Let $\phi: R \to S$ be a ring homomorphism. Let $\phi^*:Spec(S) \to Spec(R)$ is induced map of sets. Here $Spec(S)$ is the set of prime ideals of a ring. Is $\phi^*$ surjective?
I think that it's not. Here is an contr-example I thought of. Let's think of inclusion $\phi:\mathbb{Z} \hookrightarrow \mathbb{Z}[i]$. And take the preimage $\phi^{*-1}$ of ideal $<5>$. It's the ideal $<5>$ but in $\mathbb{Z}[i]$ and it's not prime.
Is it ok or I miss something?
On
Of course this map is not surjective in general, but your counterexample does not work.
$(1+2i)$ is prime in $\mathbb Z[i]$ and lies over $(5)$. Actually the famous Going-Up-Theorem states that the induced map is surjective in that case.
A better counterexample would be the map $k[X] \to k, X \mapsto 0$ for any field $k$.
Preimage means inverse image. Given a prime ideal $P\subset\mathbb{Z}[i]$, the prime ideal that $\phi^*$ sends $P$ to is $\phi^{-1}(P)\subset\mathbb{Z}$. Remember, the map $$\mathrm{Spec}(S)=\{\text{prime ideals }P\subset S\}\;\xrightarrow{\;\;\;\;\phi^*\;\;\;\;}\;\{\text{prime ideals }Q\subset R\}=\mathrm{Spec}(R)$$ is defined by sending a prime ideal $P\subset S$ to the prime ideal $$\phi^{-1}(P)=\{x\in R:\phi(x)\in P\}\subset R.$$
So the fact that $\phi(5\mathbb{Z})=5\mathbb{Z}[i]$ is not a prime ideal of $\mathbb{Z}[i]$ is irrelevant to the discussion. And in fact, the inclusion $\phi:\mathbb{Z}\hookrightarrow\mathbb{Z}[i]$ does induce a surjective map $\phi^*:\mathrm{Spec}(\mathbb{Z}[i])\to\mathrm{Spec}(\mathbb{Z})$, though that is not a trivial fact.
An example where $\phi^*$ is not surjective is the inclusion $\phi:\mathbb{Z}\hookrightarrow\mathbb{Q}$. The set of prime ideals of $\mathbb{Q}$, which is to say, $\mathrm{Spec}(\mathbb{Q})$, is just the singleton set containing the zero ideal, $\{(0)\}$. The induced map $\phi^*:\mathrm{Spec}(\mathbb{Q})\to\mathrm{Spec}(\mathbb{Z})$ is the inclusion $$\phi^*:\{(0)\}\hookrightarrow\{(0),(2),(3),(5),(7),\ldots\}$$ which is obviously not surjective.