Mixing
Constant Endomorphism
$begingroup$
Is the constant map $f$:M $longrightarrow$ M
with m $longmapsto$ a
is an endomorphism , where M is a module?
let $m ,m' in M,$ we have $f(m)=f(m')=a $ then $f(m)+f(m')=2a $
but $f(m+m')=a$
I think it isn't an homomorphism
proof-verification modules ring-homomorphism
edited Jan 19 at 22:11
Scientifica
6,79641335
asked Jan 19 at 22:00
Ali NoumSali TraoreAli NoumSali Traore
6218
$endgroup$
add a comment |
$begingroup$
Is the constant map $f$:M $longrightarrow$ M
with m $longmapsto$ a
is an endomorphism , where M is a module?
let $m ,m' in M,$ we have $f(m)=f(m')=a $ then $f(m)+f(m')=2a $
but $f(m+m')=a$
I think it isn't an homomorphism
proof-verification modules ring-homomorphism
edited Jan 19 at 22:11
Scientifica
6,79641335
asked Jan 19 at 22:00
Ali NoumSali TraoreAli NoumSali Traore
6218
$endgroup$
1
$begingroup$
The answer is trivially “no” if $a$ is nonzero. You should be able to see why. Where are you stuck?
$endgroup$
– rschwieb
Jan 19 at 22:04
$begingroup$
yes , thanks !!
$endgroup$
– Ali NoumSali Traore
Jan 19 at 22:14
add a comment |
$begingroup$
Is the constant map $f$:M $longrightarrow$ M
with m $longmapsto$ a
is an endomorphism , where M is a module?
let $m ,m' in M,$ we have $f(m)=f(m')=a $ then $f(m)+f(m')=2a $
but $f(m+m')=a$
I think it isn't an homomorphism
proof-verification modules ring-homomorphism
edited Jan 19 at 22:11
Scientifica
6,79641335
asked Jan 19 at 22:00
Ali NoumSali TraoreAli NoumSali Traore
6218
$endgroup$
Is the constant map $f$:M $longrightarrow$ M
with m $longmapsto$ a
is an endomorphism , where M is a module?
let $m ,m' in M,$ we have $f(m)=f(m')=a $ then $f(m)+f(m')=2a $
but $f(m+m')=a$
I think it isn't an homomorphism
proof-verification modules ring-homomorphism
proof-verification modules ring-homomorphism
edited Jan 19 at 22:11
Scientifica
6,79641335
asked Jan 19 at 22:00
Ali NoumSali TraoreAli NoumSali Traore
6218
edited Jan 19 at 22:11
Scientifica
6,79641335
asked Jan 19 at 22:00
Ali NoumSali TraoreAli NoumSali Traore
6218
edited Jan 19 at 22:11
Scientifica
6,79641335
edited Jan 19 at 22:11
Scientifica
6,79641335
edited Jan 19 at 22:11
Scientifica
6,79641335
6,79641335
asked Jan 19 at 22:00
Ali NoumSali TraoreAli NoumSali Traore
6218
asked Jan 19 at 22:00
Ali NoumSali TraoreAli NoumSali Traore
6218
asked Jan 19 at 22:00
Ali NoumSali TraoreAli NoumSali Traore
6218
6218
1
$begingroup$
The answer is trivially “no” if $a$ is nonzero. You should be able to see why. Where are you stuck?
$endgroup$
– rschwieb
Jan 19 at 22:04
$begingroup$
yes , thanks !!
$endgroup$
– Ali NoumSali Traore
Jan 19 at 22:14
add a comment |
1
$begingroup$
The answer is trivially “no” if $a$ is nonzero. You should be able to see why. Where are you stuck?
$endgroup$
– rschwieb
Jan 19 at 22:04
$begingroup$
yes , thanks !!
$endgroup$
– Ali NoumSali Traore
Jan 19 at 22:14
1
1
$begingroup$
The answer is trivially “no” if $a$ is nonzero. You should be able to see why. Where are you stuck?
$endgroup$
– rschwieb
Jan 19 at 22:04
$begingroup$
The answer is trivially “no” if $a$ is nonzero. You should be able to see why. Where are you stuck?
$endgroup$
– rschwieb
Jan 19 at 22:04
$begingroup$
yes , thanks !!
$endgroup$
– Ali NoumSali Traore
Jan 19 at 22:14
$begingroup$
yes , thanks !!
$endgroup$
– Ali NoumSali Traore
Jan 19 at 22:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As you wrote, you must have $f(m+m')=f(m)+f(m')$, i.e, $2a=a$. This is true only when $a=0$. So, if $aneq 0$, then as you said $f$ is not an homomorphism. On the other hand, if $a=0$, then $f$ is a homomorphism!
Another method: $f(M)$ is a module, i.e, ${a}$ is a module. Hence $a=0$.
answered Jan 19 at 22:16
ScientificaScientifica
6,79641335
$endgroup$
add a comment |
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1 Answer
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$begingroup$
As you wrote, you must have $f(m+m')=f(m)+f(m')$, i.e, $2a=a$. This is true only when $a=0$. So, if $aneq 0$, then as you said $f$ is not an homomorphism. On the other hand, if $a=0$, then $f$ is a homomorphism!
Another method: $f(M)$ is a module, i.e, ${a}$ is a module. Hence $a=0$.
answered Jan 19 at 22:16
ScientificaScientifica
6,79641335
$endgroup$
add a comment |
$begingroup$
As you wrote, you must have $f(m+m')=f(m)+f(m')$, i.e, $2a=a$. This is true only when $a=0$. So, if $aneq 0$, then as you said $f$ is not an homomorphism. On the other hand, if $a=0$, then $f$ is a homomorphism!
Another method: $f(M)$ is a module, i.e, ${a}$ is a module. Hence $a=0$.
answered Jan 19 at 22:16
ScientificaScientifica
6,79641335
$endgroup$
add a comment |
$begingroup$
As you wrote, you must have $f(m+m')=f(m)+f(m')$, i.e, $2a=a$. This is true only when $a=0$. So, if $aneq 0$, then as you said $f$ is not an homomorphism. On the other hand, if $a=0$, then $f$ is a homomorphism!
Another method: $f(M)$ is a module, i.e, ${a}$ is a module. Hence $a=0$.
answered Jan 19 at 22:16
ScientificaScientifica
6,79641335
$endgroup$
As you wrote, you must have $f(m+m')=f(m)+f(m')$, i.e, $2a=a$. This is true only when $a=0$. So, if $aneq 0$, then as you said $f$ is not an homomorphism. On the other hand, if $a=0$, then $f$ is a homomorphism!
Another method: $f(M)$ is a module, i.e, ${a}$ is a module. Hence $a=0$.
answered Jan 19 at 22:16
ScientificaScientifica
6,79641335
answered Jan 19 at 22:16
ScientificaScientifica
6,79641335
answered Jan 19 at 22:16
ScientificaScientifica
6,79641335
answered Jan 19 at 22:16
ScientificaScientifica
6,79641335
6,79641335
add a comment |
add a comment |
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$begingroup$
The answer is trivially “no” if $a$ is nonzero. You should be able to see why. Where are you stuck?
$endgroup$
– rschwieb
Jan 19 at 22:04
$begingroup$
yes , thanks !!
$endgroup$
– Ali NoumSali Traore
Jan 19 at 22:14