The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.119.5. Let $R$ be a domain. Assume every nonzero, nonunit factors into irreducibles. Then $R$ is a UFD if and only if every irreducible element is prime.

Proof. Assume $R$ is a UFD and let $x \in R$ be an irreducible element. Say $ab \in (x)$, i.e., $ab = cx$. Choose factorizations $a = a_1 \ldots a_ n$, $b = b_1 \ldots b_ m$, and $c = c_1 \ldots c_ r$. By uniqueness of the factorization

\[ a_1 \ldots a_ n b_1 \ldots b_ m = c_1 \ldots c_ r x \]

we find that $x$ is an associate of one of the elements $a_1, \ldots , b_ m$. In other words, either $a \in (x)$ or $b \in (x)$ and we conclude that $x$ is prime.

Assume every irreducible element is prime. We have to prove that factorization into irreducibles is unique up to permutation and taking associates. Say $a_1 \ldots a_ m = b_1 \ldots b_ n$ with $a_ i$ and $b_ j$ irreducible. Since $a_1$ is prime, we see that $b_ j \in (a_1)$ for some $j$. After renumbering we may assume $b_1 \in (a_1)$. Then $b_1 = a_1 u$ and since $b_1$ is irreducible we see that $u$ is a unit. Hence $a_1$ and $b_1$ are associates and $a_2 \ldots a_ n = ub_2\ldots b_ m$. By induction on $n + m$ we see that $n = m$ and $a_ i$ associate to $b_{\sigma (i)}$ for $i = 2, \ldots , n$ as desired. $\square$


Comments (0)

There are also:

  • 7 comment(s) on Section 10.119: Factorization

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 034T. Beware of the difference between the letter 'O' and the digit '0'.