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Let $\cal{C}$ be the category of $\textbf{Rings}$ and let $X$ and $Y$ be two objects in $\cal{C}$.

A coproduct of $X$ and $Y$ in $\cal{C}$ is an object $S$ of $\cal{C}$, together with morphisms $i: X \to S$ and $j: Y \to S$, with the following (universal) property:

for every object $Z$ of $\cal{C}$ and every pair of morphisms $f: X \to Z$ and $g: Y \to Z$, there exists a unique morphism $h: S \to Z$ such that $h\circ i=f$ and $h\circ j=g$.

The first issue is this: Based on the above definition we derive that $(R,\mathrm{id}_R,\mathrm{id}_R)$ is not a coproduct of $R$ and $R$.

Indeed, let $X=Y=R$ and $i=j=\mathrm{id}_R$. Then, if $(R,\mathrm{id}_R,\mathrm{id}_R)$ is not a coproduct of $R$ and $R$, by the universal property of coproduct we have that for every ring $Z$ and every morphisms $f: R \to Z$ and $g: R \to Z$ there exists a morphism $h:R\to Z$ such that $h=f=g$, which is obviously false.

The second issue is based in the following Proposition and states that $(\Bbb Z/n\Bbb Z, \mathrm{id}_{\Bbb Z/n\Bbb Z},\mathrm{id}_{\Bbb Z/n\Bbb Z})$ is a coproduct of $\Bbb Z/n\Bbb Z$ and $\Bbb Z/n\Bbb Z$:

Let $m$ and $n$ be positive integers, and let $d$ be their greatest common divisor. Let $i: \Bbb Z/m\Bbb Z → \Bbb Z/d\Bbb Z$ and $j: \Bbb Z/n\Bbb Z → \Bbb Z/d\Bbb Z$ be the canonical ring homomorphisms. Then $(\Bbb Z/d\Bbb Z, i, j)$ is a coproduct of $\Bbb Z/m\Bbb Z$ and $\Bbb Z/n\Bbb Z$ in the category $\cal{C}$.

Now taking $n=m$ we derive that $(\Bbb Z/n\Bbb Z, \mathrm{id}_{\Bbb Z/n\Bbb Z},\mathrm{id}_{\Bbb Z/n\Bbb Z})$ is a coproduct of $\Bbb Z/n\Bbb Z$ and $\Bbb Z/n\Bbb Z$.

Obviously, these issues contradict themselves. Could someone point out what' s wrong with the above statements?

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    $\begingroup$ The question in your title is independent from what you ask in the body of your post. Please fix this. $\endgroup$ – Pedro Tamaroff Mar 2 at 22:51
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    $\begingroup$ "Which is obviously false" : perhaps you should review your notion of "obviously false", so as not to include true statements. Can you tell us what a ring morphism $\mathbb{Z}\to R$ looks like ? (For any ring $R$) $\endgroup$ – Max Mar 2 at 22:54
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    $\begingroup$ Does your category consist of only commutative rings? Otherwise, the notation $\otimes$ for coproducts is rather unusual, though it doesn't make a difference for this particular example. $\endgroup$ – Eric Wofsey Mar 2 at 23:00
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    $\begingroup$ I rolled back your second edit to the title, since I'm almost positive that direct sum is not what you meant. If you prefer not to use the tensor product, you can use $\coprod$ (right click on mathjax to see the latex). I see that it's been changed back to direct sum. This doesn't make much sense. $\endgroup$ – jgon Mar 2 at 23:09
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    $\begingroup$ @jgon \amalg ($\amalg$) can be used to get an operator more suitable for infix use. \coprod is intended to be used for indexed usages like \prod or \sum. $\endgroup$ – Derek Elkins Mar 3 at 1:35
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Then, if $(R,\mathrm{id}_R,\mathrm{id}_R)$ is not a coproduct of $R$ and $R$, by the universal property of coproduct we have that for every ring $Z$ and every morphisms $f: R \to Z$ and $g: R \to Z$ there exists a morphism $h:R\to Z$ such that $h=f=g$, which is obviously false.

Why is this "obviously false"? It is indeed false for most rings, and it is not hard to come up with counterexamples. But this doesn't mean it's false for every ring, and $\mathbb{Z}/n\mathbb{Z}$ is a very special ring. Indeed, for any ring $Z$, there is at most one homomorphism $\mathbb{Z}/n\mathbb{Z}\to Z$, so in your situation above with $R=\mathbb{Z}/n\mathbb{Z}$ you must in fact have $f=g$.

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  • $\begingroup$ The fact that there is at most one homomorphism $f:\Bbb Z/n\Bbb Z→Z$ follows from the requirement of a ring homomorphism to map $1\mod n$ to $1_Z$. Why is that a requirement? $\endgroup$ – richarddedekind Mar 3 at 11:42
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No they don't, since for your ring $\mathbb Z/n$ there is only one ring endomorphism, the identity, so you cannot choose $f$ and $g$ distinct. Your argument works only if there is a non-trivial ring endomorphism $R\to R$.

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  • $\begingroup$ The fact that there is only one ring endomorphism of $\Bbb Z/n\Bbb Z$ follows from the requirement of a ring homomorphism $R\to S$ to map $1_R$ to $1_S$. Why is that a requirement? $\endgroup$ – richarddedekind Mar 3 at 11:47
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To add to Eric Wofsey and Pedro Tamaroff's excellent answers, you're missing a fundamental point about the tensor product, or perhaps being careless, and this is the source of the confusion.

Disclaimer I mostly think about commutative rings, so this will be for commutative rings rather than rings as a whole. (Particularly because your usage of the tensor product suggests commutative rings)

The answer

The tensor product is relative to a base ring, after all it's fundamentally a construction for modules that happens to be useful for rings as well.

Thus $R\otimes R$ is underspecified (without more context). You should say $R\otimes_R R$ if you mean the tensor product that's isomorphic to $R$. However, this is generally not the coproduct in the category of (commutative) rings. The correct coproduct is $R\otimes_{\Bbb{Z}} R$. For certain rings, (when $\Bbb{Z}\to R$ is an epimorphism in the category of commutative rings), the natural map $R\otimes_{\Bbb{Z}} R\to R\otimes_R R$ is an isomorphism, and this is so when $R=\Bbb{Z}/(n)$, which is why $\Bbb{Z}/(n)$ is the coproduct of two copies of itself.

For more on this, you might be interested in my answer to a related question and a link there to a related MO discussion.

Side note

The general answer here is that $R$ is a coproduct for $R$ and $R$ if and only if the natural map $R \otimes_{\Bbb{Z}} R \to R$ is an isomorphism if and only if the map $\Bbb{Z}\to R$ is an epimorphism.

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