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Let $R$ and $S$ be rings with identity (both are not necessarily commutative). The standard definition of a ring homomorphism $\varphi:R\rightarrow S$ is a function which respects the addition and multiplication operations of $R$ and $S$. It is obvious that $\varphi(1)$ is the identity element of the subring $\varphi(R)$ of $S$. But it is not necessarily the identity element of $S$. A trivial example is the homomorphism which maps each element of $R$ to $0\in S$ (assuming $S\neq\left\{ 0\right\}$). A more interesting example is the ring homomorphism $\varphi:\mathbb{Z}\rightarrow\mathbb{Z}_{6} $ defined by $\varphi(x)=3x$ mod $6$ for all $x\in\mathbb{Z}$. I have noticed that some texts, in particular, Commutative Algebra texts, require that a ring homomorphism must map $1\in R$ to $1\in S $, ruling out the previously mentioned homomorphisms. I understand that one reason this is done is so that $R-$algebras can be easily made into $R-$modules. So my question is, how restrictive is this extra condition on ring homomorphisms? Are there any important homomorphisms ruled out when this condition is made, or do pretty much all important homomorphisms satisfy this condition ?

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  • $\begingroup$ Do $R$ and $S$ have a unit? See also this question. $\endgroup$ – Dietrich Burde Aug 27 '16 at 11:57
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    $\begingroup$ Given any homomorphism like yours, you can decompose it into a standard homomorphism from $R$ to its image $f(R)$ and the inclusion map of $f(R)$ into $S$. This inclusion map is a "homomorphism" that doesn't map 1 to 1. Is it important that the definition of "subring" include "the subring must contain 1"? In most cases, it is. $\endgroup$ – mathguy Aug 27 '16 at 12:46
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A morphism should always preserve all the structure that is important. If you are discussing rings which possibly don't have units, then it doesn't make sense (it is not even well-defined) to require that $1$ maps to $1$. But if your definition of a ring requires a unit (as is the case in standard commutative algebra), then you should require it to be preserved in the definition of a morphism.

A similar example are ring morphisms which are not morphisms of local rings (i.e. they do not preserve the unique maximal ideal of the rings involved), as for example the inclusion $\mathbb{Z}_3 \to \mathbb{Q}$ (note that $\mathbb{Z}_3$ is the localization, not the quotient). If you discuss rings, this is a morphism. If you want to focus on local rings, it is not.

The reason why one draws this kind of distinction is that if you want to use things from (basic) category theory like products and coproducts (for instance), then the morphisms involved should really be morphisms of the category you are interested in, otherwise you don't get what you want.

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  • $\begingroup$ I believe that the category-theoretical reason is the most fundamental motivation to define the morphisms in a certain way. (However, it is not necessarily so easy to understand if you haven't worked at least a bit with categories...) $\endgroup$ – Daniel Robert-Nicoud Aug 27 '16 at 12:47
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    $\begingroup$ Sure, I agree. Still, the idea that morphisms should preserve "everything that matters" is understandable enough to be at least believable, as a general concept. $\endgroup$ – 57Jimmy Aug 27 '16 at 12:58
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    $\begingroup$ @Daniel: I think in this context, formal logic is the fundamental justification -- if the constant symbol $1$ is part of the language of the theory, then homomorphisms between models had better map $1$ to $1$. $\endgroup$ – user14972 Aug 31 '16 at 17:52
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Requiring that ring homomorphisms map $1$ to $1$ definitely leads to a good theory; so much so that many mathematicians consider that to be the standard for what it means to be a ring homomorphism (as well as the requirement that rings have $1$) — and they would give a different name (e.g. "rng" or "$\mathbb{Z}$-algebra") to the notion that you call a ring (and similarly, "rng homomorphism" or "algebra homomorphism" for what you call ring homomorphism).

I would even say the majority of mathematicians are of this type, and using "ring" for the more general notion is old-fashioned (but apparently still taught often enough), but my perception may just be a function of the types of mathematics that interest me.

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