How to compute associated graded algebras?
I am trying understand associated graded algebras.
Let $A=k[x,y]$ and $I=langle xy^2, x^3-y^2 rangle$. Let $R=A/I$. In the webpage, it is said that $mathrm{gr}(R) = A/(x^4, y^2)$.
Let $A=k[x_1, ldots, x_n]$ and ``$>$'' a monomial order on $A$. For a polynomial $p$, denote by $in_>(p)$ the ideal the term of $p$ with the largest monomial. For an ideal $I$ of $A$, the initial ideal of $I$ is defined as $in_{>}(I) = langle in_{>}(p) mid p in I rangle$.
I think that $gr(R)$ is isomorphic to $A/in(I)$ as a vector space, where $in(I)$ is the initial ideal of $I$. Is this correct? I obtain that $gr(R)$ is isomorphic to $k[x,y]/(xy^2, x^3, y^4)$ as a vector space. Is this correct? Thank you very much.
abstract-algebra algebraic-geometry commutative-algebra
edited Nov 30 '18 at 17:42
user26857
39.3k124083
asked Nov 29 '18 at 14:23
LJRLJR
6,55841749
add a comment |
I am trying understand associated graded algebras.
Let $A=k[x,y]$ and $I=langle xy^2, x^3-y^2 rangle$. Let $R=A/I$. In the webpage, it is said that $mathrm{gr}(R) = A/(x^4, y^2)$.
Let $A=k[x_1, ldots, x_n]$ and ``$>$'' a monomial order on $A$. For a polynomial $p$, denote by $in_>(p)$ the ideal the term of $p$ with the largest monomial. For an ideal $I$ of $A$, the initial ideal of $I$ is defined as $in_{>}(I) = langle in_{>}(p) mid p in I rangle$.
I think that $gr(R)$ is isomorphic to $A/in(I)$ as a vector space, where $in(I)$ is the initial ideal of $I$. Is this correct? I obtain that $gr(R)$ is isomorphic to $k[x,y]/(xy^2, x^3, y^4)$ as a vector space. Is this correct? Thank you very much.
abstract-algebra algebraic-geometry commutative-algebra
edited Nov 30 '18 at 17:42
user26857
39.3k124083
asked Nov 29 '18 at 14:23
LJRLJR
6,55841749
You need to use a different term order, not monomial order. The largest terms are the terms of the least degree. For $x^3-y^2$, it is $y^2$ since it is of the least degree.
– Youngsu
Nov 30 '18 at 4:21
@LJR I don't understand: do you want to compute the associated graded ring as a $k$-vector space or to figure out why the associated graded ring is $A/(x^4,y^2)$?
– user26857
Dec 1 '18 at 9:15
Looking at the history of your questions you seem to be puzzled by the isomorphism from that webpage and what you got. But if you count the dimension of both vector spaces can conclude that these are also isomorphic.
– user26857
Dec 1 '18 at 11:06
@user26857, thank you very much. I would like to figure out why the associated graded ring is $A/(x^4, y^2)$. How to compute $gr(A/I)$ as a ring? I think we can compute $gr(A/I)$ as a vector space by computing $A/in(I)$. But I don't know how to compute $gr(A/I)$ as a ring.
– LJR
Dec 1 '18 at 13:54
@LJRL I don't have a method for computing the associated graded ring. In this particular case I'd simply try to find out the relations satisfied by x and y in the associated graded ring.
– user26857
Dec 1 '18 at 15:19
add a comment |
I am trying understand associated graded algebras.
Let $A=k[x,y]$ and $I=langle xy^2, x^3-y^2 rangle$. Let $R=A/I$. In the webpage, it is said that $mathrm{gr}(R) = A/(x^4, y^2)$.
Let $A=k[x_1, ldots, x_n]$ and ``$>$'' a monomial order on $A$. For a polynomial $p$, denote by $in_>(p)$ the ideal the term of $p$ with the largest monomial. For an ideal $I$ of $A$, the initial ideal of $I$ is defined as $in_{>}(I) = langle in_{>}(p) mid p in I rangle$.
