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I'm studying an ideal $I \trianglelefteq R$ and noticed that for a certain non-injective, non-zero homomorphism $\varphi: R \rightarrow S$ I can show that $\varphi(I)=\varphi(R)$. I'm wondering if this implies that $I=R$.

It holds for the one little example I could think of. Let $R=\mathbb{Z}$, $S=\mathbb{Z}$, $I=k\mathbb{Z}$, and $\varphi_{m}(n)=mn$, $m > 1$. Suppose $\varphi_{m}(I)=\varphi_{m}(R)$. Then $\varphi_{m}(I)=\varphi_{m}(k\mathbb{Z})=mk\mathbb{Z}=\varphi_{m}(R)=\varphi_{m}(\mathbb{Z})=m\mathbb{Z}$. Thus, $mk\mathbb{Z}=m\mathbb{Z} \Rightarrow k=1 \Rightarrow I=\mathbb{Z}=R$.

(Rings not necessarily commutative and/or may not contain 1.)

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2 Answers 2

up vote 2 down vote accepted

Nah. Consider the quotient map of $\mathbb{Z}$ onto, say, $\mathbb{Z}_2$. The image of $3\mathbb{Z}$ is the whole ring. Since you went to the trouble of mentioning this shouldn't be injective, I'll note we could get non-surjectivity without breaking a sweat by making the codomain $\mathbb{Z}_2[X]$.

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Perfect. Thanks. –  Jackson Walters Aug 7 '12 at 4:32
    
@JacksonWalters A more general counterexample would be to consider $R \times R$ with projection onto the first element and simply note that $ R \times \{0\}$ is an ideal. –  JSchlather Aug 7 '12 at 4:40
    
@JacobSchlather Yup, I'm just curious now if I can generalize whatever property $\mathbb{Z}$ and its ideals have that allow my original example to work. –  Jackson Walters Aug 7 '12 at 4:58

Unless you require the homomorphism maps multiplicative unit to multiplicative unit, the trivial homomorphism from any non-trivial ring to any ring has the same image as that homomorphism applied to the zero ideal...

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Fair enough. What about with the qualifiers I added? –  Jackson Walters Aug 7 '12 at 4:24
    
What quantifiers? I can see none... –  DonAntonio Aug 7 '12 at 4:31

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