If domain $A$ of the homomorphism $f$ is a field, and if the range of $f$ has more than one element, then $f$...
$begingroup$
This is a question from Pinter's A book of Abstract Algebra:
If domain A of the homomorphism $f$ is a field, and if the range of
$f$ has more than one element, then $f$ is injective.
Here's my attempt:
Claim. If $text{im} f ne { 0}$ then $f(1)ne 0$.
Proof (by contrapositive). Suppose that $f(1)=0$. Then if $xin A$ then $f(x)= f(xcdot 1) = f(x)f(1)= f(x)cdot 0=0$. Hence, $text{im} f = { 0}$.
To show that $f$ is injective, it is enough to show that $text{ker} f ={ 0}$. Let $x in text{ker} f$. Then $f(x)=0$. We claim that $x=0$. Suppose that $xne 0$. Then there is an element $x^{-1} in A$ such that $xx^{-1} = 1$.
Now,
$f(x)=0 \ Rightarrow f(x)f(x^{-1}) = 0 cdot f(x^{-1}) \ Rightarrow f(xx^{-1})=0 \ Rightarrow f(1)=0$
Which is a contradiction!
Thus, $text{ker} f ={ 0}$ . This completes the proof.
Is this okay or am I missing something? I was wondering if there was some easier way to do this.
abstract-algebra proof-verification ring-theory field-theory
asked Dec 25 '18 at 6:23
Ashish KAshish K
890613
$endgroup$
add a comment |
$begingroup$
This is a question from Pinter's A book of Abstract Algebra:
If domain A of the homomorphism $f$ is a field, and if the range of
$f$ has more than one element, then $f$ is injective.
Here's my attempt:
Claim. If $text{im} f ne { 0}$ then $f(1)ne 0$.
Proof (by contrapositive). Suppose that $f(1)=0$. Then if $xin A$ then $f(x)= f(xcdot 1) = f(x)f(1)= f(x)cdot 0=0$. Hence, $text{im} f = { 0}$.
To show that $f$ is injective, it is enough to show that $text{ker} f ={ 0}$. Let $x in text{ker} f$. Then $f(x)=0$. We claim that $x=0$. Suppose that $xne 0$. Then there is an element $x^{-1} in A$ such that $xx^{-1} = 1$.
Now,
$f(x)=0 \ Rightarrow f(x)f(x^{-1}) = 0 cdot f(x^{-1}) \ Rightarrow f(xx^{-1})=0 \ Rightarrow f(1)=0$
Which is a contradiction!
Thus, $text{ker} f ={ 0}$ . This completes the proof.
Is this okay or am I missing something? I was wondering if there was some easier way to do this.
abstract-algebra proof-verification ring-theory field-theory
asked Dec 25 '18 at 6:23
Ashish KAshish K
890613
$endgroup$
4
$begingroup$
Are you familiar with ideals? If yes, then you can use the fact that a field has only two ideals, namely the field itself or the zero ideal. Since kernels are ideals, so for a homomorphism from a field, either the kernel is just ${0}$ or whole $Bbb{F}$. With the range being non-trivial, kernel has to be ${0}$, hence injective.
$endgroup$
– Anurag A
Dec 25 '18 at 6:29
$begingroup$
@AnuragA I did prove that fields have no nontrivial ideals (as an exercise in my book). I guess I need to verify that kernel are ideals. However, is my aforementioned proof okay?
$endgroup$
– Ashish K
Dec 25 '18 at 6:49
1
$begingroup$
Yes your proof is good.
$endgroup$
– Anurag A
Dec 25 '18 at 6:53
add a comment |
$begingroup$
This is a question from Pinter's A book of Abstract Algebra:
If domain A of the homomorphism $f$ is a field, and if the range of
$f$ has more than one element, then $f$ is injective.
Here's my attempt:
Claim. If $text{im} f ne { 0}$ then $f(1)ne 0$.
Proof (by contrapositive). Suppose that $f(1)=0$. Then if $xin A$ then $f(x)= f(xcdot 1) = f(x)f(1)= f(x)cdot 0=0$. Hence, $text{im} f = { 0}$.
To show that $f$ is injective, it is enough to show that $text{ker} f ={ 0}$. Let $x in text{ker} f$. Then $f(x)=0$. We claim that $x=0$. Suppose that $xne 0$. Then there is an element $x^{-1} in A$ such that $xx^{-1} = 1$.
Now,
$f(x)=0 \ Rightarrow f(x)f(x^{-1}) = 0 cdot f(x^{-1}) \ Rightarrow f(xx^{-1})=0 \ Rightarrow f(1)=0$
Which is a contradiction!
Thus, $text{ker} f ={ 0}$ . This completes the proof.
Is this okay or am I missing something? I was wondering if there was some easier way to do this.
abstract-algebra proof-verification ring-theory field-theory
asked Dec 25 '18 at 6:23
Ashish KAshish K
890613
$endgroup$
This is a question from Pinter's A book of Abstract Algebra:
If domain A of the homomorphism $f$ is a field, and if the range of
$f$ has more than one element, then $f$ is injective.
Here's my attempt:
Claim. If $text{im} f ne { 0}$ then $f(1)ne 0$.
