Vector space basis for a quotient ring
$begingroup$
Given the polynomial ring $mathbb{F}_2[x,y,z]$ with some ideal $I$, what is the method to find a basis (as a vector space over $mathbb{F}_2$) for the quotient ring $R/I$?
For example, take
$$
S = mathbb{F_2}[x,y,z]/(z^2+zx^{2l}−1+zx+1;y^2+yx^{2l}−1+yx+1; x^{2g}+1)
$$
where $l$ and $g$ are integers. We know that $S$ can be expressed with a Groebner basis. How do I find a linear basis or the dimension of $S$?
commutative-algebra groebner-basis groebner-generators
edited Dec 21 '18 at 9:58
user26857
39.4k124183
asked Dec 19 '18 at 14:06
cleanplaycleanplay
209110
$endgroup$
add a comment |
$begingroup$
Given the polynomial ring $mathbb{F}_2[x,y,z]$ with some ideal $I$, what is the method to find a basis (as a vector space over $mathbb{F}_2$) for the quotient ring $R/I$?
For example, take
$$
S = mathbb{F_2}[x,y,z]/(z^2+zx^{2l}−1+zx+1;y^2+yx^{2l}−1+yx+1; x^{2g}+1)
$$
where $l$ and $g$ are integers. We know that $S$ can be expressed with a Groebner basis. How do I find a linear basis or the dimension of $S$?
commutative-algebra groebner-basis groebner-generators
edited Dec 21 '18 at 9:58
user26857
39.4k124183
asked Dec 19 '18 at 14:06
cleanplaycleanplay
209110
$endgroup$
1
$begingroup$
You mean a basis as a vector space over $mathbb{F}_2$?
$endgroup$
– Daniel Schepler
Dec 19 '18 at 22:38
$begingroup$
@DanielSchepler yes, I meant as a vector space over $mathbb{F}_2$. For example, the basis for $mathbb{F}_2[x,y]/(x+y,y^2+1)$ is ${1,x}$. Let me clarify that in the question.
$endgroup$
– cleanplay
Dec 19 '18 at 22:43
add a comment |
$begingroup$
Given the polynomial ring $mathbb{F}_2[x,y,z]$ with some ideal $I$, what is the method to find a basis (as a vector space over $mathbb{F}_2$) for the quotient ring $R/I$?
For example, take
$$
S = mathbb{F_2}[x,y,z]/(z^2+zx^{2l}−1+zx+1;y^2+yx^{2l}−1+yx+1; x^{2g}+1)
$$
where $l$ and $g$ are integers. We know that $S$ can be expressed with a Groebner basis. How do I find a linear basis or the dimension of $S$?
commutative-algebra groebner-basis groebner-generators
edited Dec 21 '18 at 9:58
user26857
39.4k124183
asked Dec 19 '18 at 14:06
cleanplaycleanplay
209110
$endgroup$
Given the polynomial ring $mathbb{F}_2[x,y,z]$ with some ideal $I$, what is the method to find a basis (as a vector space over $mathbb{F}_2$) for the quotient ring $R/I$?
For example, take
$$
S = mathbb{F_2}[x,y,z]/(z^2+zx^{2l}−1+zx+1;y^2+yx^{2l}−1+yx+1; x^{2g}+1)
$$
where $l$ and $g$ are integers. We know that $S$ can be expressed with a Groebner basis. How do I find a linear basis or the dimension of $S$?
commutative-algebra groebner-basis groebner-generators
commutative-algebra groebner-basis groebner-generators
edited Dec 21 '18 at 9:58
user26857
39.4k124183
asked Dec 19 '18 at 14:06
cleanplaycleanplay
209110
edited Dec 21 '18 at 9:58
user26857
39.4k124183
asked Dec 19 '18 at 14:06
cleanplaycleanplay
209110
edited Dec 21 '18 at 9:58
user26857
39.4k124183
edited Dec 21 '18 at 9:58
user26857
39.4k124183
edited Dec 21 '18 at 9:58
user26857
39.4k124183
39.4k124183
asked Dec 19 '18 at 14:06
cleanplaycleanplay
209110
asked Dec 19 '18 at 14:06
cleanplaycleanplay
209110
asked Dec 19 '18 at 14:06
cleanplaycleanplay
209110
209110
1
$begingroup$
You mean a basis as a vector space over $mathbb{F}_2$?
$endgroup$
– Daniel Schepler
Dec 19 '18 at 22:38
$begingroup$
@DanielSchepler yes, I meant as a vector space over $mathbb{F}_2$. For example, the basis for $mathbb{F}_2[x,y]/(x+y,y^2+1)$ is ${1,x}$. Let me clarify that in the question.
