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