I am interested of finding examples of non-zero homomorphisms $f:R\to S$ of rings with unity such that $f(1_R)\neq 1_S$.
I will provide one example and I will be glad if others can also give examples.
Non-zero non-unitary ring homomorphisms [collecting examples]
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abstract-algebra
ring-theory
ring-homomorphism
2 Answers
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Let be $f:\mathbb Z \to M_2(\mathbb Z); f(a)= \left( \array{a&0\\0&0}\right) $ then $f$ is ring homomorphism and $f(1)=\left(\array{1&0\\0&0}\right) \ne \left(\array{1&0\\0&1}\right)$ .
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Let $R=\mathbb{Z}$ and $S=\mathbb{Z}\times \mathbb{Z}$. Define $f:R\to S$ by $f(n)=(n,0)$ for all $n\in\mathbb{Z}$. Then $f$ is a nonzero homomorphism but $$f(1_R)=f(1)=(1,0)\neq (1,1)=1_S.$$
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0You can imitate your example for any general ring $R$. Define $f\colon R\to R\times R$ by $f(x) =(x,0)$. – 2017-01-02
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0@P Vanchinathan Thanks. It is a nice observation. – 2017-01-02
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0You are asking for "rng" homomorphisms (by some definitions, _by definition_ $1_R\mapsto 1_S$). A further type of example is the homomorphism $f:R\to S:x\mapsto ex$, where $S=eR$ and $e$ is a nontrivial idempotent element of $R$. – 2017-01-02