Is associated graded algebra $mathrm{gr}(k[x_1, ldots, x_n]/I)$ isomorphic as a vector space to $k[x_1,...
Let $A=k[x_1, ldots, x_n]$ be the polynomial ring generated by $x_1, ldots, x_n$. Let $I$ be an ideal of $k[x_1, ldots, x_n]$ (it is possible that $I$ is not homogeneous).
The algebra $A$ is a graded algebra: $A = oplus_{i ge 0} A_i$, where $A_i$ consists of degree $i$ homogeneous polynomials.
The algebra $A/I=k[x_1, ldots, x_n]/I$ is a filtered algebra with the filtration $F_i(A/I) = (F_i(A)+I)/I$, where $F_i(A)=oplus_{j le i} A_j$.
Is associated graded algebra $$mathrm{gr}(k[x_1, ldots, x_n]/I)=mathrm{gr}(A/I)=oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$$ isomorphic as a vector space to $k[x_1, ldots, x_n]/I$?
Thank you very much.
abstract-algebra commutative-algebra
edited Dec 1 '18 at 10:22
user26857
39.3k124083
asked Nov 29 '18 at 13:58
LJRLJR
6,55841749
add a comment |
Let $A=k[x_1, ldots, x_n]$ be the polynomial ring generated by $x_1, ldots, x_n$. Let $I$ be an ideal of $k[x_1, ldots, x_n]$ (it is possible that $I$ is not homogeneous).
The algebra $A$ is a graded algebra: $A = oplus_{i ge 0} A_i$, where $A_i$ consists of degree $i$ homogeneous polynomials.
The algebra $A/I=k[x_1, ldots, x_n]/I$ is a filtered algebra with the filtration $F_i(A/I) = (F_i(A)+I)/I$, where $F_i(A)=oplus_{j le i} A_j$.
Is associated graded algebra $$mathrm{gr}(k[x_1, ldots, x_n]/I)=mathrm{gr}(A/I)=oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$$ isomorphic as a vector space to $k[x_1, ldots, x_n]/I$?
Thank you very much.
abstract-algebra commutative-algebra
edited Dec 1 '18 at 10:22
user26857
39.3k124083
asked Nov 29 '18 at 13:58
LJRLJR
6,55841749
add a comment |
Let $A=k[x_1, ldots, x_n]$ be the polynomial ring generated by $x_1, ldots, x_n$. Let $I$ be an ideal of $k[x_1, ldots, x_n]$ (it is possible that $I$ is not homogeneous).
The algebra $A$ is a graded algebra: $A = oplus_{i ge 0} A_i$, where $A_i$ consists of degree $i$ homogeneous polynomials.
The algebra $A/I=k[x_1, ldots, x_n]/I$ is a filtered algebra with the filtration $F_i(A/I) = (F_i(A)+I)/I$, where $F_i(A)=oplus_{j le i} A_j$.
Is associated graded algebra $$mathrm{gr}(k[x_1, ldots, x_n]/I)=mathrm{gr}(A/I)=oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$$ isomorphic as a vector space to $k[x_1, ldots, x_n]/I$?
Thank you very much.
abstract-algebra commutative-algebra
edited Dec 1 '18 at 10:22
user26857
39.3k124083
asked Nov 29 '18 at 13:58
LJRLJR
6,55841749
Let $A=k[x_1, ldots, x_n]$ be the polynomial ring generated by $x_1, ldots, x_n$. Let $I$ be an ideal of $k[x_1, ldots, x_n]$ (it is possible that $I$ is not homogeneous).
The algebra $A$ is a graded algebra: $A = oplus_{i ge 0} A_i$, where $A_i$ consists of degree $i$ homogeneous polynomials.
The algebra $A/I=k[x_1, ldots, x_n]/I$ is a filtered algebra with the filtration $F_i(A/I) = (F_i(A)+I)/I$, where $F_i(A)=oplus_{j le i} A_j$.
Is associated graded algebra $$mathrm{gr}(k[x_1, ldots, x_n]/I)=mathrm{gr}(A/I)=oplus_{i ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$$ isomorphic as a vector space to $k[x_1, ldots, x_n]/I$?
Thank you very much.
abstract-algebra commutative-algebra
abstract-algebra commutative-algebra
edited Dec 1 '18 at 10:22
user26857
39.3k124083
asked Nov 29 '18 at 13:58
LJRLJR
6,55841749
edited Dec 1 '18 at 10:22
user26857
39.3k124083
asked Nov 29 '18 at 13:58
LJRLJR
6,55841749
edited Dec 1 '18 at 10:22
user26857
39.3k124083
edited Dec 1 '18 at 10:22
user26857
39.3k124083
edited Dec 1 '18 at 10:22
user26857
39.3k124083
39.3k124083
asked Nov 29 '18 at 13:58
LJRLJR
6,55841749
asked Nov 29 '18 at 13:58
LJRLJR
6,55841749
asked Nov 29 '18 at 13:58
LJRLJR
6,55841749
6,55841749
add a comment |
add a comment |
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Yes.
