Basic Question on Ring Homomorphisms

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SUMMARY

The discussion centers on the concept of ring homomorphisms, specifically regarding the mapping from the zero ring \( T \) to any arbitrary ring \( R \). A.J. Berrick and M.E. Keating in their book "An Introduction to Rings and Modules With K-Theory in View" assert that the map from the zero ring to \( R \) is not a ring homomorphism unless \( R \) is also the zero ring. The confusion arises from the interpretation of the "obvious" map \( f(0_T) = 0_R \), which participants argue appears to be a homomorphism. However, it is clarified that this leads to contradictions when \( R \) is not trivial, as \( 0_R \) cannot equal \( 1_R \).

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I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help to clarify a remark of B&K regarding ring homomorphisms from the zero or trivial ring ...

The relevant text from B&K reads as follows:
https://www.physicsforums.com/attachments/6078
https://www.physicsforums.com/attachments/6079In the above text from B&K's book we read ...

"... ... This follows from the observation that the obvious map from the zero ring to $$R$$ is not a ring homomorphism (unless $$R$$ itself happens to be $$0$$). ... ... "
I do not understand the above statement that the obvious map from the zero ring to $$R$$ is not a ring homomorphism (unless $$R$$ itself happens to be $$0$$) ... ... What, indeed do B&K mean by the obvious map from the zero ring to $$R$$ ... ... ?It seems to me that the obvious map is a homomorphism ... ..Consider the rings $$T, R$$ where $$T$$ is the zero ring and $$R$$ is any arbitrary ring ... so $$T = \{ 0 \}$$ where $$0 = 1$$ ...

Then to me it seems that the "obvious" map is $$f( 0_T) = 0_R$$ ... ... which seems to me to be a ring homomorphism ...

... BUT ... this must be wrong ... but why ... ?

Can someone please clarify the above for me ...

Some help will be very much appreciated ...

Peter
 
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Hi Peter,
Peter said:
Consider the rings $$T, R$$ where $$T$$ is the zero ring and $$R$$ is any arbitrary ring ... so $$T = \{ 0 \}$$ where $$0 = 1$$ ...

The above analysis is correct.

Peter said:
Then to me it seems that the "obvious" map is $$f( 0_T) = 0_R$$ ... ... which seems to me to be a ring homomorphism ...

... BUT ... this must be wrong ... but why ... ?

The "obvious" map isn't a ring homomorphism because if $R$ is not the trivial ring, then $0_{R}\neq 1_{R}$ and so the "obvious" homomorphism necessarily leads to a contradiction:

$$1_{R}\neq 0_{R}=f(0_{T})=f(1_{T})=1_{R}$$
 

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