$endgroup$
– cleanplay
Dec 19 '18 at 22:43
add a comment |
1
$begingroup$
You mean a basis as a vector space over $mathbb{F}_2$?
$endgroup$
– Daniel Schepler
Dec 19 '18 at 22:38
$begingroup$
@DanielSchepler yes, I meant as a vector space over $mathbb{F}_2$. For example, the basis for $mathbb{F}_2[x,y]/(x+y,y^2+1)$ is ${1,x}$. Let me clarify that in the question.
$endgroup$
– cleanplay
Dec 19 '18 at 22:43
1
1
$begingroup$
You mean a basis as a vector space over $mathbb{F}_2$?
$endgroup$
– Daniel Schepler
Dec 19 '18 at 22:38
$begingroup$
You mean a basis as a vector space over $mathbb{F}_2$?
$endgroup$
– Daniel Schepler
Dec 19 '18 at 22:38
$begingroup$
@DanielSchepler yes, I meant as a vector space over $mathbb{F}_2$. For example, the basis for $mathbb{F}_2[x,y]/(x+y,y^2+1)$ is ${1,x}$. Let me clarify that in the question.
$endgroup$
– cleanplay
Dec 19 '18 at 22:43
$begingroup$
@DanielSchepler yes, I meant as a vector space over $mathbb{F}_2$. For example, the basis for $mathbb{F}_2[x,y]/(x+y,y^2+1)$ is ${1,x}$. Let me clarify that in the question.
$endgroup$
– cleanplay
Dec 19 '18 at 22:43
add a comment |
1 Answer
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$begingroup$
Suppose that $k$ is a field, and $G$ is a Groebner basis for an ideal $I$ of $R = k[x_1, x_2, ldots, x_n]$ with respect to some given monomial order. Then a basis for $R / I$ as a $k$-vector space is given by the set of monomials $M$ over $x_1, ldots, x_n$ such that for every initial monomial $N$ of an element of $G$, $N$ does not divide $M$. (The repeated division process shows that these monomials span the quotient, since any polynomial with initial element not in this set can be reduced by division by the appropriate element of $G$. Conversely, essentially by the definition of Groebner basis, you get that the elements are linearly independent.)
In the example you give, let us consider the monomial order given by lexicographic order on powers of $z$ first, then on powers of $y$, then on powers of $x$. Then since the initial monomials $z^2, y^2, x^{2g}$ are pairwise relatively prime, the given set of generators already forms a Groebner basis with respect to this monomial order. Therefore, a basis for $S$ as an $mathbb{F}_2$-vector space consists of the monomials $x^a y^b z^c$ which are not divisible by any of $z^2, y^2, x^{2g}$. This condition is equivalent to $0 le a < 2g, 0 le b < 2, 0 le c < 2$. As a corollary, $dim_{mathbb{F}_2}(S) = 8g$.
answered Dec 19 '18 at 22:55
Daniel ScheplerDaniel Schepler
9,1191721
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add a comment |
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$begingroup$
Suppose that $k$ is a field, and $G$ is a Groebner basis for an ideal $I$ of $R = k[x_1, x_2, ldots, x_n]$ with respect to some given monomial order. Then a basis for $R / I$ as a $k$-vector space is given by the set of monomials $M$ over $x_1, ldots, x_n$ such that for every initial monomial $N$ of an element of $G$, $N$ does not divide $M$. (The repeated division process shows that these monomials span the quotient, since any polynomial with initial element not in this set can be reduced by division by the appropriate element of $G$. Conversely, essentially by the definition of Groebner basis, you get that the elements are linearly independent.)
In the example you give, let us consider the monomial order given by lexicographic order on powers of $z$ first, then on powers of $y$, then on powers of $x$. Then since the initial monomials $z^2, y^2, x^{2g}$ are pairwise relatively prime, the given set of generators already forms a Groebner basis with respect to this monomial order. Therefore, a basis for $S$ as an $mathbb{F}_2$-vector space consists of the monomials $x^a y^b z^c$ which are not divisible by any of $z^2, y^2, x^{2g}$. This condition is equivalent to $0 le a < 2g, 0 le b < 2, 0 le c < 2$. As a corollary, $dim_{mathbb{F}_2}(S) = 8g$.
answered Dec 19 '18 at 22:55
Daniel ScheplerDaniel Schepler
9,1191721
$endgroup$
add a comment |
$begingroup$
Suppose that $k$ is a field, and $G$ is a Groebner basis for an ideal $I$ of $R = k[x_1, x_2, ldots, x_n]$ with respect to some given monomial order. Then a basis for $R / I$ as a $k$-vector space is given by the set of monomials $M$ over $x_1, ldots, x_n$ such that for every initial monomial $N$ of an element of $G$, $N$ does not divide $M$. (The repeated division process shows that these monomials span the quotient, since any polynomial with initial element not in this set can be reduced by division by the appropriate element of $G$. Conversely, essentially by the definition of Groebner basis, you get that the elements are linearly independent.)