I think that $gr(R)$ is isomorphic to $A/in(I)$ as a vector space, where $in(I)$ is the initial ideal of $I$. Is this correct? I obtain that $gr(R)$ is isomorphic to $k[x,y]/(xy^2, x^3, y^4)$ as a vector space. Is this correct? Thank you very much.
abstract-algebra algebraic-geometry commutative-algebra
edited Nov 30 '18 at 17:42
user26857
39.3k124083
asked Nov 29 '18 at 14:23
LJRLJR
6,55841749
I am trying understand associated graded algebras.
Let $A=k[x,y]$ and $I=langle xy^2, x^3-y^2 rangle$. Let $R=A/I$. In the webpage, it is said that $mathrm{gr}(R) = A/(x^4, y^2)$.
Let $A=k[x_1, ldots, x_n]$ and ``$>$'' a monomial order on $A$. For a polynomial $p$, denote by $in_>(p)$ the ideal the term of $p$ with the largest monomial. For an ideal $I$ of $A$, the initial ideal of $I$ is defined as $in_{>}(I) = langle in_{>}(p) mid p in I rangle$.
I think that $gr(R)$ is isomorphic to $A/in(I)$ as a vector space, where $in(I)$ is the initial ideal of $I$. Is this correct? I obtain that $gr(R)$ is isomorphic to $k[x,y]/(xy^2, x^3, y^4)$ as a vector space. Is this correct? Thank you very much.
abstract-algebra algebraic-geometry commutative-algebra
abstract-algebra algebraic-geometry commutative-algebra
edited Nov 30 '18 at 17:42
user26857
39.3k124083
asked Nov 29 '18 at 14:23
LJRLJR
6,55841749
edited Nov 30 '18 at 17:42
user26857
39.3k124083
asked Nov 29 '18 at 14:23
LJRLJR
6,55841749
edited Nov 30 '18 at 17:42
user26857
39.3k124083
edited Nov 30 '18 at 17:42
user26857
39.3k124083
edited Nov 30 '18 at 17:42
user26857
39.3k124083
39.3k124083
asked Nov 29 '18 at 14:23
LJRLJR
6,55841749
asked Nov 29 '18 at 14:23
LJRLJR
6,55841749
asked Nov 29 '18 at 14:23
LJRLJR
6,55841749
6,55841749
You need to use a different term order, not monomial order. The largest terms are the terms of the least degree. For $x^3-y^2$, it is $y^2$ since it is of the least degree.
– Youngsu
Nov 30 '18 at 4:21
@LJR I don't understand: do you want to compute the associated graded ring as a $k$-vector space or to figure out why the associated graded ring is $A/(x^4,y^2)$?
– user26857
Dec 1 '18 at 9:15
Looking at the history of your questions you seem to be puzzled by the isomorphism from that webpage and what you got. But if you count the dimension of both vector spaces can conclude that these are also isomorphic.
– user26857
Dec 1 '18 at 11:06
@user26857, thank you very much. I would like to figure out why the associated graded ring is $A/(x^4, y^2)$. How to compute $gr(A/I)$ as a ring? I think we can compute $gr(A/I)$ as a vector space by computing $A/in(I)$. But I don't know how to compute $gr(A/I)$ as a ring.
– LJR
Dec 1 '18 at 13:54
@LJRL I don't have a method for computing the associated graded ring. In this particular case I'd simply try to find out the relations satisfied by x and y in the associated graded ring.
– user26857
Dec 1 '18 at 15:19
add a comment |
You need to use a different term order, not monomial order. The largest terms are the terms of the least degree. For $x^3-y^2$, it is $y^2$ since it is of the least degree.
– Youngsu
Nov 30 '18 at 4:21
@LJR I don't understand: do you want to compute the associated graded ring as a $k$-vector space or to figure out why the associated graded ring is $A/(x^4,y^2)$?
– user26857
Dec 1 '18 at 9:15
Looking at the history of your questions you seem to be puzzled by the isomorphism from that webpage and what you got. But if you count the dimension of both vector spaces can conclude that these are also isomorphic.
