The discussion revolves around the concept of kernels in ring homomorphisms and the possibility of forming groups from multiplicative cosets of a subring. Participants explore definitions, properties, and implications of these mathematical structures, focusing on both theoretical and conceptual aspects.
Participants express differing views on the definitions and properties of kernels in homomorphisms, as well as the feasibility of forming groups from multiplicative cosets. The discussion remains unresolved, with multiple competing perspectives presented.
Limitations include the dependence on definitions of kernels and ideals, as well as the unresolved nature of whether certain sets can be classified as subrings or ideals under specific conditions.
Because there is none!Mr Davis 97 said: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?
It seems that ##N## is not even a subring, since it is not closed under addition.fresh_42 said: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.?
fresh_42 said: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?
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.Mr Davis 97 said: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?