Let $I \subseteq k[x_1, \ldots, x_n]$ be an ideal. Then there exists a finite subset $G = \{g_1, \ldots, g_t\}$ of $I$ such that $G$ is a Gröbner basis for $I$.
Given a remainder r of a polynomial p on division by a Gröbner basis G of an ideal I,
the remainder r is 0 if and only if p is in the ideal I.
Any leading coefficient of polynomial in the Gröbner basis G is required to be a unit.
Formally:
Let $G = \{g_1, \dots, g_t\}$ be a Gröbner basis for an ideal $I \subseteq k[x_1, \dots, x_n]$ and let $f \in k[x_1, \dots, x_n]$. Then $f \in I$ if and only if the remainder on division of $f$ by $G$ is zero.
Given a remainder r of a polynomial p on division by a Gröbner basis G of an ideal I,
the remainder r is 0 if and only if p is in the ideal I.
It is a variant of MonomialOrder.remainder_eq_zero_iff_mem_ideal_of_isGroebner, allowing the
finite set to contain also 0, besides polynomials with invertible leading coefficients.
Given a remainder r of a polynomial p on division by a Gröbner basis G of an ideal I,
the remainder r is 0 if and only if p is in the ideal I.
It is variant of MonomialOrder.remainder_eq_zero_iff_mem_ideal_of_isGroebner, over a field and
without hypothesis on leading coefficients in the finite set.
Let $G = \{g_1, \ldots, g_t\}$ be a Gröbner basis for an ideal $I \subseteq k[x_1, \ldots, x_n]$. Then $G$ is a basis for the vector space $I$ over $k$.
A finite set of polynomials is a Gröbner basis of an ideal if and only if it is a subset of this ideal and 0 is a remainder of each member of this ideal on division by this finite set.
Any leading coefficient of polynomial in the finite set is required to be a unit.
A finite set of polynomials is a Gröbner basis of an ideal if and only if it is a subset of this ideal and 0 is a remainder of each member of this ideal on division by this finite set.
It is a variant of MonomialOrder.isGroebnerBasis_iff_subset_ideal_and_isRemainder_zero, allowing
the finite set to contain also 0, besides polynomials with invertible leading coefficients.
A finite set of polynomials is a Gröbner basis of an ideal if and only if it is a subset of this ideal and 0 is a remainder of each member of this ideal on division by this finite set.
It is a variant of MonomialOrder.isGroebnerBasis_iff_subset_ideal_and_isRemainder_zero,
over a field and without hypothesis on leading coefficients in the finite set.
Formally:
Let $G = \{g_1, \ldots, g_t\}$ be a finite subset of $k[x_1, \ldots, x_n]$. Then $G$ is a Gröbner basis for the ideal $I = \langle G \rangle$ if and only if for every $f \in I$, the remainder of $f$ on division by $G$ is zero. whose leading coefficients are invertible with respect to a monomial order
Remainder of any polynomial on division by Gröbner basis exists and is unique.
Any leading coefficient of polynomial in the Gröbner basis is required to be a unit.
Remainder of any polynomial on division by Gröbner basis exists and is unique.
It is a variant of MonomialOrder.existsUnique_isRemainder_of_isGroebnerBasis, allowing the
Gröbner basis to contain also 0, besides polynomials with invertible leading coefficients.
Let $f, h_1, \dots, h_m \in k[\mathbf{x}] \setminus \{0\}$, and suppose $$f = c_1 h_1 + \cdots + c_m h_m, \quad \text{with } c_i \in k.$$
If $$\mathrm{lm}(h_1) = \mathrm{lm}(h_2) = \cdots = \mathrm{lm}(h_i) > \mathrm{lm}(f),$$ then $$f = \sum_{1 \leq i < j \leq m} c_{i,j} S(h_i, h_j), \quad c_{i,j} \in k.$$ Furthermore, if $S(h_i, h_j) \ne 0$, then $\mathrm{lm}(h_i) > \mathrm{lm}(S(h_i, h_j))$.
A basis $G = \{ g_1, \ldots, g_t \}$ for an ideal $I$ is a Gröbner basis if and only if $S(g_i, g_j) \to_G 0$ for all $i \neq j$.
Alias of MonomialOrder.isGroebnerBasis_iff_isRemainder_sPolynomial_zero.
A basis $G = \{ g_1, \ldots, g_t \}$ for an ideal $I$ is a Gröbner basis if and only if $S(g_i, g_j) \to_G 0$ for all $i \neq j$.
Alias of MonomialOrder.isGroebnerBasis_iff_isRemainder_sPolynomial_zero'.