# equivalent definitions for UFD

Let $R$ be an integral domain^{}. Define

$$T=\{u\in R|u\text{is invertible}\}\cup \{{p}_{1}\mathrm{\cdots}{p}_{n}\in R|{p}_{i}\text{is prime}\}.$$ |

Of course $0\notin T$ and $T$ is a multiplicative subset (recall that a prime element^{} multiplied by an invertible element is again prime). Furthermore $R$ is a UFD if and only if $T=R\backslash \{0\}$ (see the parent object for more details).

Lemma. If $a,b\in R$ are such that $ab\in T$, then both $a,b\in T$.

Proof. If $ab$ is invertible^{}, then (since $R$ is commutative^{}) both $a,b$ are invertible and thus they belong to $T$. Therefore assume that $ab$ is not invertible. Then

$$ab={p}_{1}\mathrm{\cdots}{p}_{k}$$ |

for some prime elements ${p}_{i}\in R$. We can group these prime elements in such way that ${p}_{1}\mathrm{\cdots}{p}_{n}$ divides $a$ and ${p}_{n+1}\mathrm{\cdots}{p}_{k}$ divides $b$. Thus $a=\alpha {p}_{1}\mathrm{\cdots}{p}_{n}$ and $b=\beta {p}_{n+1}\mathrm{\cdots}{p}_{k}$ for some $\alpha ,\beta \in R$. Since $R$ is an integral domain we conclude that $\alpha \beta =1$, which means that both $\alpha ,\beta $ are invertible in $R$. Therefore (for example) $\alpha {p}_{1}$ is prime and thus $a\in T$. Analogously $b\in T$, which completes^{} the proof. $\mathrm{\square}$

Theorem. (Kaplansky) An integral domain $R$ is a UFD if and only if every nonzero prime ideal^{} in $R$ contains prime element.

Proof. Without loss of generality we may assume that $R$ is not a field, because the thesis trivialy holds for fields. In this case $R$ always contains nonzero prime ideal (just take a maximal ideal^{}).

,,$\Rightarrow $” Let $P$ be a nonzero prime ideal. In particular $P$ is proper, thus there is nonzero $x\in P$ which is not invertible. By assumption^{} $x\in T$ and since $x$ is not invertible, then there are prime elements ${p}_{1},\mathrm{\dots},{p}_{k}\in R$ such that $x={p}_{1}\mathrm{\cdots}{p}_{k}\in P$. But $P$ is prime, therefore there is $i\in \{1,\mathrm{\dots},k\}$ such that ${p}_{i}\in P$, which completes this part.

,,$\Leftarrow $” Assume that $R$ is not a UFD. Thus there is a nonzero $x\in R$ such that $x\notin T$. Consider an ideal $(x)$. We will show, that $(x)\cap T=\mathrm{\varnothing}$. Assume that there is $r\in R$ such that $rx\in T$. It follows that $x\in T$ (by lemma). Contradiction^{}.

Since $(x)\cap T=\mathrm{\varnothing}$ and $T$ is a multiplicative subset, then there is a prime ideal $P$ in $R$ such that $(x)\subseteq P$ and $P\cap T=\mathrm{\varnothing}$ (please, see this entry (http://planetmath.org/MultiplicativeSetsInRingsAndPrimeIdeals) for more details). But we assumed that every nonzero prime ideal contains prime element (and $P$ is nonzero, since $x\in P$). Obtained contradiction completes the proof. $\mathrm{\square}$

Title | equivalent definitions for UFD |
---|---|

Canonical name | EquivalentDefinitionsForUFD |

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

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

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 13G05 |

Related topic | UniqueFactorizationAndIdealsInRingOfIntegers |