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Let $\phi: R \rightarrow S$ be a surjective homomorphism of rings. Prove that the image of the center of $R$ is contained in the center of $S$.

Which is from Dummit and Foote Exercise 7.3.16. I have no problem with the proof:

$\{z \in R | z\cdot r = r\cdot z \text{ for all $r \in R$ } \}$ is the center of $R$. Given $r$ in the center of $R$ then $\phi(r)\in S$ is in the center of $S$. Since $\phi$ is surjective, there exists $z \in R$ s.t. $\phi(z) =s$ for all $s\in S$, thus $s \cdot \phi(r) = \phi(z)\cdot \phi(r) = \phi(z\cdot r) = \phi(r\cdot z) = \phi(r) \cdot \phi(z) = \phi(r) \cdot s$ for all $s \in S$.

The problem I'm working on has an extra line tacked on to it:

Show that the statement is false when $\phi$ is not surjective.

Surjectivity is a sufficient condition and allowed us to prove the first part. If any false cases existed, then necessarily the homomorphism must not be surjective. Am I supposed to prove a counterexample? What if $R \subsetneq S$ and $i:R \rightarrow S$ is the inclusion mapping. Then $i$ is not surjective and the image of the center of $R$ must necessarily be contained in the center of $S$. So there exists a non surjective homomorphism whose image of the ring center is central.

Since the image of the center of $R$ necessarily commutes with every element in $\phi(R)$, it must be contained in the center of $\phi(R)$. The only rings I could think of, given the limited types I've been introduced to, that were noncommutative and had pockets of elements that were commutative between themselves were matrix rings and quaternions. Quaternions were a little harder to get what I wanted.

All I could come up with is $\phi: \mathbb{Z} \rightarrow M_2(\mathbb{Z})$, where $M_2(\mathbb{Z})$ is the set of all $2\times 2$ matrices with entries in $\mathbb{Z}$ and $$a \mapsto \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} $$

Then $\phi(\mathbb{Z}) = \left\{ \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix}\Bigg| a\in \mathbb{Z} \right\} \nsubseteq \left\{ \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}\Bigg| a\in \mathbb{Z} \right\}$. I extrapolated from here that the center of $M_2(\mathbb{Z})$ is $\left\{\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}\Bigg| a\in \mathbb{Z} \right\}$.

Does this work as a counterexample? If so, is there a less trivial, more interesting example than this?

no lemon no melon
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    Your $\phi$ would not be considered a homomorphism of rings by most authors because it does not map $1$ to $1$. I don't know about Dummit & Foote because I have never read it. Anyway, I would use the following $\phi:\Bbb{C}\to M_2(\Bbb{R})$: $$\phi(a+bi)=\pmatrix{a&b\cr -b&a\cr}.$$ The center of $\Bbb{C}$ is itself, but it is easy to find for example diagonal matrices that don't commute with $\phi(i)$, so $\phi(i)$ is not in the center of $M_2(\Bbb{R})$. It is not too difficult to show that the center of $M_2(\Bbb{R})$ consists of the scalar multiples of $I_2$. – Jyrki Lahtonen Feb 06 '21 at 05:04
  • I have Abstract Algebra by Dummit and Foote, and it is not specified, and Abstract Algebra by Beachy and Blair and it is specified... and a plethora of other Abstract Algebra texts that I have but have just skimmed through. That's weird, I wonder why they don't specify it in D&F. The course I'm taking is using D&F and it was never mentioned in lectures either, I guess we're not using that part of the definition. Your example looks way more interesting than mine so thank you for that! – no lemon no melon Feb 06 '21 at 05:25
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    To clarify, your inclusion map example is correct, but you are just trying to find a counterexample, i.e. $i$ does not speak to the verity of the statement when $\varphi$ is not surjective. – Noah Solomon Feb 06 '21 at 05:49
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    @NoahSolomon the way I was interpreting it, is it’s not necessarily true, there are cases where the function is not surjective and the image of the ring center is central, and the image of the ring center is not central. However, yes the statement (*)“$\phi$ is not surjective, then the image of the ring center is central” is false, which fair enough can easily be interpreted from the above statement. But nonetheless I still found it confusing, maybe “Find a counterexample when $\phi$ is not surjective, thus “(*)” statement is false” would of been better at 1AM. Anyways big thanks! – no lemon no melon Feb 06 '21 at 14:27
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    If Dummit & Foote don't require a homomorphism of rings to map $1$ to $1$, then your example is fine. Their convention of choice would not surprise me given that they have adopted a number of other conventions I personally disprove of (so I have heard), but that is irrelevant to you for the purposes of this exercise. – Jyrki Lahtonen Feb 07 '21 at 19:26
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    Perhaps a better way to think about this is to prove that the assumption that $\phi$ is surjective cannot be relaxed, i.e., to exhibit rings $R$ and $S$ and a ring homomorphism $\phi : R \to S$ such that the image of the center of $R$ under $\phi$ is not contained in the center of $S.$ For this purpose, you do not need both of your rings to be unital. – Dylan C. Beck Feb 19 '21 at 16:55

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