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Say we have a ring $R$ and define the homomorphism $\varphi : \mathbb{Z} \rightarrow R$ to be $\varphi (z) = z \cdot 1$. For all $z \in \mathbb{Z}$, $z \cdot 1$ is in the center of $R$ (i.e. for all $r \in R, zr = rz$).

Say we define $$(z) = \{rz\ |\ r \in R\}$$ to be the principal ideal generated by $z$.

Do we know necessarily that $(z) \subseteq \mathbb{Z}$? My thought is that if we use $\mathbb{Q}$ as our ring, $(z)$ may contain rationals.

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  • $\begingroup$ First of all we need $\Bbb Z\subseteq R$, dont't we? But as your examplke with $R=\Bbb Q$ shows, we cannot assume $(z)\subseteq \Bbb Z$ without knowing specifics abour $R$. However, in case you take this from some context, please check if it is not implicitly assume that $R=\Bbb Z$. $\endgroup$
    – Hagen von Eitzen
    Sep 15, 2016 at 20:22
  • $\begingroup$ How exactly are you defining multiplication of integers by elements of $ R $? Are we to assume that a "compatible" (whatever that means) multiplication exists? In that case, your $ R = \mathbb Q $ is a counterexample to your own claim, as you've noticed... $\endgroup$
    – Ege Erdil
    Sep 15, 2016 at 20:23
  • $\begingroup$ You are right, but notice that $\mathbb{Q}$ is a field, so any of its ideals (principal or not) are either the trivial ideal $(0)$ or the entire field $(1)$. $\endgroup$
    – Fimpellizzeri
    Sep 15, 2016 at 20:25
  • $\begingroup$ I've made some edits to define multiplication of integers by elements of $R$. $\endgroup$
    – stackedtritones
    Sep 15, 2016 at 20:29
  • $\begingroup$ The question does not specify that $R = \mathbb{Z}$, but this is apparently the case. I'm not sure how to understand this, especially since the homomorphism is not necessarily surjective. $\endgroup$
    – stackedtritones
    Sep 15, 2016 at 20:52

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$(z)\subset \mathbb {Z }$ does not really make any sense, as $\{ rz| z \in R \} \subset R$. Even the identification of $\mathbb{Z}$ with the image of the natural map $\mathbb{Z}\to R$ is not really a thing as the image might just be $\mathbb{Z}/2\mathbb{Z}$. And you saw it yourself with the map $\mathbb{Z}\to \mathbb{Q}$. Here you see that by the properties of an ideal $z \in (z)$ therefore $ 1= \frac {1}{ z}·z \in (z)$, but if 1 is in an ideal everything is, so we have $(z)=\mathbb{Q}$.

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