I Kernel of a ring homomrphism

Mr Davis 97

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?

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fresh_42

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2018 Award
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.?

Mr Davis 97

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?

fresh_42

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2018 Award
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?

Mr Davis 97

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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2018 Award
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.

"Kernel of a ring homomrphism"

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