As a vector space, is $A/I$ isomorphic to $A/mathrm{in}(I)$?
$begingroup$
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 '18 at 17:30
user26857
39.3k124083
asked Nov 30 '18 at 9:37
LJRLJR
6,57341749
$endgroup$
add a comment |
$begingroup$
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 '18 at 17:30
user26857
39.3k124083
asked Nov 30 '18 at 9:37
LJRLJR
6,57341749
$endgroup$
$begingroup$
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).
$endgroup$
– darij grinberg
Nov 30 '18 at 9:49
add a comment |
$begingroup$
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 '18 at 17:30
user26857
39.3k124083
asked Nov 30 '18 at 9:37
LJRLJR
6,57341749
$endgroup$
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 '18 at 17:30
user26857
39.3k124083
asked Nov 30 '18 at 9:37
LJRLJR
6,57341749
edited Nov 30 '18 at 17:30
user26857
39.3k124083
asked Nov 30 '18 at 9:37
LJRLJR
6,57341749
edited Nov 30 '18 at 17:30
user26857
39.3k124083
edited Nov 30 '18 at 17:30
user26857
39.3k124083
edited Nov 30 '18 at 17:30
user26857
39.3k124083
39.3k124083
asked Nov 30 '18 at 9:37
LJRLJR
6,57341749
asked Nov 30 '18 at 9:37
LJRLJR
6,57341749
asked Nov 30 '18 at 9:37
LJRLJR
6,57341749
6,57341749
$begingroup$
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).
$endgroup$
– darij grinberg
Nov 30 '18 at 9:49
add a comment |
$begingroup$
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).
$endgroup$
– darij grinberg
Nov 30 '18 at 9:49
$begingroup$
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).
$endgroup$
– darij grinberg
Nov 30 '18 at 9:49
$begingroup$
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).
$endgroup$
– darij grinberg
Nov 30 '18 at 9:49
add a comment |
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$begingroup$
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).
$endgroup$
– darij grinberg
Nov 30 '18 at 9:49