Fundamental homomorphism theorem (epimorphism)
up vote
0
down vote
favorite
Let φ : R → S be a ring epimorphism. Prove that R/kerφ ∼= S.
Is this the fundamental homomorphism theorem? I thought the FHT started with a ring homomorphism and not an epimorphism. Does this change the proof of the theorem ?
ring-theory
asked Nov 12 at 16:24
Johnmallu
74
add a comment |
up vote
0
down vote
favorite
Let φ : R → S be a ring epimorphism. Prove that R/kerφ ∼= S.
Is this the fundamental homomorphism theorem? I thought the FHT started with a ring homomorphism and not an epimorphism. Does this change the proof of the theorem ?
ring-theory
asked Nov 12 at 16:24
Johnmallu
74
If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
– xbh
Nov 12 at 16:43
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let φ : R → S be a ring epimorphism. Prove that R/kerφ ∼= S.
Is this the fundamental homomorphism theorem? I thought the FHT started with a ring homomorphism and not an epimorphism. Does this change the proof of the theorem ?
ring-theory
asked Nov 12 at 16:24
Johnmallu
74
Let φ : R → S be a ring epimorphism. Prove that R/kerφ ∼= S.
Is this the fundamental homomorphism theorem? I thought the FHT started with a ring homomorphism and not an epimorphism. Does this change the proof of the theorem ?
ring-theory
ring-theory
asked Nov 12 at 16:24
Johnmallu
74
asked Nov 12 at 16:24
Johnmallu
74
asked Nov 12 at 16:24
Johnmallu
74
asked Nov 12 at 16:24
Johnmallu
74
asked Nov 12 at 16:24
Johnmallu
74
74
If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
– xbh
Nov 12 at 16:43
add a comment |
If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
– xbh
Nov 12 at 16:43
If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
– xbh
Nov 12 at 16:43
If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
– xbh
Nov 12 at 16:43
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
You can state the theorem without the epimorphism assumption replacing $S$ by $mathrm{Im} varphi$. So both formulation are equivalent.
answered Nov 12 at 16:47
Can I play with Mathness
3894
Okay thank you !
– Johnmallu
Nov 12 at 16:52
add a comment |
up vote
0
down vote
The general formula for the homomorphism theorem is
$$R/ker(phi)cong mathrm{Im}(phi)$$
In the special case, $phi$ is an epimorphism, then $mathrm{Im}(phi)=S$, but be careful, you cannot deduce the general formula from the special case.
answered Nov 12 at 18:45
Fakemistake
1,582714
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You can state the theorem without the epimorphism assumption replacing $S$ by $mathrm{Im} varphi$. So both formulation are equivalent.
answered Nov 12 at 16:47
Can I play with Mathness
3894
Okay thank you !
– Johnmallu
Nov 12 at 16:52
add a comment |
up vote
0
down vote
accepted
You can state the theorem without the epimorphism assumption replacing $S$ by $mathrm{Im} varphi$. So both formulation are equivalent.
answered Nov 12 at 16:47
Can I play with Mathness
3894
Okay thank you !
– Johnmallu
Nov 12 at 16:52
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You can state the theorem without the epimorphism assumption replacing $S$ by $mathrm{Im} varphi$. So both formulation are equivalent.
answered Nov 12 at 16:47
Can I play with Mathness
3894
You can state the theorem without the epimorphism assumption replacing $S$ by $mathrm{Im} varphi$. So both formulation are equivalent.
answered Nov 12 at 16:47
Can I play with Mathness
3894
answered Nov 12 at 16:47
Can I play with Mathness
3894
answered Nov 12 at 16:47
Can I play with Mathness
3894
answered Nov 12 at 16:47
Can I play with Mathness
3894
3894
Okay thank you !
– Johnmallu
Nov 12 at 16:52
add a comment |
Okay thank you !
– Johnmallu
Nov 12 at 16:52
Okay thank you !
– Johnmallu
Nov 12 at 16:52
Okay thank you !
– Johnmallu
Nov 12 at 16:52
add a comment |
up vote
0
down vote
The general formula for the homomorphism theorem is
$$R/ker(phi)cong mathrm{Im}(phi)$$
In the special case, $phi$ is an epimorphism, then $mathrm{Im}(phi)=S$, but be careful, you cannot deduce the general formula from the special case.
answered Nov 12 at 18:45
Fakemistake
1,582714
add a comment |
up vote
0
down vote
The general formula for the homomorphism theorem is
$$R/ker(phi)cong mathrm{Im}(phi)$$
In the special case, $phi$ is an epimorphism, then $mathrm{Im}(phi)=S$, but be careful, you cannot deduce the general formula from the special case.
answered Nov 12 at 18:45
Fakemistake
1,582714
add a comment |
up vote
0
down vote
up vote
0
down vote
The general formula for the homomorphism theorem is
$$R/ker(phi)cong mathrm{Im}(phi)$$
In the special case, $phi$ is an epimorphism, then $mathrm{Im}(phi)=S$, but be careful, you cannot deduce the general formula from the special case.
answered Nov 12 at 18:45
Fakemistake
1,582714
The general formula for the homomorphism theorem is
$$R/ker(phi)cong mathrm{Im}(phi)$$
In the special case, $phi$ is an epimorphism, then $mathrm{Im}(phi)=S$, but be careful, you cannot deduce the general formula from the special case.
answered Nov 12 at 18:45
Fakemistake
1,582714
answered Nov 12 at 18:45
Fakemistake
1,582714
answered Nov 12 at 18:45
Fakemistake
1,582714
answered Nov 12 at 18:45
Fakemistake
1,582714
1,582714
add a comment |
add a comment |
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If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
– xbh
Nov 12 at 16:43