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?
The kernel of a ring homomorphism is the set of all elements in the domain that map to the additive identity (zero) in the codomain. It is denoted as Ker(f) or ker(f).
The kernel of a ring homomorphism is a subset of the elements in the domain of the homomorphism. It consists of all elements that are mapped to the additive identity in the codomain.
The kernel plays a crucial role in ring homomorphisms as it helps to determine the structure and properties of the homomorphism. It also helps to identify whether the homomorphism is injective, surjective, or bijective.
The kernel of a ring homomorphism is analogous to the null space in linear algebra. In both cases, it represents the set of elements that are mapped to the additive identity, or the zero vector, in the codomain.
Yes, the kernel of a ring homomorphism can be empty if the homomorphism is injective. This means that there are no elements in the domain that map to the additive identity in the codomain, resulting in an empty kernel.