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I have the category of pointed rings. Objects are all pairs $(R, r)$ where $R$ - ring (with 1) and r is the element of R. Morphisms are homomorphism of rings. Morphism $(R, r) \longrightarrow (R', r')$ exist if exist homomorphism p: $R \longrightarrow R'$ and $p(r)=r'$.

Can somebody help to find initial object or proof that it doesn't exist? I think that it doesn't exist but can't find a way to proof.

Thanks.

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    $\begingroup$ So what are the morphisms here? What is a morphism from $(R, r)$ to $(S, s)$ in this category? Is it merely a ring homomorphism from $R$ to $S$? (No; why?) $\endgroup$ – M. Vinay Mar 20 '19 at 7:40
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    $\begingroup$ Okay, you answered that part in your updated question. That leads to the answer. $\endgroup$ – M. Vinay Mar 20 '19 at 7:41
  • $\begingroup$ At least intuitively, it seems to make more sense that it would be the ring formed by the integers with the usual multiplication/etc., with any integer as the fixed point. After all, we can't have a ring with no elements, and the initial object in the category of rings is the usual ring of the integers, so I'm not exactly sure how there would be a difference if you fix a particular point. (This as opposed to the category of sets - with initial object the empty set - and the category of pointed sets - with initial object a singleton set.) Maybe I'm overlooking something obvious though. $\endgroup$ – Eevee Trainer Mar 20 '19 at 7:41
  • $\begingroup$ Homomorphism can't just change one element to another. It must save structure unlike pointed sets $\endgroup$ – Ivan Sharapenkov Mar 20 '19 at 7:53
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    $\begingroup$ I would presume it is $(\mathbb{Z}\langle x \rangle , x)$ given that the property he wants is more or less the universal property of the free algebra over $\mathbb{Z}$. (every ring with $1$ has a canonical map from $\mathbb{Z}$ into it and the marked point gives the evaluation of $x$, so the univ property of the free algebra gives us a unique morhism from $\mathbb{Z}\langle x \rangle$ into any pointed ring sending $x$ to $r$) $\endgroup$ – Enkidu Mar 20 '19 at 9:10
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It is up to unique isomorphism $(\mathbb{Z}[x],x)$ as this has the universal property that any assignement $x \mapsto r\in R$ gives a unique morphism $$\mathbb{Z}[x] \to R$$ hence any choice of $(r,R)$ gives you such a unique map and hence it is the initial object.

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  • $\begingroup$ Thank you very much. $\endgroup$ – Ivan Sharapenkov Mar 20 '19 at 15:12
  • $\begingroup$ you are welcome, sorry that i did not post it as answer immediately but as comment, those homological algebra arguments usually blow up quite fast, so I just like to give pointers, but turned out here it was actually a 3 liner. Also I was confuesd as being the first one to see this. $\endgroup$ – Enkidu Mar 20 '19 at 15:17

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