# multiplicative sets in rings and prime ideals

Proposition^{}. Let $R$ be a commutative ring, $S\subseteq R$ a mutliplicative subset of $R$ such that $0\notin S$. Then there exists prime ideal^{} $P\subseteq R$ such that $P\cap S=\mathrm{\varnothing}$.

Proof. Consider the family $\mathcal{A}=\{I\subseteq R|I\text{is an ideal and}I\cap S=\mathrm{\varnothing}\}$. Of course $\mathcal{A}\ne \mathrm{\varnothing}$, because the zero ideal^{} $0\in \mathcal{A}$. We will show, that $\mathcal{A}$ is inductive (i.e. satisfies Zorn’s Lemma’s assumptions^{}) with respect to inclusion.

Let ${\{{I}_{k}\}}_{k\in K}$ be a chain in $\mathcal{A}$ (i.e. for any $a,b\in K$ either ${I}_{a}\subseteq {I}_{b}$ or ${I}_{b}\subseteq {I}_{a}$). Consider $I={\bigcup}_{k\in K}{I}_{k}$. Obviously $I$ is an ideal. Furthermore, if $x\in I\cap S$, then there is $k\in K$ such that $x\in {I}_{k}\cap S=\mathrm{\varnothing}$. Thus $I\cap S=\mathrm{\varnothing}$, so $I\in \mathcal{A}$. Lastely each ${I}_{k}\subseteq I$, which completes^{} this part of proof.

By Zorn’s Lemma there is a maximal element^{} $P\in \mathcal{A}$. We will show that this ideal is prime. Let $x,y\in R$ be such that $xy\in P$. Assume that neither $x\notin P$ nor $y\notin P$. Then $P\subset P+(x)$ and $P\subset P+(y)$ and these inclusions are proper. Therefore both $P+(x)$ and $P+(y)$ do not belong to $\mathcal{A}$ (because $P$ is maximal). This implies that there exist $a\in (P+(x))\cap S$ and $b\in (P+(y))\cap S$. Thus

$$a={m}_{1}+{r}_{1}x\in S;b={m}_{2}+{r}_{2}y\in S;$$ |

where ${m}_{1},{m}_{2}\in P$ and ${r}_{1},{r}_{2}\in R$. Note that $ab\in S$. We calculate

$$ab=({m}_{1}+{r}_{1}x)({m}_{2}+{r}_{2}y)={m}_{1}{m}_{2}+{m}_{2}{r}_{1}x+{m}_{1}{r}_{2}y+xy{r}_{1}{r}_{2}.$$ |

Of course ${m}_{1}{m}_{2},{m}_{2}{r}_{1}x,{m}_{1}{r}_{2}y\in P$, because ${m}_{1},{m}_{2}\in P$ and $xy{r}_{1}{r}_{2}\in P$ by our assumption that $xy\in P$. This shows, that $ab\in P$. But $ab\in S$ and $P\in \mathcal{A}$. Contradiction^{}. $\mathrm{\square}$

Corollary. Let $R$ be a commutative ring, $I$ an ideal in $R$ and $S\subseteq R$ a multiplicative subset such that $I\cap S=\mathrm{\varnothing}$. Then there exists prime ideal $P$ in $R$ such that $I\subseteq P$ and $P\cap S=\mathrm{\varnothing}$.

Proof. Let $\pi :R\to R/I$ be the projection. Then $\pi (S)\subseteq R/I$ is a multiplicative subset in $R/I$ such that $0+I\notin \pi (S)$ (because $I\cap S=\mathrm{\varnothing}$). Thus, by proposition, there exists a prime ideal $P$ in $R/I$ such that $P\cap \pi (S)=\mathrm{\varnothing}$. Of course the preimage^{} of a prime ideal is again a prime ideal. Furthermore $I\subseteq {\pi}^{-1}(P)$. Finaly ${\pi}^{-1}(P)\cap S=\mathrm{\varnothing}$, because $P\cap \pi (S)=\mathrm{\varnothing}$. This completes the proof.

Title | multiplicative sets in rings and prime ideals |
---|---|

Canonical name | MultiplicativeSetsInRingsAndPrimeIdeals |

Date of creation | 2013-03-22 19:03:58 |

Last modified on | 2013-03-22 19:03:58 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 5 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 16U20 |

Classification | msc 13B30 |