Here are the definitions that we use for this problem:
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A morphism $\varphi : X \to Y$ between two varieties is said to be dominant if the image of $\varphi$ is dense in $Y$ (c.f. Hartshorne exercise 1.3.17)
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We say $X$ is an affine variety if it is an irreducible closed subset of $\Bbb{A}^n$. (see the definition after example 1.1.4 of Hartshorne)
Now let $X$ and $Y$ be affine varieties and $\varphi : X \to Y$ a dominant morphism. Now this means that for all $p \in X$, $\varphi^\ast_p : \mathcal{O}_{\varphi(p),Y} \to \mathcal{O}_{p,X}$ is injective (c.f. Hartshorne exercise 1.3.3 (c)). Because we have assumed that $X,Y$ are affine varieties we identify $\mathcal{O}_{p,X} \cong A(X)_{\mathfrak{m}_p}$ and similarly for $Y$ to get that
$$(\varphi^\ast)_p : A(Y)_{\mathfrak{m}_{\varphi(p)}} \to A(X)_{\mathfrak{m}_p}$$
is injective for all $p\in X$.
My question is: Is it true that $\varphi^\ast : A(Y) \to A(X)$ is injective?
Here are some partial results I can get. Choose a maximal ideal $\mathfrak{m}_q \subseteq A(Y)$. If $\ker \varphi^\ast$ is completely contained in $\mathfrak{m}_q$ then
$$(\mathfrak{m}_q)_{\mathfrak{m}_q} (\ker \varphi^\ast)_{\mathfrak{m}_q} = (\ker \varphi^\ast)_{\mathfrak{m}_q}.$$
The ring $A(Y)_{\mathfrak{m}_q}$ is local and Noetherian so $(\ker \varphi^\ast)_{\mathfrak{m}_q}$ is finitely generated as a module over this ring. By Nakayama's lemma it follows $(\ker \varphi^\ast)_{\mathfrak{m}_q} =0$.
Edit: Adeel's answer below shows that under my assumptions we have $\ker \varphi \subseteq \text{nilrad} A(Y)$, so my proof is complete.
Best Answer
Proposition: If $X = \mathrm{Spec}(A)$ and $Y = \mathrm{Spec}(B)$ are affine schemes, a morphism $f : X \to Y$ is dominant if and only if the kernel of the corresponding homomorphism $\varphi : B \to A$ is contained in the nilradical of $B$.
Proof: Recall that there is an equality $$ V(\varphi^{-1}(I)) = \overline{f(V(I))} $$ for ideals $I \subset A$; in particular, $$ V(\varphi^{-1}(0)) = \overline{f(X)} = Y $$ if and only if $f$ is dominant. This is equivalent to the ideal $\varphi^{-1}(0) = \ker(\varphi)$ being contained in every prime ideal, i.e. in the intersection of all the prime ideals; but this intersection is precisely the nilradical of $B$.
In particular when Y is reduced (e.g. a variety), the homomorphism is injective.