The discussion centers on the properties of the kernel of a ring homomorphism and the concept of multiplicative cosets in relation to subrings. It is established that the kernel consists of elements mapping to the additive identity (0) rather than the multiplicative identity (1), as not all rings possess a multiplicative identity. The participants explore the implications of defining a set of elements that map to 1 and question whether multiplicative cosets of a subring can form a group. It is concluded that such cosets may not satisfy the necessary group properties without imposing significant restrictions.
PREREQUISITESMathematicians, algebraists, and students studying abstract algebra, particularly those interested in ring theory and group theory.
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?