# equivalent definitions for UFD

Let $R$ be an integral domain. Define

 $T=\{u\in R\ |\ u\mbox{ is invertible}\}\cup\{p_{1}\cdots p_{n}\in R\ |\ p_{i}% \mbox{ is prime}\}.$

Of course $0\not\in T$ 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

 $ab=p_{1}\cdots p_{k}$

for some prime elements $p_{i}\in R$. We can group these prime elements in such way that $p_{1}\cdots p_{n}$ divides $a$ and $p_{n+1}\cdots p_{k}$ divides $b$. Thus $a=\alpha p_{1}\cdots p_{n}$ and $b=\beta p_{n+1}\cdots p_{k}$ 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. $\square$

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 $p_{1},\ldots,p_{k}\in R$ such that $x=p_{1}\cdots p_{k}\in P$. But $P$ is prime, therefore there is $i\in\{1,\ldots,k\}$ 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 $x\not\in T$. Consider an ideal $(x)$. We will show, that $(x)\cap T=\emptyset$. Assume that there is $r\in R$ such that $rx\in T$. It follows that $x\in T$ (by lemma). Contradiction.

Since $(x)\cap T=\emptyset$ and $T$ is a multiplicative subset, then there is a prime ideal $P$ in $R$ such that $(x)\subseteq P$ and $P\cap T=\emptyset$ (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. $\square$

Title equivalent definitions for UFD EquivalentDefinitionsForUFD 2013-03-22 19:04:04 2013-03-22 19:04:04 joking (16130) joking (16130) 4 joking (16130) Theorem msc 13G05 UniqueFactorizationAndIdealsInRingOfIntegers