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There's a Lemma about monomial ideals that says:

  • "Let $I=\left$ be a monomial ideal. Then a monomial $x^β$ lies in $I$ if and only if $x^β$ is divisible by $x^α$ for some $α ∈ A$."

My question here is: take for example a monomial ideal $I=\left=\left$. If I want to know if a polynomial/monomial $g \in I$, it's enough to show that $g$ is divisible by ONE of the $f_s$ ($s=\{1,2,3\}$)?

Or does $g$ has to be divisible for ALL of the $f_s$?

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    If $g$ is a monomial it's enough tho show that some $f_s\mid g$, not all. If instead $g$ is a polynomial, you must show that every monomial which appears in $g$ belongs to $I$. – 2017-01-15

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A polynomial $g$ is in the ideal $I=\langle f_1, f_2, f_3\rangle$, if you can write $g = a_1 f_1 + a_2 f_2 + a_3 f_3$ with $a_i$ taken from the ring of polynomials.

If $g$ is divisible by one of the polynomials (say $f_1$) with quoutient $c$, it can be written as $g = cf_1$ rioght? Since $0$ is always a part of the ring, you can say $a_2 = a_3 =0$ and have a presentation, that gives you $g \in I$.

Keep in mind, that the conversion is not always true, Even, if $g$ is not divisible by $f_1$ to $f_3$ it can still be part of $I$. For that, read on Groebner bases.

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    First of all thanks for answer, but you say that even if $g$ is not divisible by $f_1$ to $f_3$ it can still be part of $I$. But the Lemma says "If and only If". So if g is not divisible can I not conclude that $g$ is not in $I$? @Laray – 2017-01-14
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    Sorry, I was thinking about binomial (or other) ideals. There you have to take Groebner Bases into account. On monomial Ideals, the base is a Groebner base. You are right, it's "if and only if" on monmial ideals – 2017-01-15