defining a dominant rational map from an algebra homomorphism (Theorem I 4.4 in Hartshorne)

defining a dominant rational map from an algebra homomorphism (Theorem I
4.4 in Hartshorne)

The following shows up in the proof of theorem I 4.4 in Hartshorne.
Let $X,Y$ be varieties and $\theta: K(Y) \rightarrow K(X)$ a homomorphism
of $k$-algebras. We want to construct a dominant rational map $X
\rightarrow Y$. We can assume that $Y$ is affine with coordinate ring
$A(Y)$. Let $y_1,\cdots,y_n$ be $k$-algebra generators of $A(Y)$. Then
$\theta(y_1),\cdots,\theta(y_n) \in K(X)$ and we can find an open set $U$
of $X$ such that $\theta(y_i)$ is regular on $U$. Then by restricting
$\theta$ on $A(Y)$ we obtain a $k$-algebra homomorphism $A(Y) \rightarrow
\mathcal{O}(U)$.
So far, i have clear understanding.
But then Hartshorne says that the homomorphism $A(Y) \rightarrow
\mathcal{O}(U)$ is injective, and this is where i am stuck. Here is my
effort: Let $p(y)$ be inside the kernel. Then $\theta|_{A(Y)}(p(y))=0
\Rightarrow p(\theta(y))=0 \Rightarrow
p(\theta(y_1)(P),\cdots,\theta(y_n)(P))=0, \forall P \in X \Rightarrow
(\theta(y_1)(P),\cdots,\theta(y_n)(P)) \in \mathcal{Z}(p)$.
The goal should be to show that $p$ actually vanishes on $Y$, which would
be true if every point of $Y$ can be written in the form
$(\theta(y_1)(P),\cdots,\theta(y_n)(P))$. But why would this be true?