Proof (by contrapositive). Suppose that $f(1)=0$. Then if $xin A$ then $f(x)= f(xcdot 1) = f(x)f(1)= f(x)cdot 0=0$. Hence, $text{im} f = { 0}$.
To show that $f$ is injective, it is enough to show that $text{ker} f ={ 0}$. Let $x in text{ker} f$. Then $f(x)=0$. We claim that $x=0$. Suppose that $xne 0$. Then there is an element $x^{-1} in A$ such that $xx^{-1} = 1$.
Now,
$f(x)=0 \ Rightarrow f(x)f(x^{-1}) = 0 cdot f(x^{-1}) \ Rightarrow f(xx^{-1})=0 \ Rightarrow f(1)=0$
Which is a contradiction!
Thus, $text{ker} f ={ 0}$ . This completes the proof.
Is this okay or am I missing something? I was wondering if there was some easier way to do this.
abstract-algebra proof-verification ring-theory field-theory
abstract-algebra proof-verification ring-theory field-theory
asked Dec 25 '18 at 6:23
Ashish KAshish K
890613
asked Dec 25 '18 at 6:23
Ashish KAshish K
890613
asked Dec 25 '18 at 6:23
Ashish KAshish K
890613
asked Dec 25 '18 at 6:23
Ashish KAshish K
890613
asked Dec 25 '18 at 6:23
Ashish KAshish K
890613
890613
4
$begingroup$
Are you familiar with ideals? If yes, then you can use the fact that a field has only two ideals, namely the field itself or the zero ideal. Since kernels are ideals, so for a homomorphism from a field, either the kernel is just ${0}$ or whole $Bbb{F}$. With the range being non-trivial, kernel has to be ${0}$, hence injective.
$endgroup$
– Anurag A
Dec 25 '18 at 6:29
$begingroup$
@AnuragA I did prove that fields have no nontrivial ideals (as an exercise in my book). I guess I need to verify that kernel are ideals. However, is my aforementioned proof okay?
$endgroup$
– Ashish K
Dec 25 '18 at 6:49
1
$begingroup$
Yes your proof is good.
$endgroup$
– Anurag A
Dec 25 '18 at 6:53
add a comment |
4
$begingroup$
Are you familiar with ideals? If yes, then you can use the fact that a field has only two ideals, namely the field itself or the zero ideal. Since kernels are ideals, so for a homomorphism from a field, either the kernel is just ${0}$ or whole $Bbb{F}$. With the range being non-trivial, kernel has to be ${0}$, hence injective.
$endgroup$
– Anurag A
Dec 25 '18 at 6:29
$begingroup$
@AnuragA I did prove that fields have no nontrivial ideals (as an exercise in my book). I guess I need to verify that kernel are ideals. However, is my aforementioned proof okay?
$endgroup$
– Ashish K
Dec 25 '18 at 6:49
1
$begingroup$
Yes your proof is good.
$endgroup$
– Anurag A
Dec 25 '18 at 6:53
4
4
$begingroup$
Are you familiar with ideals? If yes, then you can use the fact that a field has only two ideals, namely the field itself or the zero ideal. Since kernels are ideals, so for a homomorphism from a field, either the kernel is just ${0}$ or whole $Bbb{F}$. With the range being non-trivial, kernel has to be ${0}$, hence injective.
$endgroup$
– Anurag A
Dec 25 '18 at 6:29
$begingroup$
Are you familiar with ideals? If yes, then you can use the fact that a field has only two ideals, namely the field itself or the zero ideal. Since kernels are ideals, so for a homomorphism from a field, either the kernel is just ${0}$ or whole $Bbb{F}$. With the range being non-trivial, kernel has to be ${0}$, hence injective.
$endgroup$
– Anurag A
Dec 25 '18 at 6:29
$begingroup$
@AnuragA I did prove that fields have no nontrivial ideals (as an exercise in my book). I guess I need to verify that kernel are ideals. However, is my aforementioned proof okay?
$endgroup$
– Ashish K
Dec 25 '18 at 6:49
$begingroup$
@AnuragA I did prove that fields have no nontrivial ideals (as an exercise in my book). I guess I need to verify that kernel are ideals. However, is my aforementioned proof okay?
$endgroup$
– Ashish K
Dec 25 '18 at 6:49
1
1
$begingroup$
Yes your proof is good.
$endgroup$
– Anurag A
Dec 25 '18 at 6:53
$begingroup$
Yes your proof is good.
$endgroup$
– Anurag A
Dec 25 '18 at 6:53
add a comment |
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4
$begingroup$
Are you familiar with ideals? If yes, then you can use the fact that a field has only two ideals, namely the field itself or the zero ideal. Since kernels are ideals, so for a homomorphism from a field, either the kernel is just ${0}$ or whole $Bbb{F}$. With the range being non-trivial, kernel has to be ${0}$, hence injective.
$endgroup$
– Anurag A
Dec 25 '18 at 6:29
$begingroup$
@AnuragA I did prove that fields have no nontrivial ideals (as an exercise in my book). I guess I need to verify that kernel are ideals. However, is my aforementioned proof okay?
$endgroup$
– Ashish K
Dec 25 '18 at 6:49
1
$begingroup$
Yes your proof is good.
$endgroup$
– Anurag A
Dec 25 '18 at 6:53