Quotient of monomial ideals
I've done a lot of calculations but they all seem to lead nowhere:
Consider the ring of multivariate polynomials with field coefficients $K[X_1,dots,X_n]$, and two monomial ideals, say $alpha=left<X^aright>_{ain A}$ and $beta=left<X^bright>_{bin B}$, I need to prove their quotient $(alpha : beta)$ is also monomial.
Considering that $(alpha : beta)=cap_{bin B}(alpha :X^b)$ and that (I've proved) intersection of monomial ideals is monomial too, I only need to fix a $bin B$ and prove the corresponding $(alpha :X^b)$ is monomial (correct me if I'm wrong). Trying to figure out what can be its generating set I've managed to prove, as long as my calculations are right, that for every polynomial $F$ in such ideal, being $N(F)$ its Newton's diagram we have:
$$b+N(F)subseteq A+mathbb{N}^n$$
but I cannot see if this helps me to conclude. I really appreciate any other idea on how to proceed, but I cannot see a more straightforward road.
polynomials ring-theory commutative-algebra monomial-ideals
add a comment |
I've done a lot of calculations but they all seem to lead nowhere:
Consider the ring of multivariate polynomials with field coefficients $K[X_1,dots,X_n]$, and two monomial ideals, say $alpha=left<X^aright>_{ain A}$ and $beta=left<X^bright>_{bin B}$, I need to prove their quotient $(alpha : beta)$ is also monomial.
Considering that $(alpha : beta)=cap_{bin B}(alpha :X^b)$ and that (I've proved) intersection of monomial ideals is monomial too, I only need to fix a $bin B$ and prove the corresponding $(alpha :X^b)$ is monomial (correct me if I'm wrong). Trying to figure out what can be its generating set I've managed to prove, as long as my calculations are right, that for every polynomial $F$ in such ideal, being $N(F)$ its Newton's diagram we have:
$$b+N(F)subseteq A+mathbb{N}^n$$
but I cannot see if this helps me to conclude. I really appreciate any other idea on how to proceed, but I cannot see a more straightforward road.
polynomials ring-theory commutative-algebra monomial-ideals
If $I$ is an ideal generated by some monomials $m_i$, and $n$ is a monomial, then $(I:n)$ is generated by $m_i/gcd(m_i,n)$.
– user26857
Dec 7 at 20:27
en.wikipedia.org/wiki/Ideal_quotient
– user26857
Dec 7 at 20:29
add a comment |
I've done a lot of calculations but they all seem to lead nowhere:
Consider the ring of multivariate polynomials with field coefficients $K[X_1,dots,X_n]$, and two monomial ideals, say $alpha=left<X^aright>_{ain A}$ and $beta=left<X^bright>_{bin B}$, I need to prove their quotient $(alpha : beta)$ is also monomial.
Considering that $(alpha : beta)=cap_{bin B}(alpha :X^b)$ and that (I've proved) intersection of monomial ideals is monomial too, I only need to fix a $bin B$ and prove the corresponding $(alpha :X^b)$ is monomial (correct me if I'm wrong). Trying to figure out what can be its generating set I've managed to prove, as long as my calculations are right, that for every polynomial $F$ in such ideal, being $N(F)$ its Newton's diagram we have:
$$b+N(F)subseteq A+mathbb{N}^n$$
but I cannot see if this helps me to conclude. I really appreciate any other idea on how to proceed, but I cannot see a more straightforward road.
polynomials ring-theory commutative-algebra monomial-ideals
I've done a lot of calculations but they all seem to lead nowhere:
Consider the ring of multivariate polynomials with field coefficients $K[X_1,dots,X_n]$, and two monomial ideals, say $alpha=left<X^aright>_{ain A}$ and $beta=left<X^bright>_{bin B}$, I need to prove their quotient $(alpha : beta)$ is also monomial.
Considering that $(alpha : beta)=cap_{bin B}(alpha :X^b)$ and that (I've proved) intersection of monomial ideals is monomial too, I only need to fix a $bin B$ and prove the corresponding $(alpha :X^b)$ is monomial (correct me if I'm wrong). Trying to figure out what can be its generating set I've managed to prove, as long as my calculations are right, that for every polynomial $F$ in such ideal, being $N(F)$ its Newton's diagram we have:
$$b+N(F)subseteq A+mathbb{N}^n$$
but I cannot see if this helps me to conclude. I really appreciate any other idea on how to proceed, but I cannot see a more straightforward road.
polynomials ring-theory commutative-algebra monomial-ideals
polynomials ring-theory commutative-algebra monomial-ideals
edited Dec 7 at 20:32
user26857
39.2k123882
39.2k123882
asked Dec 6 at 17:24
Renato Faraone
2,31411627
2,31411627
If $I$ is an ideal generated by some monomials $m_i$, and $n$ is a monomial, then $(I:n)$ is generated by $m_i/gcd(m_i,n)$.
– user26857
Dec 7 at 20:27
en.wikipedia.org/wiki/Ideal_quotient
– user26857
Dec 7 at 20:29
add a comment |
If $I$ is an ideal generated by some monomials $m_i$, and $n$ is a monomial, then $(I:n)$ is generated by $m_i/gcd(m_i,n)$.
– user26857
Dec 7 at 20:27
en.wikipedia.org/wiki/Ideal_quotient
– user26857
Dec 7 at 20:29
If $I$ is an ideal generated by some monomials $m_i$, and $n$ is a monomial, then $(I:n)$ is generated by $m_i/gcd(m_i,n)$.
– user26857
Dec 7 at 20:27
If $I$ is an ideal generated by some monomials $m_i$, and $n$ is a monomial, then $(I:n)$ is generated by $m_i/gcd(m_i,n)$.
– user26857
Dec 7 at 20:27
en.wikipedia.org/wiki/Ideal_quotient
– user26857
Dec 7 at 20:29
en.wikipedia.org/wiki/Ideal_quotient
– user26857
Dec 7 at 20:29
add a comment |
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If $I$ is an ideal generated by some monomials $m_i$, and $n$ is a monomial, then $(I:n)$ is generated by $m_i/gcd(m_i,n)$.
– user26857
Dec 7 at 20:27
en.wikipedia.org/wiki/Ideal_quotient
– user26857
Dec 7 at 20:29