How do you prove Let φ : R → S be a EPIMORPHISM of rings. Then the image of φ is isomorphic to the factor...
$begingroup$
Below I have proved this question for a ring homomorphism.
MY QUESTION: How do you prove Let φ : R → S be a EPIMORPHISM of rings. Then the image of φ is isomorphic to the factor ring R/ ker φ
Let φ : R → S be a homomorphism of rings. Then the image of φ is isomorphic to the factor ring R/ ker φ
Proof:
Let I denote the kernel of φ, so I is a two-sided ideal of R. Define a function
φ ̄ : R/I → Imφ by:
φ ̄ (a + I) = φ(a) for a ∈ R.
φ ̄ is well-defined (i.e. the image of a + I does not depend on a choice of coset representative). Suppose that a + I = a1 + I for some a, a1 ∈ R. Then a−a1 ∈IbyLemma3.3.2. Henceφ(a−a1)=0S =φ(a)−φ(a1). Thus φ(a) = φ(a1 ) as required.
φ ̄ is a ring homomorphism.
Suppose a + I, b + I are elements of R/I. Then
̄ φ((a +I) +(b +I)) = φ ̄ ((a +b) +I)
= φ(a+b)
= φ(a) + φ(b)
= φ ̄(a+I)+φ ̄(b+I).
So φ is additive. Also
̄ φ((a+I)(b+I)) = φ ̄(ab+I) = φ(ab)
= φ(a)φ(b)
= φ ̄(a+I)φ ̄(b+I)
φ ̄ is injective.
Suppose a+I ∈ kerφ. Thenφ(a+I) = 0S soφ(a) = 0S. This means a ∈ ker φ, so a ∈ I. Then a + I = I = 0R + I, a + I is the zero element of R/I. Thus ker φ ̄ contains only the zero element of R/Iφ ̄ is surjective.
Let s∈Imφ. Thens=φ(r)for some r∈R. Thus s=φ(r+I)and every element of Imφ is the image under φ ̄ of some coset of I in R.
Thus φ ̄ : R/ ker φ −→ Imφ is a ring isomorphism, and Imφ is isomorphic to the factor ring R/ ker φ.
ring-theory
asked Dec 2 '18 at 16:02
JacobKnightJacobKnight
163
$endgroup$
add a comment |
$begingroup$
Below I have proved this question for a ring homomorphism.
MY QUESTION: How do you prove Let φ : R → S be a EPIMORPHISM of rings. Then the image of φ is isomorphic to the factor ring R/ ker φ
Let φ : R → S be a homomorphism of rings. Then the image of φ is isomorphic to the factor ring R/ ker φ
Proof:
Let I denote the kernel of φ, so I is a two-sided ideal of R. Define a function
φ ̄ : R/I → Imφ by:
φ ̄ (a + I) = φ(a) for a ∈ R.
φ ̄ is well-defined (i.e. the image of a + I does not depend on a choice of coset representative). Suppose that a + I = a1 + I for some a, a1 ∈ R. Then a−a1 ∈IbyLemma3.3.2. Henceφ(a−a1)=0S =φ(a)−φ(a1). Thus φ(a) = φ(a1 ) as required.
φ ̄ is a ring homomorphism.
Suppose a + I, b + I are elements of R/I. Then
̄ φ((a +I) +(b +I)) = φ ̄ ((a +b) +I)
= φ(a+b)
= φ(a) + φ(b)
= φ ̄(a+I)+φ ̄(b+I).
So φ is additive. Also
̄ φ((a+I)(b+I)) = φ ̄(ab+I) = φ(ab)
= φ(a)φ(b)
= φ ̄(a+I)φ ̄(b+I)
φ ̄ is injective.
Suppose a+I ∈ kerφ. Thenφ(a+I) = 0S soφ(a) = 0S. This means a ∈ ker φ, so a ∈ I. Then a + I = I = 0R + I, a + I is the zero element of R/I. Thus ker φ ̄ contains only the zero element of R/Iφ ̄ is surjective.
