\mathbb{C}[x_1,\ldots,x_n]$ induced by mapping $x_i$ to $\lambda x_i$. In particular, you have that $\psi_{\lambda}\circ\psi_{\mu} = \psi_{\lambda\mu}$. 
Note that $\psi_{\lambda}$ must induce a permutation on $P_1,\ldots,P_k$, since $\psi_{\lambda}(P_i) = P_j$ for some $j$.
Since $\psi_{\lambda}$ induces a permutation of $P_1,\ldots,P_k$, we get a homomorphism from $\mathbb{C}-\{0\}$ to $S_k$, the permutation group of $k$ elements: given $\lambda$, let $\sigma_{\lambda}(i) = j$ if and only if $\psi_{\lambda}(P_i) = P_j$. But now write $\lambda = \mu^{k!}$ for some complex number $k$. Then 
$$\sigma_{\lambda} = \sigma_{\mu^{k!}} = (\sigma_{\mu})^{k!} = 1_{S_k}$$ 
(since $|S_k|=k!$, so any element raised to the $k!$ power is the identity). Therefore, $\sigma_{\lambda}$ is the identity.
(That is, $\mathbb{C}-\{0\}$ under multiplication is a divisible group, so every homomorphism into a finite group is trivial, so the image of $\mathbb{C}-\{0\}$ in $S_k$ induced by the action of $\psi_{\lambda}$ on $P_1,\ldots,P_k$ must be trivial, so the action is trivial).
