# examples of radicals of ideals in commutative rings

Let $R$ be a commutative ring. Recall, that ideals $I,J$ in $R$ are called coprime^{} iff $I+J=R$. It can be shown, that if $I,J$ are coprime, then $IJ=I\cap J$. Elements ${x}_{1},\mathrm{\dots},{x}_{n}\in R$ are called pairwise coprime iff $({x}_{i})+({x}_{j})=R$ for $i\ne j$. It follows by induction, that for pairwise coprime ${x}_{1},\mathrm{\dots},{x}_{n}\in R$ we have $({x}_{1}\mathrm{\cdots}{x}_{n})=({x}_{1})\cap \mathrm{\cdots}\cap ({x}_{n})$,

Let $x\in R$ be such that

$$x={p}_{1}^{{\alpha}_{1}}\mathrm{\cdots}{p}_{n}^{{\alpha}_{n}},$$ |

for some prime elements^{} ${p}_{i}\in R$, ${\alpha}_{i}\in \mathbb{N}$ and assume that ${p}_{1},\mathrm{\dots},{p}_{n}$ are coprime. Denote by

$$\overline{x}={p}_{1}\mathrm{\cdots}{p}_{n}.$$ |

We shall denote by $r(I)$ the radical^{} of an ideal $I\subseteq R$.

Proposition. $r\left((x)\right)=(\overline{x})$.

Proof. ,,$\supseteq $” Let $\alpha =\mathrm{max}({\alpha}_{1},\mathrm{\dots},{\alpha}_{n})$. Then we have

$${\overline{x}}^{\alpha}={({p}_{1}\mathrm{\cdots}{p}_{n})}^{\alpha}={p}_{1}^{\alpha}\mathrm{\cdots}{p}_{n}^{\alpha}={p}_{1}^{\alpha -{\alpha}_{1}}\mathrm{\cdots}{p}_{n}^{\alpha -{\alpha}_{n}}{p}_{1}^{{\alpha}_{1}}\mathrm{\cdots}{p}_{n}^{{\alpha}_{n}}=yx$$ |

and thus ${\overline{x}}^{\alpha}\in (x)$. This shows the first inclusion.

,,$\subseteq $” Assume that $y\in r\left((x)\right)$ and $y\ne 0$. Then there is $n\in \mathbb{N}$ such that ${y}^{n}\in (x)$. Thus $x$ divides ${y}^{n}$. Of course for any $i\in \{1,\mathrm{\dots},n\}$ we have that ${p}_{i}$ divides $x$. Thus ${p}_{i}$ divides ${y}^{n}$ and since ${p}_{i}$ is prime, we obtain that ${p}_{i}$ divides $y$. Now for $i\ne j$ elements ${p}_{i}$ and ${p}_{j}$ are coprime, thus $\overline{x}$ divides $y$ and therefore $y\in (\overline{x})$, which completes^{} the proof. $\mathrm{\square}$

Remark. If we assume that $R$ is a PID (and thus UFD), then the previous proposition gives us the full characterization of radicals of ideals in $R$. In particular an ideal in PID is radical if and only if it is generated by an element of the form ${p}_{1}\mathrm{\cdots}{p}_{n}$, where for $i\ne j$ elements ${p}_{i}$ and ${p}_{j}$ are not associated primes.

Examples. Consider ring of integers^{} $\mathbb{Z}$. Then we have:

$$r\left((12)\right)=(6);$$ |

$$r\left((9)\right)=(3);$$ |

$$r\left((7)\right)=(7);$$ |

$$r\left((1125)\right)=(15).$$ |

Title | examples of radicals of ideals in commutative rings |
---|---|

Canonical name | ExamplesOfRadicalsOfIdealsInCommutativeRings |

Date of creation | 2013-03-22 19:04:34 |

Last modified on | 2013-03-22 19:04:34 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 5 |

Author | joking (16130) |

Entry type | Example |

Classification | msc 16N40 |

Classification | msc 14A05 |

Classification | msc 13-00 |