Basic Question on Ring Homomorphisms

[Math Amateur]

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

Then to me it seems that the "obvious" map is \(\displaystyle f( 0_T) = 0_R\) ... ... 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

 

[GJA]

Hi Peter,

Consider the rings \(\displaystyle T, R\) where \(\displaystyle T\) is the zero ring and \(\displaystyle R\) is any arbitrary ring ... so \(\displaystyle T = \{ 0 \}\) where \(\displaystyle 0 = 1\) ...

The above analysis is correct.

Then to me it seems that the "obvious" map is \(\displaystyle f( 0_T) = 0_R\) ... ... 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:

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