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Let $R=\mathbb{Q}[x,y,z]$ and $I$ an ideal of $R$. Suppose that $G$ is a Grobner basis for $I$ with a certain monomial order. We know that $f,g \in R$ are such that $\bar{f}^G=0$ and $ \bar{g}^G=3x+3$.

Which one of these statements is correct:

1) $fg \in I$.

2) $\frac{1}{3}g \in I$.

I don't understand how to use the definitions/properties of Grobner basis and Ideals to solve this problem. For those who dont know the notation, $\bar{f}^F$ means the remainder on division of $f$ by the ordered s-tuple $F=(f_1,...,f_s)$.

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    Hint: $p\in I$ iff $\overline{p}^G=0$. Then use properties of ideals. – 2017-01-12
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    That's the easy part, my problem is because $f$ is multiplying with $g$. – 2017-01-13
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    If a certain polynomial $h$ is in $I$ then multiplying any other polynomial $j$ with $h$ ($hf$) is also in $I$ ?? – 2017-01-13
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    @Daniel I think I managed to Solve this problem. Not $100$% sure though. - We have $\bar{f}^G=0$ so $f \in I$ (by definition). And because $f \in I$ we have $fg \in I$. - If we had $\frac{1}{3}g \in I$ , we would also have $3\frac{1}{3}g \in I$ and we would have to have $\bar{g}^G=0$ but we don't. So this affirmation is false. My question is: if a certain polynomial is in a Ideal $I$ we can conclude that this certain polynomial is still in $I$ even if he is multiplying with any other polynomials and constants? – 2017-01-13
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    @MathScientist Recall the definition of ideal. – 2017-01-15
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    @user26857 Yes, I've done it. I understand the problem now. We have $f \in I$ and $g \in R$ therefore $fg \in I$ by definition of Ideal. We have $g$ not in $I$ , so $\frac{1}{3}g$ not in $I$ – 2017-01-15

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I already solved this problem some time ago. I'm answering myself so I can close this post.

  • $\bar{f}^G=0$ implies that $f \in I$.It's said that $g \in R$ therefore $fg \in I$ (by definition of Ideal)

  • $ \bar{g}^G=3x+3 \neq 0$ implies that we don't have $g$ in $I$. So $\frac{1}{3}g$ can't be either.