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Good day,

Let $M$ be a multiplicative set and $I$ any Ideal, both in a noetherian Ring $R$.

I want to know whether there is a way to decide whether $M \cap I = \emptyset$ and, if not, to find an element $m \in M \cap I$.

I know that $R \setminus M$ has to be a primeideal $P$ (see Prime Ideals and multiplicative sets) .

Now, i figured $I \cap M = I \cap (R \setminus P) \Leftrightarrow I \subseteq P$.

I can decide that if i know the generators to $I$ and $P$. Then i can calculate a groebner basis for $P$ and check whether all generators of $I$ are in $P$, and if not, i can use one of them as the desired $m$.

However, is there a way (or a special case) in which i can find generators for $P$ if i only have it given in form of a set $R \setminus M$?

best regards

kolja
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  • First of all, the complement of a multiplicative set is not necessarily a prime ideal. For instance, {1} is a multiplicative set in every ring, but (for instance) $\mathbb{Z} \setminus {1}$ isn't an ideal of $\mathbb{Z}$. What is true is that if $M$ is multiplicative and $M \setminus R$ is an ideal, then this ideal is prime. – Daniel McLaury Dec 22 '15 at 16:59
  • Second, since this appears to be a computational question, how are you going to specify an arbitrary multiplicative set to the computer? For instance, the ideal $(2) \subset \mathbb{Z}$ is prime, so its complement $\mathbb{Z} \setminus (2)$ which consists of all the odd numbers is a multiplicative set. Now how are you going to express that in a finite amount of space? It's not finitely generated as a multiplicative set. – Daniel McLaury Dec 22 '15 at 17:01
  • Thank you for your answer. Yes, you are right, i need to specify this more. By this point i am in theoretical work and need to check for which $M$ i can solve this problem, so with your answer, it definitely has to be an $M$ which is finitely generated, for example $M=\lbrace {f_1}^{a_1},...,{f_n}^{a_n}$ \rbrace . Is there an easy way to know for which multiplicative sets $R \setminus M$ is an Ideal? – kolja Dec 22 '15 at 17:59
  • clarification: Of course $f_1,...,f_n$ are supposed to be elements in $R$. – kolja Dec 22 '15 at 18:47
  • In most familiar cases, e.g. $R = F[x]$ or $R = \mathbb{Z}$, the complement of a finitely-generated multiplicative set will never be an ideal.

    If we want to look at your original question of finding an element in the intersection of a (finitely-generated) multiplicative set and an ideal in a noetherian ring, then that's probably doable, but you'd also have to say how you're encoding the ring. (Or we could describe a general, nonconstructive process for taking a ring and building an algorithm that works for that ring alone.)

     – Daniel McLaury Dec 22 '15 at 19:49
  • That said, maybe it's better if you start by explaining what you ultimately want to do, because it sounds like maybe this was something you'd intended as an intermediate step but which might not actually be a good intermediate. – Daniel McLaury Dec 22 '15 at 19:50
  • I see I had some misconceptions, my head is quite clearer now. I thought about the problem and now I know some ways to solve it: As you said, when $M=R \setminus P$ with $P$ being a prime ideal, then I can just check whether $I \subseteq P = \emptyset$ (for example with groebner bases), and, if not, i know $M \cap I$ is not empty and i can use an element which is in $I$ but not in $P$ as $m$. – kolja Dec 23 '15 at 07:27
  • There is also the case with. $M=\lbrace f^n \mid b \in \mathbb{N}$ for a $f \in R$. I can Calculate the radical ideal of $I$ with Singular (a computeralgebra system in which i do implement the rings and ideals) and check whether $f \in Rad(I)$, and, if yes, i know that i can choose a $f^n$ for some $n$ as $m$ and can calculate it in finitely many steps. – kolja Dec 23 '15 at 07:30
  • The remaining problem which is my main interest now is the case $M=\lbrace {f_1}^{m_1}...{f_n}^{m_n} \mid m_i \in \mathbb{N}_0,1 \leq i \leq n \rbrace$. The Radical Ideal of $I$ alone won't do here, at least i could not figure how yet. Solving the problem whether i can , if possible, find an element $m$ in $M \cap I$ for that case (with, again, $R$ being commutatice and Noetherian, so $I$ is definitely finitely generated) would give me enough tools to solve the bigger task i am working at at the moment. – kolja Dec 23 '15 at 07:34
  • The thing which i ultimately want to do is to solve linear equations in a localization of $M$. I did all the steps to break this problem down to what i have asked. There are some assumptions on the Ring $R$. The assumptions lead to the fact that i can calculate the module of solutions of $0=b_0 a+ b_1 r_1 +... + b_n r_n$ over $R$ (with $b_0,...,b_n$ being the variables). So, i have all solutations for this equation, but i need a solution with $b_0 \in M$ (because with all the other transformations i have done, after finding it, i can find a solution for an equation in the localization). – kolja Dec 23 '15 at 08:00
  • Now, be $h \in H$ with $H$ being the (existing in a notherian commutative ring) generating system of those solutions. to know whether there is a solution with $b_0 \in M$, it is the same as asking whether there are coefficients with $m=\sum\limits_{h\in H} c_h h_0$ (so whether the first component of the solutions can add up to an Element in $M$), and the first components of the $h \in H$ generate the Ideal $I$. And this is where my initial questions starts, namely, in which cases i can decided whether $I \cap M = \emptyset$ and subsequently if not, whether i can find an element in it. – kolja Dec 23 '15 at 08:05
  • (clarification, i want to solve equations over a localization of $R$, not $M$) – kolja Dec 23 '15 at 08:11

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