Counterexample of $f^* : \operatorname{Spec}B \to \operatorname{Spec}A$ injective implies every prime ideal of $B$ extended

Question

This is exercise 3.20, (ii) of Atiyah & Macdonald. $f^*$ is the induced map on $\operatorname{Spec}$ of $f: A \to B$, a ring homomorphism.

I have seen a counterexample on MathSE, stating that $k[t^2,t^3] \subset k[t]$ is one. But I cannot seem to understand why this is a counterexample. In this case $f$ must be the inclusion $k[t^2, t^3] \to k[t]$. How is the induced map $f^*$ injective in the first place? Also, how could I know about the prime ideals of $k[t]$ when I don't even know what $k$ is?

I have made this a separate question rather than a comment, since the original post is quite old. I would appreciate both hints or full descrpitions.

EDIT I have just realized that if $k$ is assumed to be algebraically closed, then the question becomes rather trivial. I am not deleting this post for future reference.

Answer 1

We will check that $k[t^2, t^3] \subseteq k[t]$ induces an injection on spectra for any ring $k$. (On the other hand, clear the prime ideal $t$ is not extended from $k[t^2, t^3]$.)

Set $I = (t^2, t^3)$.

The map $k[t^2, t^3] \rightarrow k[t]$ induced by the inclusion issuch that $(t^2, t^3) k[t] \subseteq k[t^2, t^3]$

Note that $I k[t] \subseteq k[t^2, t^3]$ and that $Ik[t] \cap k[t^2, t^3] = I$. Thus $k \subseteq k[t]/Ik[t]$.

This extension is just the result of adjoining a nilpotent, and we get a retract $k[t]/Ik[t] \rightarrow k$ by killing $t$. Since modding by nilpotents does not affect the spectrum, deduce that $k \subseteq k[t]/Ik[t]$ induces a homeomorphism on spectra.

Now suppose that $P_1, P_2$ are two primes of $k[t]$ contracting $Q$.

Assuming for the sake of contradiction that $P_1 \not= P_2$ without loss of generality we would be able to pick $f \in P_1 \setminus P_2$. Then $If \subseteq P_1 \cap k[t^2,t^3] = Q \subseteq P_2$. By primeness of $P_2$, deduce $I \subseteq P_2$, and therefore $I \subseteq Q$.

But then $P_1/I$ and $P_2/I$ would be too distinct primes of $k[t]/Ik[t]$ contracting to the prime $Q/I$ of $k$. This contradicts our earlier observation that the inclusion $k \subseteq k[t]/Ik[t]$ induces a homeomorphism on spectra.