find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z

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

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

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

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

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

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

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

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

What about trivial homomorphism?
– Thomas Shelby
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

@ThomasShelby okss
– lomber
Nov 21 at 9:31

add a comment |

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