In the example you give, let us consider the monomial order given by lexicographic order on powers of $z$ first, then on powers of $y$, then on powers of $x$. Then since the initial monomials $z^2, y^2, x^{2g}$ are pairwise relatively prime, the given set of generators already forms a Groebner basis with respect to this monomial order. Therefore, a basis for $S$ as an $mathbb{F}_2$-vector space consists of the monomials $x^a y^b z^c$ which are not divisible by any of $z^2, y^2, x^{2g}$. This condition is equivalent to $0 le a < 2g, 0 le b < 2, 0 le c < 2$. As a corollary, $dim_{mathbb{F}_2}(S) = 8g$.
answered Dec 19 '18 at 22:55
Daniel ScheplerDaniel Schepler
9,1191721
$endgroup$
add a comment |
$begingroup$
Suppose that $k$ is a field, and $G$ is a Groebner basis for an ideal $I$ of $R = k[x_1, x_2, ldots, x_n]$ with respect to some given monomial order. Then a basis for $R / I$ as a $k$-vector space is given by the set of monomials $M$ over $x_1, ldots, x_n$ such that for every initial monomial $N$ of an element of $G$, $N$ does not divide $M$. (The repeated division process shows that these monomials span the quotient, since any polynomial with initial element not in this set can be reduced by division by the appropriate element of $G$. Conversely, essentially by the definition of Groebner basis, you get that the elements are linearly independent.)
In the example you give, let us consider the monomial order given by lexicographic order on powers of $z$ first, then on powers of $y$, then on powers of $x$. Then since the initial monomials $z^2, y^2, x^{2g}$ are pairwise relatively prime, the given set of generators already forms a Groebner basis with respect to this monomial order. Therefore, a basis for $S$ as an $mathbb{F}_2$-vector space consists of the monomials $x^a y^b z^c$ which are not divisible by any of $z^2, y^2, x^{2g}$. This condition is equivalent to $0 le a < 2g, 0 le b < 2, 0 le c < 2$. As a corollary, $dim_{mathbb{F}_2}(S) = 8g$.
answered Dec 19 '18 at 22:55
Daniel ScheplerDaniel Schepler
9,1191721
$endgroup$
Suppose that $k$ is a field, and $G$ is a Groebner basis for an ideal $I$ of $R = k[x_1, x_2, ldots, x_n]$ with respect to some given monomial order. Then a basis for $R / I$ as a $k$-vector space is given by the set of monomials $M$ over $x_1, ldots, x_n$ such that for every initial monomial $N$ of an element of $G$, $N$ does not divide $M$. (The repeated division process shows that these monomials span the quotient, since any polynomial with initial element not in this set can be reduced by division by the appropriate element of $G$. Conversely, essentially by the definition of Groebner basis, you get that the elements are linearly independent.)
In the example you give, let us consider the monomial order given by lexicographic order on powers of $z$ first, then on powers of $y$, then on powers of $x$. Then since the initial monomials $z^2, y^2, x^{2g}$ are pairwise relatively prime, the given set of generators already forms a Groebner basis with respect to this monomial order. Therefore, a basis for $S$ as an $mathbb{F}_2$-vector space consists of the monomials $x^a y^b z^c$ which are not divisible by any of $z^2, y^2, x^{2g}$. This condition is equivalent to $0 le a < 2g, 0 le b < 2, 0 le c < 2$. As a corollary, $dim_{mathbb{F}_2}(S) = 8g$.
answered Dec 19 '18 at 22:55
Daniel ScheplerDaniel Schepler
9,1191721
edited Dec 21 '18 at 13:20
answered Dec 19 '18 at 22:55
Daniel ScheplerDaniel Schepler
9,1191721
answered Dec 19 '18 at 22:55
Daniel ScheplerDaniel Schepler
9,1191721
answered Dec 19 '18 at 22:55
Daniel ScheplerDaniel Schepler
9,1191721
9,1191721
add a comment |
add a comment |
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$begingroup$
You mean a basis as a vector space over $mathbb{F}_2$?
$endgroup$
– Daniel Schepler
Dec 19 '18 at 22:38
$begingroup$
@DanielSchepler yes, I meant as a vector space over $mathbb{F}_2$. For example, the basis for $mathbb{F}_2[x,y]/(x+y,y^2+1)$ is ${1,x}$. Let me clarify that in the question.
$endgroup$
– cleanplay
Dec 19 '18 at 22:43