How to prove an isomorphism related to a filtration.
Let $A$ be a filtered algebra with a filtration $F_0(A) subset F_1(A) subset cdots subset A$. Let $I$ be a two sided ideal of $A$. The algebra $A/I$ has a filtration $F_i(A/I)=F_i(A)/(F_i(A)cap I)$. In this post it is said that
begin{align}
F_i(A)/(F_{i-1}(A)+F_i(A) cap I) cong frac{ F_i(A)/(F_i(A) cap I) }{F_{i-1}(A)/(F_{i-1}(A) cap I)}.
end{align}
How to prove this identity? The third isomorphism theorem says that $(A/I)/(B/I)cong A/B$. But here $F_{i}(A) cap I ne F_{i-1}(A) cap I$. We cannot use the third isomorphism theorem directly. Thank you very much.
abstract-algebra ring-theory filtrations
edited Dec 1 at 10:57
user26857
39.2k123882
asked Nov 15 at 17:16
LJR
6,55641649
add a comment |
Let $A$ be a filtered algebra with a filtration $F_0(A) subset F_1(A) subset cdots subset A$. Let $I$ be a two sided ideal of $A$. The algebra $A/I$ has a filtration $F_i(A/I)=F_i(A)/(F_i(A)cap I)$. In this post it is said that
begin{align}
F_i(A)/(F_{i-1}(A)+F_i(A) cap I) cong frac{ F_i(A)/(F_i(A) cap I) }{F_{i-1}(A)/(F_{i-1}(A) cap I)}.
end{align}
How to prove this identity? The third isomorphism theorem says that $(A/I)/(B/I)cong A/B$. But here $F_{i}(A) cap I ne F_{i-1}(A) cap I$. We cannot use the third isomorphism theorem directly. Thank you very much.
abstract-algebra ring-theory filtrations
edited Dec 1 at 10:57
user26857
39.2k123882
asked Nov 15 at 17:16
LJR
6,55641649
add a comment |
Let $A$ be a filtered algebra with a filtration $F_0(A) subset F_1(A) subset cdots subset A$. Let $I$ be a two sided ideal of $A$. The algebra $A/I$ has a filtration $F_i(A/I)=F_i(A)/(F_i(A)cap I)$. In this post it is said that
begin{align}
F_i(A)/(F_{i-1}(A)+F_i(A) cap I) cong frac{ F_i(A)/(F_i(A) cap I) }{F_{i-1}(A)/(F_{i-1}(A) cap I)}.
end{align}
How to prove this identity? The third isomorphism theorem says that $(A/I)/(B/I)cong A/B$. But here $F_{i}(A) cap I ne F_{i-1}(A) cap I$. We cannot use the third isomorphism theorem directly. Thank you very much.
abstract-algebra ring-theory filtrations
edited Dec 1 at 10:57
user26857
39.2k123882
asked Nov 15 at 17:16
LJR
6,55641649
Let $A$ be a filtered algebra with a filtration $F_0(A) subset F_1(A) subset cdots subset A$. Let $I$ be a two sided ideal of $A$. The algebra $A/I$ has a filtration $F_i(A/I)=F_i(A)/(F_i(A)cap I)$. In this post it is said that
begin{align}
F_i(A)/(F_{i-1}(A)+F_i(A) cap I) cong frac{ F_i(A)/(F_i(A) cap I) }{F_{i-1}(A)/(F_{i-1}(A) cap I)}.
end{align}
How to prove this identity? The third isomorphism theorem says that $(A/I)/(B/I)cong A/B$. But here $F_{i}(A) cap I ne F_{i-1}(A) cap I$. We cannot use the third isomorphism theorem directly. Thank you very much.
abstract-algebra ring-theory filtrations
abstract-algebra ring-theory filtrations
edited Dec 1 at 10:57
user26857
39.2k123882
asked Nov 15 at 17:16
LJR
6,55641649
edited Dec 1 at 10:57
user26857
39.2k123882
asked Nov 15 at 17:16
LJR
6,55641649
edited Dec 1 at 10:57
user26857
39.2k123882
edited Dec 1 at 10:57
user26857
39.2k123882
edited Dec 1 at 10:57
user26857
39.2k123882
39.2k123882
asked Nov 15 at 17:16
LJR
6,55641649
asked Nov 15 at 17:16
LJR
6,55641649
asked Nov 15 at 17:16
LJR
6,55641649
6,55641649
add a comment |
add a comment |
1 Answer
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$frac{ F_i(A)/(F_i(A) cap I) }{F_{i-1}(A)/(F_{i-1}(A) cap I)}$ is not well-defined since $F_{i-1}(A)/(F_{i-1}(A) cap I)$ is not a subspace of $F_i(A)/(F_i(A) cap I)$.
A proof of $mathrm{gr}(A/I)= oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$ is in the following.
begin{align}
& F_i(A/I) = F_i(A)/(F_{i}(A) cap I) = (F_i(A) + I)/I,
end{align}
begin{align}
& mathrm{gr}(A/I) = oplus_{i ge 0} F_i(A/I)/F_{i-1}(A/I) \
& = oplus_{i ge 0} ((F_i(A) + I)/I)/((F_{i-1}(A) + I)/I) \
& = oplus_{i ge 0} (F_i(A) + I)/(F_{i-1}(A) + I) \
& = oplus_{i ge 0} F_i(A)/(F_i(A) cap ( F_{i-1}(A) + I )) \
& = oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I),
end{align}
where the second and third isomorphism theorems are used.
edited Dec 1 at 10:58
user26857
39.2k123882
answered Nov 29 at 14:07
LJR
6,55641649
add a comment |
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$frac{ F_i(A)/(F_i(A) cap I) }{F_{i-1}(A)/(F_{i-1}(A) cap I)}$ is not well-defined since $F_{i-1}(A)/(F_{i-1}(A) cap I)$ is not a subspace of $F_i(A)/(F_i(A) cap I)$.
