# Basic Question on Ring Homomorphisms

• MHB
• Math Amateur
In summary, B&K state that the "obvious" map from the zero ring to R is not a ring homomorphism unless R itself is the trivial ring. They then provide an explanation for this statement, clarifying that if R is not the trivial ring, then the "obvious" map leads to a contradiction.
Math Amateur
Gold Member
MHB
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

Hi Peter,
Peter said:
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.

Peter said:
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:

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

## What is a ring homomorphism?

A ring homomorphism is a function between two rings that preserves the operations of addition and multiplication. This means that for any two elements in the first ring, their sum and product will also be preserved in the second ring.

## What are the properties of a ring homomorphism?

The properties of a ring homomorphism include preserving the ring's identity elements, distributive property, and inverses for addition. It also preserves the multiplicative identity element and the associativity property for multiplication.

## What is the difference between a ring homomorphism and a ring isomorphism?

A ring isomorphism is a special type of ring homomorphism where the function is also bijective, meaning it is both one-to-one and onto. This results in a one-to-one correspondence between the elements of the two rings, making them essentially equivalent in structure.

## What is the kernel of a ring homomorphism?

The kernel of a ring homomorphism is the set of elements in the first ring that are mapped to the additive identity element in the second ring. In other words, it is the set of elements that get mapped to 0 in the second ring when the homomorphism is applied.

## How are ring homomorphisms used in algebra and number theory?

Ring homomorphisms are used in algebra and number theory to study the structure and properties of rings. They are especially useful in understanding how two different rings may be related and can be used to prove theorems and solve problems involving rings and their elements.

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