As a vector space, is $A/I$ isomorphic to $A/mathrm{in}(I)$?
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Let $A=k[x_1, ldots, x_n]$ and "$>$" a monomial order on $A$. For a polynomial $p$, denote by $mathrm{in}_>(p)$ the term of $p$ with the largest monomial. For an ideal $I$ of $A$, the initial ideal of $I$ is defined as $mathrm{in}_{>}(I) = langle mathrm{in}_{>}(p) mid p in I rangle$.
As a vector space, is $A/I$ isomorphic to $A/mathrm{in}_{>}(I)$?
Thank you very much.
algebraic-geometry commutative-algebra
edited 10 hours ago
user26857
39.1k123882
asked 18 hours ago
LJR
6,53141648
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Let $A=k[x_1, ldots, x_n]$ and "$>$" a monomial order on $A$. For a polynomial $p$, denote by $mathrm{in}_>(p)$ the term of $p$ with the largest monomial. For an ideal $I$ of $A$, the initial ideal of $I$ is defined as $mathrm{in}_{>}(I) = langle mathrm{in}_{>}(p) mid p in I rangle$.
As a vector space, is $A/I$ isomorphic to $A/mathrm{in}_{>}(I)$?
Thank you very much.
algebraic-geometry commutative-algebra
edited 10 hours ago
user26857
39.1k123882
asked 18 hours ago
LJR
6,53141648
Yes. All you need to show is that the projections of the reduced monomials (i.e., the monomials not in $in_>(I)$) form a basis of $A/I$. This is the Macaulay-Buchberger basis theorem (see, e.g., Proposition 3.10 in arXiv:1704.00839v6 detailed version for a proof).
– darij grinberg
18 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A=k[x_1, ldots, x_n]$ and "$>$" a monomial order on $A$. For a polynomial $p$, denote by $mathrm{in}_>(p)$ the term of $p$ with the largest monomial. For an ideal $I$ of $A$, the initial ideal of $I$ is defined as $mathrm{in}_{>}(I) = langle mathrm{in}_{>}(p) mid p in I rangle$.
As a vector space, is $A/I$ isomorphic to $A/mathrm{in}_{>}(I)$?
Thank you very much.
algebraic-geometry commutative-algebra
edited 10 hours ago
user26857
39.1k123882
asked 18 hours ago
LJR
6,53141648
Let $A=k[x_1, ldots, x_n]$ and "$>$" a monomial order on $A$. For a polynomial $p$, denote by $mathrm{in}_>(p)$ the term of $p$ with the largest monomial. For an ideal $I$ of $A$, the initial ideal of $I$ is defined as $mathrm{in}_{>}(I) = langle mathrm{in}_{>}(p) mid p in I rangle$.
As a vector space, is $A/I$ isomorphic to $A/mathrm{in}_{>}(I)$?
Thank you very much.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
edited 10 hours ago
user26857
39.1k123882
asked 18 hours ago
LJR
6,53141648
edited 10 hours ago
user26857
39.1k123882
asked 18 hours ago
LJR
6,53141648
edited 10 hours ago
user26857
39.1k123882
edited 10 hours ago
user26857
39.1k123882
edited 10 hours ago
user26857
39.1k123882
39.1k123882
asked 18 hours ago
LJR
6,53141648
asked 18 hours ago
LJR
6,53141648
asked 18 hours ago
LJR
6,53141648
6,53141648
Yes. All you need to show is that the projections of the reduced monomials (i.e., the monomials not in $in_>(I)$) form a basis of $A/I$. This is the Macaulay-Buchberger basis theorem (see, e.g., Proposition 3.10 in arXiv:1704.00839v6 detailed version for a proof).
– darij grinberg
18 hours ago
add a comment |
Yes. All you need to show is that the projections of the reduced monomials (i.e., the monomials not in $in_>(I)$) form a basis of $A/I$. This is the Macaulay-Buchberger basis theorem (see, e.g., Proposition 3.10 in arXiv:1704.00839v6 detailed version for a proof).
– darij grinberg
18 hours ago
Yes. All you need to show is that the projections of the reduced monomials (i.e., the monomials not in $in_>(I)$) form a basis of $A/I$. This is the Macaulay-Buchberger basis theorem (see, e.g., Proposition 3.10 in arXiv:1704.00839v6 detailed version for a proof).
– darij grinberg
18 hours ago
Yes. All you need to show is that the projections of the reduced monomials (i.e., the monomials not in $in_>(I)$) form a basis of $A/I$. This is the Macaulay-Buchberger basis theorem (see, e.g., Proposition 3.10 in arXiv:1704.00839v6 detailed version for a proof).
– darij grinberg
18 hours ago
add a comment |
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Yes. All you need to show is that the projections of the reduced monomials (i.e., the monomials not in $in_>(I)$) form a basis of $A/I$. This is the Macaulay-Buchberger basis theorem (see, e.g., Proposition 3.10 in arXiv:1704.00839v6 detailed version for a proof).
– darij grinberg
18 hours ago