I Kernel of a ring homomrphism

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We know that kernel of a homomorphism consists of all the elements that map to the additive identity, 0. Here is my naive question: Why don't we define the kernel as all of the elements that map to the multiplicative identity, 1? Why isn't there a name for the set of all elements that map to the multiplicative identity?
 

fresh_42

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We know that kernel of a homomorphism consists of all the elements that map to the additive identity, 0. Here is my naive question: Why don't we define the kernel as all of the elements that map to the multiplicative identity, 1? Why isn't there a name for the set of all elements that map to the multiplicative identity?
Because there is none!

The kernel of a homomorphism is what is mapped to the neutral element, whatever this is. In multiplicative groups such as automorphism groups, it is ##1##, and in rings it is ##0##. Not all rings have a ##1## and most of all: most elements do not multiply to ##1##. You can investigate this on your own. Take a ring homomorphism (for the sake of simplicity a commutative ring with ##1## and an endomorphism) ##\varphi\, : \,R \longrightarrow R## and see which properties ##N:=\{\,r\in R\,|\,\varphi(r)=1\,\}## has. Is it an ideal, so that we can factor ##R/N## what we usually want to do with a kernel? Is it at least a subring? What happens if we multiply it with ideals, etc.?
 
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Because there is none!

The kernel of a homomorphism is what is mapped to the neutral element, whatever this is. In multiplicative groups such as automorphism groups, it is ##1##, and in rings it is ##0##. Not all rings have a ##1## and most of all: most elements do not multiply to ##1##. You can investigate this on your own. Take a ring homomorphism (for the sake of simplicity a commutative ring with ##1## and an endomorphism) ##\varphi\, : \,R \longrightarrow R## and see which properties ##N:=\{\,r\in R\,|\,\varphi(r)=1\,\}## has. Is it an ideal, so that we can factor ##R/N## what we usually want to do with a kernel? Is it at least a subring? What happens if we multiply it with ideals, etc.?
It seems that ##N## is not even a subring, since it is not closed under addition.

A related question I have is it possible to form a ring structure out of the multiplicative cosets of a subring?
 

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Can you explain what you mean by multiplicative cosets? Let's say we have a subring ##S \subseteq R##. Then we can build ##R/S## as a set. If we want to define ##(p+S)\cdot (q+S) = pq + pS +Sq +S = pq+S## we need, ##pS\, , \,Sq \subseteq S##, that is ##S## should be an ideal. In this case ##R/S## is a ring. Now what did you want to do?
 
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Can you explain what you mean by multiplicative cosets? Let's say we have a subring ##S \subseteq R##. Then we can build ##R/S## as a set. If we want to define ##(p+S)\cdot (q+S) = pq + pS +Sq +S = pq+S## we need, ##pS\, , \,Sq \subseteq S##, that is ##S## should be an ideal. In this case ##R/S## is a ring. Now what did you want to do?
Suppose that ##S## is a subring of ##R##. Then if ##r\in R## we can look at the set ##rS = \{rs\mid s\in S\}##. can these multiplicative cosets form a group?
 

fresh_42

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Suppose that ##S## is a subring of ##R##. Then if ##r\in R## we can look at the set ##rS = \{rs\mid s\in S\}##. can these multiplicative cosets form a group?
What should be the inverse to ##0\cdot S\,##? I guess you will have to enforce so many restrictions, that you will end up with multiplicative groups.
 

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