Showing that the definition of a morphism from $V_1$ to $V_2$ in Silverman is well-defined

Question

Silverman (in Arithmetic of Elliptic Curves) defines a morphism between projective varieties $V_1$ and $V_2$ to be a map $\phi = [\phi_0, \dots, \phi_n]$ such that

1. the $\phi_i(X) \in \overline{K}[X] = \overline{K}[X_0, \dots, X_n]$ are homogeneous polynomials, not all in $I(V_1)$, having the same degree;
2. for every $f \in I(V_2)$, $f(\phi_0(X), \dots, \phi_n(X)) \in I(V_1)$.

Then says that if $P$ is a point of $V_1$ where all of the $\phi_i$ vanish, and we have a set of polynomials $\psi_0, \dots, \psi_n$ such that

1. $\psi_0, \dots, \psi_n$ have the same degree;
2. $\phi_i \psi_j \equiv \psi_i \phi_j \pmod{I(V_1)}$ for all $0 \leq i, j \leq n$;
3. $\psi_i(P) \neq 0$ for some $i$,

we take $\phi(P) = [\psi_0(P), \dots, \psi_n(P)]$.

I'm having trouble verifying from the definitions that $[\psi_0(P), \dots, \psi_n(P)]$ is guaranteed to be in $V_2$. I think that the requirements on the $\psi_i$ and $\phi_i$ should force $f(\psi_0(X), \dots, \psi_n(X)) \in I(V_1)$ just algebraically, which would be enough, but I can't figure out how to use $\phi_i \psi_j \equiv \psi_i \phi_j \pmod{I(V_1)}$ for all $0 \leq i, j \leq n$ to get this to come out. Can anyone help me find the trick?

Answer 1

Figured this out with the help of a peer so I'll post this to get the question off of the unanswered queue.

Since $I(V_2)$ is a homogeneous ideal, we need only check that $f(\psi_0, \dots, \psi_n) \in I(V_1)$ for the homogeneous generators of $I(V_2)$. Letting $f$ be such a generator, and $i$ be some index such that $\phi_i$ is not in $I(V_1)$, we have that $\psi_i^{\deg f} f(\phi_0, \dots, \phi_n) \in I(V_1).$ But since $f$ is homogeneous, we get that $\psi_i^{\deg f} f(\phi_0, \dots, \phi_n) = f(\psi_i \phi_0, \dots, \psi_i \phi_n) \in I(V_1)$. Looking modulo $I(V_1)$, we have that $f(\psi_i \phi_0, \dots, \psi_i \phi_n) \equiv f(\psi_0 \phi_i, \dots, \psi_n \phi_i) \equiv \phi_i^n f(\psi_0, \dots, \psi_n) \equiv 0$. But $I(V_1)$ is prime, and $\phi_i \not\in I(V_1)$, so $f(\psi_0, \dots, \psi_n) \in I(V_1)$. Since this is true for the homogeneous generators of $I(V_2)$, it holds for all $f \in I(V_2)$, so $[\psi_0(P), \dots, \psi_n(P)] \in V_2$.