Question regarding the image of the unity $e$ of the ring $R$.
I was reading Ring Homomorphism .
$phi : R to R'$ is a ring homomorphism and $e , e'$ are the unities of $R $ and $R'$ respectively.
I understood that $phi (e) $ may not be unity of $R'$.
I think the following statements are true.
( 1 ) $phi (e) $ is always unity of $phi (R) $ .
(2) If $phi $ is on to then $phi (e) = e'$
(3) If $phi $ is non trivial and $R'$ is a field then $phi (e) = e'$.
Can someone check if there is any mistake in my understanding?
abstract-algebra ring-theory field-theory
asked Dec 11 '18 at 4:24
cmi
1,003212
|
show 2 more comments
I was reading Ring Homomorphism .
$phi : R to R'$ is a ring homomorphism and $e , e'$ are the unities of $R $ and $R'$ respectively.
I understood that $phi (e) $ may not be unity of $R'$.
I think the following statements are true.
( 1 ) $phi (e) $ is always unity of $phi (R) $ .
(2) If $phi $ is on to then $phi (e) = e'$
(3) If $phi $ is non trivial and $R'$ is a field then $phi (e) = e'$.
Can someone check if there is any mistake in my understanding?
abstract-algebra ring-theory field-theory
asked Dec 11 '18 at 4:24
cmi
1,003212
$phi (R) $ is a ring with unity $phi (e)$@BadamBaplan
– cmi
Dec 11 '18 at 4:47
Did you understand or not?@BadamBaplan
– cmi
Dec 11 '18 at 4:56
$phi (R)$ is always a subring and $phi(e)$ may not be the unity of $R'$. Proof or the first statement is very easy and see this math.stackexchange.com/questions/179842/… for the scond one.@BadamBaplan
– cmi
Dec 11 '18 at 5:08
No no unity@BadamBaplan
– cmi
Dec 11 '18 at 5:09
Yes $phi(R)$ is always a ring contained in $R$ with unit $phi(e)$. You seem to understand all of these things fine, so I'm not sure what you're looking for. Usually we require 'subrings' to contain the unity of their parent rings. I was just pointing out the connection that $phi(R)$ is a subring in that sense iff $phi(e) = e'$
– Badam Baplan
Dec 11 '18 at 5:31
|
show 2 more comments
I was reading Ring Homomorphism .
$phi : R to R'$ is a ring homomorphism and $e , e'$ are the unities of $R $ and $R'$ respectively.
I understood that $phi (e) $ may not be unity of $R'$.
I think the following statements are true.
( 1 ) $phi (e) $ is always unity of $phi (R) $ .
(2) If $phi $ is on to then $phi (e) = e'$
(3) If $phi $ is non trivial and $R'$ is a field then $phi (e) = e'$.
Can someone check if there is any mistake in my understanding?
abstract-algebra ring-theory field-theory
asked Dec 11 '18 at 4:24
cmi
1,003212
I was reading Ring Homomorphism .
$phi : R to R'$ is a ring homomorphism and $e , e'$ are the unities of $R $ and $R'$ respectively.
I understood that $phi (e) $ may not be unity of $R'$.
I think the following statements are true.
( 1 ) $phi (e) $ is always unity of $phi (R) $ .
(2) If $phi $ is on to then $phi (e) = e'$
(3) If $phi $ is non trivial and $R'$ is a field then $phi (e) = e'$.
Can someone check if there is any mistake in my understanding?
