Factorization Theorem
Let $f_1:A longrightarrow A_1 , f_2 :A longrightarrow A_2$ homomorphism of rings such that $ker(f_1)subseteq ker(f_2)$. Prove that exist a homomorphism $f:A_1 longrightarrow A_2 $ such that $f_2=f circ f_1$.
Is a generalization of factorization theorem. In the book the hint is that define $f$ as $f(f_1(a))=f_2(a)$. But in that case $f: f_1(A)subseteq A_1 longrightarrow A_2$ and the proof that is a homomorphism is not difficult. My question is if you can define a $f$ that goes from $A_1$ to $A_2$. If $f_1$ is a epimorphism is very easy because $f_1(A)=A$
abstract-algebra
asked Dec 10 '18 at 1:25
Juan Daniel Valdivia Fuentes
104
add a comment |
Let $f_1:A longrightarrow A_1 , f_2 :A longrightarrow A_2$ homomorphism of rings such that $ker(f_1)subseteq ker(f_2)$. Prove that exist a homomorphism $f:A_1 longrightarrow A_2 $ such that $f_2=f circ f_1$.
Is a generalization of factorization theorem. In the book the hint is that define $f$ as $f(f_1(a))=f_2(a)$. But in that case $f: f_1(A)subseteq A_1 longrightarrow A_2$ and the proof that is a homomorphism is not difficult. My question is if you can define a $f$ that goes from $A_1$ to $A_2$. If $f_1$ is a epimorphism is very easy because $f_1(A)=A$
abstract-algebra
asked Dec 10 '18 at 1:25
Juan Daniel Valdivia Fuentes
104
add a comment |
Let $f_1:A longrightarrow A_1 , f_2 :A longrightarrow A_2$ homomorphism of rings such that $ker(f_1)subseteq ker(f_2)$. Prove that exist a homomorphism $f:A_1 longrightarrow A_2 $ such that $f_2=f circ f_1$.
Is a generalization of factorization theorem. In the book the hint is that define $f$ as $f(f_1(a))=f_2(a)$. But in that case $f: f_1(A)subseteq A_1 longrightarrow A_2$ and the proof that is a homomorphism is not difficult. My question is if you can define a $f$ that goes from $A_1$ to $A_2$. If $f_1$ is a epimorphism is very easy because $f_1(A)=A$
abstract-algebra
asked Dec 10 '18 at 1:25
Juan Daniel Valdivia Fuentes
104
Let $f_1:A longrightarrow A_1 , f_2 :A longrightarrow A_2$ homomorphism of rings such that $ker(f_1)subseteq ker(f_2)$. Prove that exist a homomorphism $f:A_1 longrightarrow A_2 $ such that $f_2=f circ f_1$.
Is a generalization of factorization theorem. In the book the hint is that define $f$ as $f(f_1(a))=f_2(a)$. But in that case $f: f_1(A)subseteq A_1 longrightarrow A_2$ and the proof that is a homomorphism is not difficult. My question is if you can define a $f$ that goes from $A_1$ to $A_2$. If $f_1$ is a epimorphism is very easy because $f_1(A)=A$
abstract-algebra
abstract-algebra
asked Dec 10 '18 at 1:25
Juan Daniel Valdivia Fuentes
104
asked Dec 10 '18 at 1:25
Juan Daniel Valdivia Fuentes
104
asked Dec 10 '18 at 1:25
Juan Daniel Valdivia Fuentes
104
asked Dec 10 '18 at 1:25
Juan Daniel Valdivia Fuentes
104
asked Dec 10 '18 at 1:25
Juan Daniel Valdivia Fuentes
104
104
add a comment |
add a comment |
1 Answer
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I'm not quite sure what your question is, but I think you're saying that you think the statement (which you appear to be saying is given in your book) is false.
In that case you'd be correct.
Let $f_1 : Bbb{Z}hookrightarrow Bbb{Q}$ be the inclusion and let $f_2 : Bbb{Z}to Bbb{Z}$ be the identity map. Then both $f_1$ and $f_2$ are injective, so $ker f_1 =ker f_2 = 0$. However, there is no map at all from $Bbb{Q}$ to $Bbb{Z}$, let alone one with $fcirc f_1=f_2$.
Note that in this case $f_1$ is actually an epimorphism, since if $g,h : Bbb{Q}to R$ are ring homomorphisms that agree when restricted to $Bbb{Z}$, they must agree everywhere since $$g(n/m)=g(n)/g(m)=h(n)/h(m)=h(n/m).$$
Thus it doesn't suffice for $f_1$ to be an epimorphism. It does however suffice for $f_1$ to be surjective (note that in rings this is a distinct notion).
See here for more on epimorphisms of rings.
answered Dec 10 '18 at 1:32
jgon
12.8k21940
And if $f_1$ is a epimorphism is true?.
– Juan Daniel Valdivia Fuentes
Dec 10 '18 at 1:36
@JuanDanielValdiviaFuentes I just edited to address that.
– jgon
Dec 10 '18 at 1:36
add a comment |
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1 Answer
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1 Answer
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I'm not quite sure what your question is, but I think you're saying that you think the statement (which you appear to be saying is given in your book) is false.
In that case you'd be correct.
Let $f_1 : Bbb{Z}hookrightarrow Bbb{Q}$ be the inclusion and let $f_2 : Bbb{Z}to Bbb{Z}$ be the identity map. Then both $f_1$ and $f_2$ are injective, so $ker f_1 =ker f_2 = 0$. However, there is no map at all from $Bbb{Q}$ to $Bbb{Z}$, let alone one with $fcirc f_1=f_2$.