Let s∈Imφ. Thens=φ(r)for some r∈R. Thus s=φ(r+I)and every element of Imφ is the image under φ ̄ of some coset of I in R.
Thus φ ̄ : R/ ker φ −→ Imφ is a ring isomorphism, and Imφ is isomorphic to the factor ring R/ ker φ.
ring-theory
asked Dec 2 '18 at 16:02
JacobKnightJacobKnight
163
$endgroup$
$begingroup$
An epimorphism of rings is just a surjective homomorphism of rings, so you've already proven it. Note that in this case $operatorname{Im} varphi = S$.
$endgroup$
– ÍgjøgnumMeg
Dec 2 '18 at 16:45
$begingroup$
Thank you I was just checking that I wasn't assuming that incorrectly @ÍgjøgnumMeg
$endgroup$
– JacobKnight
Dec 4 '18 at 15:22
add a comment |
$begingroup$
Below I have proved this question for a ring homomorphism.
MY QUESTION: How do you prove Let φ : R → S be a EPIMORPHISM of rings. Then the image of φ is isomorphic to the factor ring R/ ker φ
Let φ : R → S be a homomorphism of rings. Then the image of φ is isomorphic to the factor ring R/ ker φ
Proof:
Let I denote the kernel of φ, so I is a two-sided ideal of R. Define a function
φ ̄ : R/I → Imφ by:
φ ̄ (a + I) = φ(a) for a ∈ R.
φ ̄ is well-defined (i.e. the image of a + I does not depend on a choice of coset representative). Suppose that a + I = a1 + I for some a, a1 ∈ R. Then a−a1 ∈IbyLemma3.3.2. Henceφ(a−a1)=0S =φ(a)−φ(a1). Thus φ(a) = φ(a1 ) as required.
φ ̄ is a ring homomorphism.
Suppose a + I, b + I are elements of R/I. Then
̄ φ((a +I) +(b +I)) = φ ̄ ((a +b) +I)
= φ(a+b)
= φ(a) + φ(b)
= φ ̄(a+I)+φ ̄(b+I).
So φ is additive. Also
̄ φ((a+I)(b+I)) = φ ̄(ab+I) = φ(ab)
= φ(a)φ(b)
= φ ̄(a+I)φ ̄(b+I)
φ ̄ is injective.
Suppose a+I ∈ kerφ. Thenφ(a+I) = 0S soφ(a) = 0S. This means a ∈ ker φ, so a ∈ I. Then a + I = I = 0R + I, a + I is the zero element of R/I. Thus ker φ ̄ contains only the zero element of R/Iφ ̄ is surjective.
Let s∈Imφ. Thens=φ(r)for some r∈R. Thus s=φ(r+I)and every element of Imφ is the image under φ ̄ of some coset of I in R.
Thus φ ̄ : R/ ker φ −→ Imφ is a ring isomorphism, and Imφ is isomorphic to the factor ring R/ ker φ.
ring-theory
asked Dec 2 '18 at 16:02
JacobKnightJacobKnight
163
$endgroup$
Below I have proved this question for a ring homomorphism.
MY QUESTION: How do you prove Let φ : R → S be a EPIMORPHISM of rings. Then the image of φ is isomorphic to the factor ring R/ ker φ
Let φ : R → S be a homomorphism of rings. Then the image of φ is isomorphic to the factor ring R/ ker φ
Proof:
Let I denote the kernel of φ, so I is a two-sided ideal of R. Define a function
φ ̄ : R/I → Imφ by:
φ ̄ (a + I) = φ(a) for a ∈ R.
φ ̄ is well-defined (i.e. the image of a + I does not depend on a choice of coset representative). Suppose that a + I = a1 + I for some a, a1 ∈ R. Then a−a1 ∈IbyLemma3.3.2. Henceφ(a−a1)=0S =φ(a)−φ(a1). Thus φ(a) = φ(a1 ) as required.