A proof of $mathrm{gr}(A/I)= oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$ is in the following.
begin{align}
& F_i(A/I) = F_i(A)/(F_{i}(A) cap I) = (F_i(A) + I)/I,
end{align}
begin{align}
& mathrm{gr}(A/I) = oplus_{i ge 0} F_i(A/I)/F_{i-1}(A/I) \
& = oplus_{i ge 0} ((F_i(A) + I)/I)/((F_{i-1}(A) + I)/I) \
& = oplus_{i ge 0} (F_i(A) + I)/(F_{i-1}(A) + I) \
& = oplus_{i ge 0} F_i(A)/(F_i(A) cap ( F_{i-1}(A) + I )) \
& = oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I),
end{align}
where the second and third isomorphism theorems are used.
edited Dec 1 at 10:58
user26857
39.2k123882
answered Nov 29 at 14:07
LJR
6,55641649
add a comment |
$frac{ F_i(A)/(F_i(A) cap I) }{F_{i-1}(A)/(F_{i-1}(A) cap I)}$ is not well-defined since $F_{i-1}(A)/(F_{i-1}(A) cap I)$ is not a subspace of $F_i(A)/(F_i(A) cap I)$.
A proof of $mathrm{gr}(A/I)= oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$ is in the following.
begin{align}
& F_i(A/I) = F_i(A)/(F_{i}(A) cap I) = (F_i(A) + I)/I,
end{align}
begin{align}
& mathrm{gr}(A/I) = oplus_{i ge 0} F_i(A/I)/F_{i-1}(A/I) \
& = oplus_{i ge 0} ((F_i(A) + I)/I)/((F_{i-1}(A) + I)/I) \
& = oplus_{i ge 0} (F_i(A) + I)/(F_{i-1}(A) + I) \
& = oplus_{i ge 0} F_i(A)/(F_i(A) cap ( F_{i-1}(A) + I )) \
& = oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I),
end{align}
where the second and third isomorphism theorems are used.
edited Dec 1 at 10:58
user26857
39.2k123882
answered Nov 29 at 14:07
LJR
6,55641649
add a comment |
$frac{ F_i(A)/(F_i(A) cap I) }{F_{i-1}(A)/(F_{i-1}(A) cap I)}$ is not well-defined since $F_{i-1}(A)/(F_{i-1}(A) cap I)$ is not a subspace of $F_i(A)/(F_i(A) cap I)$.
A proof of $mathrm{gr}(A/I)= oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$ is in the following.
begin{align}
& F_i(A/I) = F_i(A)/(F_{i}(A) cap I) = (F_i(A) + I)/I,
end{align}
begin{align}
& mathrm{gr}(A/I) = oplus_{i ge 0} F_i(A/I)/F_{i-1}(A/I) \
& = oplus_{i ge 0} ((F_i(A) + I)/I)/((F_{i-1}(A) + I)/I) \
& = oplus_{i ge 0} (F_i(A) + I)/(F_{i-1}(A) + I) \
& = oplus_{i ge 0} F_i(A)/(F_i(A) cap ( F_{i-1}(A) + I )) \
& = oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I),
end{align}
where the second and third isomorphism theorems are used.
edited Dec 1 at 10:58
user26857
39.2k123882
answered Nov 29 at 14:07
LJR
6,55641649
$frac{ F_i(A)/(F_i(A) cap I) }{F_{i-1}(A)/(F_{i-1}(A) cap I)}$ is not well-defined since $F_{i-1}(A)/(F_{i-1}(A) cap I)$ is not a subspace of $F_i(A)/(F_i(A) cap I)$.
A proof of $mathrm{gr}(A/I)= oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$ is in the following.
begin{align}
& F_i(A/I) = F_i(A)/(F_{i}(A) cap I) = (F_i(A) + I)/I,
end{align}
begin{align}
& mathrm{gr}(A/I) = oplus_{i ge 0} F_i(A/I)/F_{i-1}(A/I) \
& = oplus_{i ge 0} ((F_i(A) + I)/I)/((F_{i-1}(A) + I)/I) \
& = oplus_{i ge 0} (F_i(A) + I)/(F_{i-1}(A) + I) \
& = oplus_{i ge 0} F_i(A)/(F_i(A) cap ( F_{i-1}(A) + I )) \
& = oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I),
end{align}
where the second and third isomorphism theorems are used.
edited Dec 1 at 10:58
user26857
39.2k123882
answered Nov 29 at 14:07
LJR
6,55641649
edited Dec 1 at 10:58
user26857
39.2k123882
edited Dec 1 at 10:58
user26857
39.2k123882
edited Dec 1 at 10:58
user26857
39.2k123882
39.2k123882
answered Nov 29 at 14:07
LJR
6,55641649
answered Nov 29 at 14:07
LJR
6,55641649
answered Nov 29 at 14:07
LJR
6,55641649
6,55641649
add a comment |
add a comment |
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