find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
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find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
My answer : one
let $f$ be a ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
we know that $mathbb{Z_2} times mathbb{Z_2}= K_4$ where $K_4$ is Klien Group
$K_4 = {a^2 =b^2 =c^2 =e text{and} c=ab}$
There is only one idempotent element $e$ and there are $3$ nilpotent elements $a,b,c$ .
Now in $mathbb{Z_4} $ we have
$0^2=0,$
$1^2=1,$
$2^2=0,$
$3^2= 1$
so, $0$ and $1$ are idempotent elements
we know that in a ring Homomorphism idempotent map to idempotent and nilpotent map to nilpotents.
so here I get only one ring homorphism by the diagram given below
$K_4 mathbb{Z_4}$
$e rightarrow 0$
$ 1$
$(a,b,c) rightarrow 2$
$ 3$
Is its correct ?
Is their any mistake in my solution?
thanks u
abstract-algebra
asked Nov 21 at 1:22
lomber
765219
add a comment |
up vote
0
down vote
favorite
find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
My answer : one
let $f$ be a ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
we know that $mathbb{Z_2} times mathbb{Z_2}= K_4$ where $K_4$ is Klien Group
$K_4 = {a^2 =b^2 =c^2 =e text{and} c=ab}$
There is only one idempotent element $e$ and there are $3$ nilpotent elements $a,b,c$ .
Now in $mathbb{Z_4} $ we have
$0^2=0,$
$1^2=1,$
$2^2=0,$
$3^2= 1$
so, $0$ and $1$ are idempotent elements
we know that in a ring Homomorphism idempotent map to idempotent and nilpotent map to nilpotents.
so here I get only one ring homorphism by the diagram given below
$K_4 mathbb{Z_4}$
$e rightarrow 0$
$ 1$
$(a,b,c) rightarrow 2$
$ 3$
Is its correct ?
Is their any mistake in my solution?
thanks u
abstract-algebra
asked Nov 21 at 1:22
lomber
765219
1
What about trivial homomorphism?
– Thomas Shelby
Nov 21 at 1:47
1
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 at 1:47
@RobArthan thanks u
– lomber
Nov 21 at 9:31
@ThomasShelby okss
– lomber
Nov 21 at 9:31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
My answer : one
let $f$ be a ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
we know that $mathbb{Z_2} times mathbb{Z_2}= K_4$ where $K_4$ is Klien Group
$K_4 = {a^2 =b^2 =c^2 =e text{and} c=ab}$
There is only one idempotent element $e$ and there are $3$ nilpotent elements $a,b,c$ .
Now in $mathbb{Z_4} $ we have
$0^2=0,$
$1^2=1,$
$2^2=0,$
$3^2= 1$
so, $0$ and $1$ are idempotent elements
we know that in a ring Homomorphism idempotent map to idempotent and nilpotent map to nilpotents.
so here I get only one ring homorphism by the diagram given below
$K_4 mathbb{Z_4}$
$e rightarrow 0$
$ 1$
$(a,b,c) rightarrow 2$
$ 3$
Is its correct ?
Is their any mistake in my solution?
thanks u
abstract-algebra
asked Nov 21 at 1:22
lomber
765219
find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
My answer : one
let $f$ be a ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
we know that $mathbb{Z_2} times mathbb{Z_2}= K_4$ where $K_4$ is Klien Group
$K_4 = {a^2 =b^2 =c^2 =e text{and} c=ab}$
There is only one idempotent element $e$ and there are $3$ nilpotent elements $a,b,c$ .
Now in $mathbb{Z_4} $ we have
$0^2=0,$
$1^2=1,$
$2^2=0,$
$3^2= 1$
so, $0$ and $1$ are idempotent elements
we know that in a ring Homomorphism idempotent map to idempotent and nilpotent map to nilpotents.
so here I get only one ring homorphism by the diagram given below
$K_4 mathbb{Z_4}$
$e rightarrow 0$
$ 1$
$(a,b,c) rightarrow 2$
$ 3$
Is its correct ?
Is their any mistake in my solution?
thanks u
abstract-algebra
abstract-algebra
asked Nov 21 at 1:22
lomber
765219
asked Nov 21 at 1:22
lomber
765219
asked Nov 21 at 1:22
lomber
765219
asked Nov 21 at 1:22
lomber
765219
asked Nov 21 at 1:22
lomber
765219
765219
1
What about trivial homomorphism?
– Thomas Shelby
Nov 21 at 1:47
1
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 at 1:47
@RobArthan thanks u
– lomber
Nov 21 at 9:31
@ThomasShelby okss
– lomber
Nov 21 at 9:31
add a comment |
1
What about trivial homomorphism?
– Thomas Shelby
Nov 21 at 1:47
1
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 at 1:47
@RobArthan thanks u
– lomber
Nov 21 at 9:31
@ThomasShelby okss
– lomber
Nov 21 at 9:31
1
1
What about trivial homomorphism?
– Thomas Shelby
Nov 21 at 1:47
What about trivial homomorphism?
– Thomas Shelby
Nov 21 at 1:47
1
1
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 at 1:47
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 at 1:47
@RobArthan thanks u
– lomber
Nov 21 at 9:31
@RobArthan thanks u
– lomber
Nov 21 at 9:31
@ThomasShelby okss
– lomber
Nov 21 at 9:31
@ThomasShelby okss
– lomber
Nov 21 at 9:31
add a comment |
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1
What about trivial homomorphism?
– Thomas Shelby
Nov 21 at 1:47
1
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 at 1:47
@RobArthan thanks u
– lomber
Nov 21 at 9:31
@ThomasShelby okss
– lomber
Nov 21 at 9:31