I'm reading Qing Liu's Algebraic Geometry and Arithmetic Curves. The proof of the following proposition says that (a) and (b) implies (c). But I can't prove (c) from (a) and (b).
Proposition 1.11.
Let $k$ be a field, and $I$ a proper ideal of $k[X_1,\ldots,X_n]$. Then there exist a polynomial sub-$k$-algebra $k[S_1,\ldots,S_n]$ of $k[X_1,\ldots,X_n]$ and an integer $0\leq r\leq n$ such that:
(a) $k[X_1,\ldots,X_n]$ is finite over $k[S_1,\ldots,S_n]$;
(b) $k[S_1,\ldots,S_n]\cap I = (S_1,\ldots,S_r)$ (this is the zero ideal if $r=0$);
(c) $k[S_{r+1},\ldots,S_n]\to k[X_1,\ldots,X_n]/I$ is finite injective.
So the map in (c) is the canonical map $f:P(S_{r+1},\ldots,S_n)\mapsto \overline{P(S_{r+1},\ldots,S_n)}$ right? Assuming (a) and (b), I proved that $f$ is finite. But I can't prove the injectivity. Here's my attempt for proving the injectivity:
Fix $P,Q\in k[S_{r+1},\ldots,S_n]$ such that $f(P) = f(Q)$. $f(P-Q)=0$ so $P-Q\in I$. By part (b), we have $P-Q\in (S_1,\ldots,S_r)$. Write $P = \sum_{\nu\in\mathbb N^{n-r}} \alpha_\nu S^\nu$ and $Q = \sum_{\nu\in\mathbb N^{n-r}} \beta_\nu S^\nu$. We have $\sum_{\nu\in\mathbb N^{n-r}} (\alpha_\nu-\beta_\nu) S^\nu = P-Q \in (S_1,\ldots,S_r)$. Since $P,Q\in k[S_{r+1},\ldots,S_n]$, all the terms in $P-Q$ doesn't have any of $S_1,\ldots,S_r$. So all $\alpha_\nu-\beta_\nu$ should be $0$ if the following is true:
$\sum_{\nu\in\mathbb N^n}\gamma_\nu S^\nu = 0$ implies that $\gamma_\nu = 0$ for all $\nu\in\mathbb N^n$.
But I don't know if the above statement is true.
The ring $k[S_1,\ldots, S_n]$ is exactly the formal linear combinations of terms of the form $\sum_{\nu \in \mathbb{N}^n} \gamma_\nu S^\nu$. So indeed, if a polynomial is zero, then all its coefficients are zero. This is what it means for $k[S_1,\ldots, S_n]$ to be the polynomial ring with generators $\{S_1,\ldots, S_n\}$.