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$\varphi:R\rightarrow S$ is said to be a ring homomorphism if, $R,S$ are rings and $\varphi$ is a map such that:

$\varphi(r_{1}+r_{2})= \varphi(r_{1})+\varphi(r_{2})$,

$\varphi(r_{1}.r_{2})= \varphi(r_{1}).\varphi(r_{2})$.

$\varphi(1_{R})= 1_{S}$

In the definition of a ring isomorphism, the third condition isn't stated, but here, we need the third condition. My question is, why do we need to state explicitly that the identity in $R$ is mapped to the identity in $S$. Can't we just deduce it from the definitions of a ring homomorphism?

EDIT: In reference to Anurag A's comment that rings are sometimes defined without the existence of unity, all the rings I am looking at will have unity in them, so if the answers are restricted to the case for when they do have unity, that would be extremely helpful

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  • $\begingroup$ For an example where it fails consider the integers mod $6$ and the inclusion of the ideal generated by $3$. $\endgroup$ – Tobias Kildetoft Apr 27 '15 at 17:51
  • $\begingroup$ Some authors take the 3rd point as part of definition but some don't because many books do not include unity to be a necessary part of the ring definition itself. For example, authors who do not include unity as part of the definition of a ring will say $\mathbb{2Z}$ is a ring whereas some others will say it is not. $\endgroup$ – Anurag A Apr 27 '15 at 17:51
  • $\begingroup$ As I see it, you have that $\varphi{1_R} = 1_S$ is the identity of the ring $S$, that's becuase you must include the third condition in the definition of ring homomorphism $\endgroup$ – Alexei0709 Apr 27 '15 at 17:52
  • $\begingroup$ Depends on what properties we want for homomorphisms the third condition may or may not be included as part of the definition. For example the zero map can be valid homomorphism if we don't have the third condition. So one cannot say that it can be derived from the first two. $\endgroup$ – Anurag A Apr 27 '15 at 18:03
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No, $ℤ → ℤ × ℤ,~x ↦ (x,0)$ is not considered to be a ring homomorphism as it doesn’t preserve the identity element, yet is additive and multiplicative.

But an additive and multiplicative map of rings with identity elements $φ \colon R → S$ is at least guaranteed to be a ring homomorphism if it is surjective. That’s why you don’t need to require the identity elements to be preserved in the definition of a ring isomorphism – it’s automatically given.

This is because $\mathrm{img}~φ ⊂ S$ is a ring with identity $φ(1_R)$ as you can easily check for yourself. If $φ$ is surjective, it follows that $φ(1_R)$ is an identity for $S$ and from the uniqueness of identity elements $φ(1_R) = 1_S$.

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The unit of a ring is unique, so if $\varphi: R\to S$ preserves addition and multiplication, and has an inverse, then it must send $1_R$ to $1_S$.

As others have pointed out, there is no similar argument if $\varphi$ is not assumed to have an inverse. To take an extreme case, the zero map $R\to S$ satisfies the first two properties but not the third (unless $S=0$).

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We cannot deduce that $\varphi(1_R)=1_S$ for a map $\varphi: R\to S$ that preserves addition and multiplication. The main obstruction is the existence of idempotents in $S$.

Indeed, $\varphi(1_R)=\varphi(1_R^2)=\varphi(1_R)^2$ and so $\varphi(1_R)$ is an idempotent in $S$.

If $S$ has no nontrivial idempotents, then this forces $\varphi(1_R)=0$ or $\varphi(1_R)=1_S$.

$\varphi(1_R)=0$ implies $\varphi(r)=0$ for all $r \in R$. Otherwise, $\varphi(1_R)=1_S$.

So all is well when $\varphi$ is not the zero map and $S$ has no nontrivial idempotents.

If $S$ is commutative and has a nontrivial idempotent $e$, then $\varphi(x)=ex$ is a map $S \to S$ that preserves addition and multiplication but does not send $1_S$ to $1_S$.

A typical example is $S=\mathbb Z \times \mathbb Z$ and $e=(1,0)$.

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