– user26857
Dec 1 '18 at 11:06
@user26857, thank you very much. I would like to figure out why the associated graded ring is $A/(x^4, y^2)$. How to compute $gr(A/I)$ as a ring? I think we can compute $gr(A/I)$ as a vector space by computing $A/in(I)$. But I don't know how to compute $gr(A/I)$ as a ring.
– LJR
Dec 1 '18 at 13:54
@LJRL I don't have a method for computing the associated graded ring. In this particular case I'd simply try to find out the relations satisfied by x and y in the associated graded ring.
– user26857
Dec 1 '18 at 15:19
You need to use a different term order, not monomial order. The largest terms are the terms of the least degree. For $x^3-y^2$, it is $y^2$ since it is of the least degree.
– Youngsu
Nov 30 '18 at 4:21
You need to use a different term order, not monomial order. The largest terms are the terms of the least degree. For $x^3-y^2$, it is $y^2$ since it is of the least degree.
– Youngsu
Nov 30 '18 at 4:21
@LJR I don't understand: do you want to compute the associated graded ring as a $k$-vector space or to figure out why the associated graded ring is $A/(x^4,y^2)$?
– user26857
Dec 1 '18 at 9:15
@LJR I don't understand: do you want to compute the associated graded ring as a $k$-vector space or to figure out why the associated graded ring is $A/(x^4,y^2)$?
– user26857
Dec 1 '18 at 9:15
Looking at the history of your questions you seem to be puzzled by the isomorphism from that webpage and what you got. But if you count the dimension of both vector spaces can conclude that these are also isomorphic.
– user26857
Dec 1 '18 at 11:06
Looking at the history of your questions you seem to be puzzled by the isomorphism from that webpage and what you got. But if you count the dimension of both vector spaces can conclude that these are also isomorphic.
– user26857
Dec 1 '18 at 11:06
@user26857, thank you very much. I would like to figure out why the associated graded ring is $A/(x^4, y^2)$. How to compute $gr(A/I)$ as a ring? I think we can compute $gr(A/I)$ as a vector space by computing $A/in(I)$. But I don't know how to compute $gr(A/I)$ as a ring.
– LJR
Dec 1 '18 at 13:54
@user26857, thank you very much. I would like to figure out why the associated graded ring is $A/(x^4, y^2)$. How to compute $gr(A/I)$ as a ring? I think we can compute $gr(A/I)$ as a vector space by computing $A/in(I)$. But I don't know how to compute $gr(A/I)$ as a ring.
– LJR
Dec 1 '18 at 13:54
@LJRL I don't have a method for computing the associated graded ring. In this particular case I'd simply try to find out the relations satisfied by x and y in the associated graded ring.
– user26857
Dec 1 '18 at 15:19
@LJRL I don't have a method for computing the associated graded ring. In this particular case I'd simply try to find out the relations satisfied by x and y in the associated graded ring.
– user26857
Dec 1 '18 at 15:19
add a comment |
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You need to use a different term order, not monomial order. The largest terms are the terms of the least degree. For $x^3-y^2$, it is $y^2$ since it is of the least degree.
– Youngsu
Nov 30 '18 at 4:21
@LJR I don't understand: do you want to compute the associated graded ring as a $k$-vector space or to figure out why the associated graded ring is $A/(x^4,y^2)$?
– user26857
Dec 1 '18 at 9:15
Looking at the history of your questions you seem to be puzzled by the isomorphism from that webpage and what you got. But if you count the dimension of both vector spaces can conclude that these are also isomorphic.
– user26857
Dec 1 '18 at 11:06
@user26857, thank you very much. I would like to figure out why the associated graded ring is $A/(x^4, y^2)$. How to compute $gr(A/I)$ as a ring? I think we can compute $gr(A/I)$ as a vector space by computing $A/in(I)$. But I don't know how to compute $gr(A/I)$ as a ring.
– LJR
Dec 1 '18 at 13:54
@LJRL I don't have a method for computing the associated graded ring. In this particular case I'd simply try to find out the relations satisfied by x and y in the associated graded ring.
– user26857
Dec 1 '18 at 15:19