Observe that $mathrm{gr}(A/I)$ is also a filtered algebra, with $F_i(mathrm{gr}(A/I)) = oplus_{i=0}^r~mathrm{gr}(A/I)_i$.
Further observe that $$dim F_i(mathrm{gr}(A/I) = sum_{i=0}^r dim mathrm{gr}(A/I)_i = sum_{i=0}^r (dim F_i(A/I) - dim F_{i-1}(A/I)) = dim F_i(A/I).$$
Therefore $mathrm{gr}(A/I)$ is isomorphic to $A/I$ as a vector space.
edited Dec 1 '18 at 10:24
user26857
39.3k124083
answered Nov 29 '18 at 15:53
ChristopherChristopher
6,45711628
add a comment |
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1 Answer
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Yes.
Observe that $mathrm{gr}(A/I)$ is also a filtered algebra, with $F_i(mathrm{gr}(A/I)) = oplus_{i=0}^r~mathrm{gr}(A/I)_i$.
Further observe that $$dim F_i(mathrm{gr}(A/I) = sum_{i=0}^r dim mathrm{gr}(A/I)_i = sum_{i=0}^r (dim F_i(A/I) - dim F_{i-1}(A/I)) = dim F_i(A/I).$$
Therefore $mathrm{gr}(A/I)$ is isomorphic to $A/I$ as a vector space.
edited Dec 1 '18 at 10:24
user26857
39.3k124083
answered Nov 29 '18 at 15:53
ChristopherChristopher
6,45711628
add a comment |
Yes.
Observe that $mathrm{gr}(A/I)$ is also a filtered algebra, with $F_i(mathrm{gr}(A/I)) = oplus_{i=0}^r~mathrm{gr}(A/I)_i$.
Further observe that $$dim F_i(mathrm{gr}(A/I) = sum_{i=0}^r dim mathrm{gr}(A/I)_i = sum_{i=0}^r (dim F_i(A/I) - dim F_{i-1}(A/I)) = dim F_i(A/I).$$
Therefore $mathrm{gr}(A/I)$ is isomorphic to $A/I$ as a vector space.
edited Dec 1 '18 at 10:24
user26857
39.3k124083
answered Nov 29 '18 at 15:53
ChristopherChristopher
6,45711628
add a comment |
Yes.
Observe that $mathrm{gr}(A/I)$ is also a filtered algebra, with $F_i(mathrm{gr}(A/I)) = oplus_{i=0}^r~mathrm{gr}(A/I)_i$.
Further observe that $$dim F_i(mathrm{gr}(A/I) = sum_{i=0}^r dim mathrm{gr}(A/I)_i = sum_{i=0}^r (dim F_i(A/I) - dim F_{i-1}(A/I)) = dim F_i(A/I).$$
Therefore $mathrm{gr}(A/I)$ is isomorphic to $A/I$ as a vector space.
edited Dec 1 '18 at 10:24
user26857
39.3k124083
answered Nov 29 '18 at 15:53
ChristopherChristopher
6,45711628
Yes.
Observe that $mathrm{gr}(A/I)$ is also a filtered algebra, with $F_i(mathrm{gr}(A/I)) = oplus_{i=0}^r~mathrm{gr}(A/I)_i$.
Further observe that $$dim F_i(mathrm{gr}(A/I) = sum_{i=0}^r dim mathrm{gr}(A/I)_i = sum_{i=0}^r (dim F_i(A/I) - dim F_{i-1}(A/I)) = dim F_i(A/I).$$
Therefore $mathrm{gr}(A/I)$ is isomorphic to $A/I$ as a vector space.
edited Dec 1 '18 at 10:24
user26857
39.3k124083
answered Nov 29 '18 at 15:53
ChristopherChristopher
6,45711628
edited Dec 1 '18 at 10:24
user26857
39.3k124083
edited Dec 1 '18 at 10:24
user26857
39.3k124083
edited Dec 1 '18 at 10:24
user26857
39.3k124083
39.3k124083
answered Nov 29 '18 at 15:53
ChristopherChristopher
6,45711628
answered Nov 29 '18 at 15:53
ChristopherChristopher
6,45711628
answered Nov 29 '18 at 15:53
ChristopherChristopher
6,45711628
6,45711628
add a comment |
add a comment |
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