φ ̄ is a ring homomorphism.
Suppose a + I, b + I are elements of R/I. Then
̄ φ((a +I) +(b +I)) = φ ̄ ((a +b) +I)
= φ(a+b)
= φ(a) + φ(b)
= φ ̄(a+I)+φ ̄(b+I).
So φ is additive. Also
̄ φ((a+I)(b+I)) = φ ̄(ab+I) = φ(ab)
= φ(a)φ(b)
= φ ̄(a+I)φ ̄(b+I)
φ ̄ is injective.
Suppose a+I ∈ kerφ. Thenφ(a+I) = 0S soφ(a) = 0S. This means a ∈ ker φ, so a ∈ I. Then a + I = I = 0R + I, a + I is the zero element of R/I. Thus ker φ ̄ contains only the zero element of R/Iφ ̄ is surjective.
Let s∈Imφ. Thens=φ(r)for some r∈R. Thus s=φ(r+I)and every element of Imφ is the image under φ ̄ of some coset of I in R.
Thus φ ̄ : R/ ker φ −→ Imφ is a ring isomorphism, and Imφ is isomorphic to the factor ring R/ ker φ.
ring-theory
ring-theory
asked Dec 2 '18 at 16:02
JacobKnightJacobKnight
163
asked Dec 2 '18 at 16:02
JacobKnightJacobKnight
163
asked Dec 2 '18 at 16:02
JacobKnightJacobKnight
163
asked Dec 2 '18 at 16:02
JacobKnightJacobKnight
163
asked Dec 2 '18 at 16:02
JacobKnightJacobKnight
163
163
$begingroup$
An epimorphism of rings is just a surjective homomorphism of rings, so you've already proven it. Note that in this case $operatorname{Im} varphi = S$.
$endgroup$
– ÍgjøgnumMeg
Dec 2 '18 at 16:45
$begingroup$
Thank you I was just checking that I wasn't assuming that incorrectly @ÍgjøgnumMeg
$endgroup$
– JacobKnight
Dec 4 '18 at 15:22
add a comment |
$begingroup$
An epimorphism of rings is just a surjective homomorphism of rings, so you've already proven it. Note that in this case $operatorname{Im} varphi = S$.
$endgroup$
– ÍgjøgnumMeg
Dec 2 '18 at 16:45
$begingroup$
Thank you I was just checking that I wasn't assuming that incorrectly @ÍgjøgnumMeg
$endgroup$
– JacobKnight
Dec 4 '18 at 15:22
$begingroup$
An epimorphism of rings is just a surjective homomorphism of rings, so you've already proven it. Note that in this case $operatorname{Im} varphi = S$.
$endgroup$
– ÍgjøgnumMeg
Dec 2 '18 at 16:45
$begingroup$
An epimorphism of rings is just a surjective homomorphism of rings, so you've already proven it. Note that in this case $operatorname{Im} varphi = S$.
$endgroup$
– ÍgjøgnumMeg
Dec 2 '18 at 16:45
$begingroup$
Thank you I was just checking that I wasn't assuming that incorrectly @ÍgjøgnumMeg
$endgroup$
– JacobKnight
Dec 4 '18 at 15:22
$begingroup$
Thank you I was just checking that I wasn't assuming that incorrectly @ÍgjøgnumMeg
$endgroup$
– JacobKnight
Dec 4 '18 at 15:22
add a comment |
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$begingroup$
An epimorphism of rings is just a surjective homomorphism of rings, so you've already proven it. Note that in this case $operatorname{Im} varphi = S$.
$endgroup$
– ÍgjøgnumMeg
Dec 2 '18 at 16:45
$begingroup$
Thank you I was just checking that I wasn't assuming that incorrectly @ÍgjøgnumMeg
$endgroup$
– JacobKnight
Dec 4 '18 at 15:22