I need to show that if $S$ and $R$ are domains, $\varphi \colon R \longrightarrow S$ is a ring homomorphism and $f\colon \text{Spec}(S) \longrightarrow \text{Spec}(R)$ is the induced map, then:
If $f$ is finite and dominant, then $f$ is surjective.
$f$ is finite, by definition, if $\varphi$ is finite. Finite ring homomorphisms are integral, so $S$ is integral over $R$. By Going Up, we know that for any prime ideal $P$ in $R$ there exists a prime ideal $Q$ in $S$ such that the contraction of $Q$ is equal to $P$. Then, we would obtain that $f$ is surjective.
I am not using that $f$ is dominant, so something must be wrong. Can someone help me?
Thanks in advance.
You used that $f$ is dominant implicitly when you used Going Up. Since $R,S$ are domains, the condition that $f$ is dominant is equivalent to $\varphi$ being injective.
There are finite, non-injective ring maps $\varphi : R \to S$ between integral domains that induce non-surjective maps on spectra, for example the quotient map $R \to R/\mathfrak{p}$ where $\mathfrak{p}$ is any nonzero prime ideal of $R$.