As a vector space, is $A/I$ isomorphic to $A/mathrm{in}(I)$?
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 Nov 30 at 17:30
user26857
39.2k123882
asked Nov 30 at 9:37
LJR
6,55641649
add a comment |
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 Nov 30 at 17:30
user26857
39.2k123882
asked Nov 30 at 9:37
LJR
6,55641649
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
Nov 30 at 9:49
add a comment |
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 Nov 30 at 17:30
user26857
39.2k123882
asked Nov 30 at 9:37
LJR
6,55641649
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 Nov 30 at 17:30
user26857
39.2k123882
asked Nov 30 at 9:37
LJR
6,55641649
edited Nov 30 at 17:30
user26857
39.2k123882
asked Nov 30 at 9:37
LJR
6,55641649
edited Nov 30 at 17:30
user26857
39.2k123882
edited Nov 30 at 17:30
user26857
39.2k123882
edited Nov 30 at 17:30
user26857
39.2k123882
39.2k123882
asked Nov 30 at 9:37
LJR
6,55641649
asked Nov 30 at 9:37
LJR
6,55641649
asked Nov 30 at 9:37
LJR
6,55641649
6,55641649
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
Nov 30 at 9:49
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
Nov 30 at 9:49
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
Nov 30 at 9:49
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
Nov 30 at 9:49
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
Nov 30 at 9:49