abstract-algebra ring-theory field-theory
abstract-algebra ring-theory field-theory
asked Dec 11 '18 at 4:24
cmi
1,003212
asked Dec 11 '18 at 4:24
cmi
1,003212
edited Dec 11 '18 at 4:41
asked Dec 11 '18 at 4:24
cmi
1,003212
asked Dec 11 '18 at 4:24
cmi
1,003212
asked Dec 11 '18 at 4:24
cmi
1,003212
1,003212
$phi (R) $ is a ring with unity $phi (e)$@BadamBaplan
– cmi
Dec 11 '18 at 4:47
Did you understand or not?@BadamBaplan
– cmi
Dec 11 '18 at 4:56
$phi (R)$ is always a subring and $phi(e)$ may not be the unity of $R'$. Proof or the first statement is very easy and see this math.stackexchange.com/questions/179842/… for the scond one.@BadamBaplan
– cmi
Dec 11 '18 at 5:08
No no unity@BadamBaplan
– cmi
Dec 11 '18 at 5:09
Yes $phi(R)$ is always a ring contained in $R$ with unit $phi(e)$. You seem to understand all of these things fine, so I'm not sure what you're looking for. Usually we require 'subrings' to contain the unity of their parent rings. I was just pointing out the connection that $phi(R)$ is a subring in that sense iff $phi(e) = e'$
– Badam Baplan
Dec 11 '18 at 5:31
|
show 2 more comments
$phi (R) $ is a ring with unity $phi (e)$@BadamBaplan
– cmi
Dec 11 '18 at 4:47
Did you understand or not?@BadamBaplan
– cmi
Dec 11 '18 at 4:56
$phi (R)$ is always a subring and $phi(e)$ may not be the unity of $R'$. Proof or the first statement is very easy and see this math.stackexchange.com/questions/179842/… for the scond one.@BadamBaplan
– cmi
Dec 11 '18 at 5:08
No no unity@BadamBaplan
– cmi
Dec 11 '18 at 5:09
Yes $phi(R)$ is always a ring contained in $R$ with unit $phi(e)$. You seem to understand all of these things fine, so I'm not sure what you're looking for. Usually we require 'subrings' to contain the unity of their parent rings. I was just pointing out the connection that $phi(R)$ is a subring in that sense iff $phi(e) = e'$
– Badam Baplan
Dec 11 '18 at 5:31
$phi (R) $ is a ring with unity $phi (e)$@BadamBaplan
– cmi
Dec 11 '18 at 4:47
$phi (R) $ is a ring with unity $phi (e)$@BadamBaplan
– cmi
Dec 11 '18 at 4:47
Did you understand or not?@BadamBaplan
– cmi
Dec 11 '18 at 4:56
Did you understand or not?@BadamBaplan
– cmi
Dec 11 '18 at 4:56
$phi (R)$ is always a subring and $phi(e)$ may not be the unity of $R'$. Proof or the first statement is very easy and see this math.stackexchange.com/questions/179842/… for the scond one.@BadamBaplan
– cmi
Dec 11 '18 at 5:08
$phi (R)$ is always a subring and $phi(e)$ may not be the unity of $R'$. Proof or the first statement is very easy and see this math.stackexchange.com/questions/179842/… for the scond one.@BadamBaplan
– cmi
Dec 11 '18 at 5:08
No no unity@BadamBaplan
– cmi
Dec 11 '18 at 5:09
No no unity@BadamBaplan
– cmi
Dec 11 '18 at 5:09
Yes $phi(R)$ is always a ring contained in $R$ with unit $phi(e)$. You seem to understand all of these things fine, so I'm not sure what you're looking for. Usually we require 'subrings' to contain the unity of their parent rings. I was just pointing out the connection that $phi(R)$ is a subring in that sense iff $phi(e) = e'$
– Badam Baplan
Dec 11 '18 at 5:31
Yes $phi(R)$ is always a ring contained in $R$ with unit $phi(e)$. You seem to understand all of these things fine, so I'm not sure what you're looking for. Usually we require 'subrings' to contain the unity of their parent rings. I was just pointing out the connection that $phi(R)$ is a subring in that sense iff $phi(e) = e'$
– Badam Baplan
Dec 11 '18 at 5:31
|
show 2 more comments
1 Answer
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Yes your 3 statements are true and easy to prove ,I assume you can prove them .
answered Dec 11 '18 at 5:24
StuartMN
1,406410
add a comment |
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Yes your 3 statements are true and easy to prove ,I assume you can prove them .
answered Dec 11 '18 at 5:24
StuartMN
1,406410
add a comment |
Yes your 3 statements are true and easy to prove ,I assume you can prove them .
answered Dec 11 '18 at 5:24
StuartMN
1,406410
add a comment |
Yes your 3 statements are true and easy to prove ,I assume you can prove them .
answered Dec 11 '18 at 5:24
StuartMN
1,406410
Yes your 3 statements are true and easy to prove ,I assume you can prove them .
answered Dec 11 '18 at 5:24
StuartMN
1,406410
answered Dec 11 '18 at 5:24
StuartMN
1,406410
answered Dec 11 '18 at 5:24
StuartMN
1,406410
answered Dec 11 '18 at 5:24
StuartMN
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$phi (R) $ is a ring with unity $phi (e)$@BadamBaplan
– cmi
Dec 11 '18 at 4:47
Did you understand or not?@BadamBaplan
– cmi
Dec 11 '18 at 4:56
$phi (R)$ is always a subring and $phi(e)$ may not be the unity of $R'$. Proof or the first statement is very easy and see this math.stackexchange.com/questions/179842/… for the scond one.@BadamBaplan
– cmi
Dec 11 '18 at 5:08
No no unity@BadamBaplan
– cmi
Dec 11 '18 at 5:09
Yes $phi(R)$ is always a ring contained in $R$ with unit $phi(e)$. You seem to understand all of these things fine, so I'm not sure what you're looking for. Usually we require 'subrings' to contain the unity of their parent rings. I was just pointing out the connection that $phi(R)$ is a subring in that sense iff $phi(e) = e'$
– Badam Baplan
Dec 11 '18 at 5:31