Preimage of prime ideal under induced map

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?

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.

• Thank you! Is there a more or less simple non-trivial contr-example? Something more interesting than $\mathbb{Q}$? – Stan Feb 2 '15 at 9:53
• $\mathbb C[X] \hookrightarrow \mathbb C[X,Y]/(XY-1)$ is the common "non-trivial" counterexample. You have to think of the induced map as the projection from the hyperbola to the x-axis. The origin does not occur in the image. – MooS Feb 2 '15 at 10:01
• Thnx again. And let me ask one more question just to make sure I understand it all correct. My contrexample is still suitable for showing that $\phi^*$ is not injective? (as well as $\phi : \mathbb{Z} \to \mathbb{Z}, x \mapsto 2x$, here $<3>$ has at least two preimages under $\phi^*$: $<3>$ and $<2>$) – Stan Feb 2 '15 at 17:33
• @stasetz: $\newcommand{\Z}{\mathbb{Z}}$ Your map $\phi:\Z\to\Z$ sending $x$ to $2x$ is not even a ring homomorphism, since part of the definition of a ring homomorphism $f:R\to S$ is that $f(1_R)=1_S$. Moreover, for any sets $X$, $Y$ and function $f:X\to Y$, the preimage of a subset $A\subseteq Y$ is always one single subset of $X$. In the case of your function $\phi:\Z\to\Z$, for the subset $(3)\subset\Z$ it's \begin{align*}\phi^{-1}(3)&=\{x\in\Z: \phi(x)\in(3)\}\\&=\{x\in\Z:2x\text{ is a multiple of 3}\}\\&=\{x\in\Z:x\text{ is a multiple of 3}\}\\&=(3) \end{align*} – Sal Feb 2 '15 at 17:47
• really sorry for such stupid propositions :) Now, applying this way of reasoning for $\phi : \mathbb{Z} \hookrightarrow \mathbb{Z}[i]$ I see that $5 \in \phi^{-1}(<1+2i>)$ and $5 \in \phi^{-1}(<1-2i>)$ and as these preimages should be prime ideals it turns out that they are the same and we get two ideals in $\mathbb{Z}[i]$ going to the same ideal <5> under $\phi^*$, right? – Stan Feb 2 '15 at 18:15

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 $k$.

• Thank you! Can't rate it now as I don't have enough rep for it :) – Stan Feb 2 '15 at 10:05