Note that in this case $f_1$ is actually an epimorphism, since if $g,h : Bbb{Q}to R$ are ring homomorphisms that agree when restricted to $Bbb{Z}$, they must agree everywhere since $$g(n/m)=g(n)/g(m)=h(n)/h(m)=h(n/m).$$
Thus it doesn't suffice for $f_1$ to be an epimorphism. It does however suffice for $f_1$ to be surjective (note that in rings this is a distinct notion).
See here for more on epimorphisms of rings.
answered Dec 10 '18 at 1:32
jgon
12.8k21940
And if $f_1$ is a epimorphism is true?.
– Juan Daniel Valdivia Fuentes
Dec 10 '18 at 1:36
@JuanDanielValdiviaFuentes I just edited to address that.
– jgon
Dec 10 '18 at 1:36
add a comment |
I'm not quite sure what your question is, but I think you're saying that you think the statement (which you appear to be saying is given in your book) is false.
In that case you'd be correct.
Let $f_1 : Bbb{Z}hookrightarrow Bbb{Q}$ be the inclusion and let $f_2 : Bbb{Z}to Bbb{Z}$ be the identity map. Then both $f_1$ and $f_2$ are injective, so $ker f_1 =ker f_2 = 0$. However, there is no map at all from $Bbb{Q}$ to $Bbb{Z}$, let alone one with $fcirc f_1=f_2$.
Note that in this case $f_1$ is actually an epimorphism, since if $g,h : Bbb{Q}to R$ are ring homomorphisms that agree when restricted to $Bbb{Z}$, they must agree everywhere since $$g(n/m)=g(n)/g(m)=h(n)/h(m)=h(n/m).$$
Thus it doesn't suffice for $f_1$ to be an epimorphism. It does however suffice for $f_1$ to be surjective (note that in rings this is a distinct notion).
See here for more on epimorphisms of rings.
answered Dec 10 '18 at 1:32
jgon
12.8k21940
And if $f_1$ is a epimorphism is true?.
– Juan Daniel Valdivia Fuentes
Dec 10 '18 at 1:36
@JuanDanielValdiviaFuentes I just edited to address that.
– jgon
Dec 10 '18 at 1:36
add a comment |
I'm not quite sure what your question is, but I think you're saying that you think the statement (which you appear to be saying is given in your book) is false.
In that case you'd be correct.
Let $f_1 : Bbb{Z}hookrightarrow Bbb{Q}$ be the inclusion and let $f_2 : Bbb{Z}to Bbb{Z}$ be the identity map. Then both $f_1$ and $f_2$ are injective, so $ker f_1 =ker f_2 = 0$. However, there is no map at all from $Bbb{Q}$ to $Bbb{Z}$, let alone one with $fcirc f_1=f_2$.
Note that in this case $f_1$ is actually an epimorphism, since if $g,h : Bbb{Q}to R$ are ring homomorphisms that agree when restricted to $Bbb{Z}$, they must agree everywhere since $$g(n/m)=g(n)/g(m)=h(n)/h(m)=h(n/m).$$
Thus it doesn't suffice for $f_1$ to be an epimorphism. It does however suffice for $f_1$ to be surjective (note that in rings this is a distinct notion).
See here for more on epimorphisms of rings.
answered Dec 10 '18 at 1:32
jgon
12.8k21940
I'm not quite sure what your question is, but I think you're saying that you think the statement (which you appear to be saying is given in your book) is false.
In that case you'd be correct.
Let $f_1 : Bbb{Z}hookrightarrow Bbb{Q}$ be the inclusion and let $f_2 : Bbb{Z}to Bbb{Z}$ be the identity map. Then both $f_1$ and $f_2$ are injective, so $ker f_1 =ker f_2 = 0$. However, there is no map at all from $Bbb{Q}$ to $Bbb{Z}$, let alone one with $fcirc f_1=f_2$.
Note that in this case $f_1$ is actually an epimorphism, since if $g,h : Bbb{Q}to R$ are ring homomorphisms that agree when restricted to $Bbb{Z}$, they must agree everywhere since $$g(n/m)=g(n)/g(m)=h(n)/h(m)=h(n/m).$$
Thus it doesn't suffice for $f_1$ to be an epimorphism. It does however suffice for $f_1$ to be surjective (note that in rings this is a distinct notion).
See here for more on epimorphisms of rings.
answered Dec 10 '18 at 1:32
jgon
12.8k21940
edited Dec 10 '18 at 1:36
answered Dec 10 '18 at 1:32
jgon
12.8k21940
answered Dec 10 '18 at 1:32
jgon
12.8k21940
answered Dec 10 '18 at 1:32
jgon
12.8k21940
12.8k21940
And if $f_1$ is a epimorphism is true?.
– Juan Daniel Valdivia Fuentes
Dec 10 '18 at 1:36
@JuanDanielValdiviaFuentes I just edited to address that.
– jgon
Dec 10 '18 at 1:36
add a comment |
And if $f_1$ is a epimorphism is true?.
– Juan Daniel Valdivia Fuentes
Dec 10 '18 at 1:36
@JuanDanielValdiviaFuentes I just edited to address that.
– jgon
Dec 10 '18 at 1:36
And if $f_1$ is a epimorphism is true?.
– Juan Daniel Valdivia Fuentes
Dec 10 '18 at 1:36
And if $f_1$ is a epimorphism is true?.
– Juan Daniel Valdivia Fuentes
Dec 10 '18 at 1:36
@JuanDanielValdiviaFuentes I just edited to address that.
– jgon
Dec 10 '18 at 1:36
@JuanDanielValdiviaFuentes I just edited to address that.
– jgon
Dec 10 '18 at 1:36
add a comment |
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