23

Any irreducible element of a factorial ring $D$ is a prime element of $D$.

Proof. Let $p$ be an arbitrary irreducible element of $ D$. Thus $ p$ is a non-unit. If $ ab \in (p)\smallsetminus\{0\}$, then $ ab = cp$ with $ c \in D$. We write $ a,\,b,\,c$ as products of irreducibles: $$\displaystyle a \;=\; p_1\cdots p_l, \quad b \;=\; q_1\cdots q_m, \quad c \;=\; r_1\cdots r_n.$$ Here, one of those first two products may be empty, i.e., it may be a unit. We have $$\displaystyle p_1\cdots p_l\,q_1\cdots q_m \;=\; r_1\cdots r_n\,p\tag{1}$$

Due to the uniqueness of prime factorization, every factor $ r_k$ is an associate of certain of the $l+m$ irreducibles on the left hand side of $(1)$. Accordingly, $p$ has to be an associate of one of the $ p_i$'s or $ q_j$'s. It means that either $ a \in (p)$ or $ b \in (p)$. Thus, $ (p)$ is a prime ideal of $ D$, and its generator must be a prime element.

It may be too simple, but why $ a \in (p)$ instead of $p_1 \in (p)$? Is it because $p$ has to be an associate of one of the $ p_i$'s or $ q_j$'s? Let's say $p_2$ is an associate of $p$. So, $p_2=pw$, $w\in R$. Since $a=p_1p_2\cdots p_l$ then $a=p_1pwp_3\cdots p_l$ and $a=p(p_1p_3\cdots p_lw)$, $p_1p_3\cdots p_lw \in R$ so $a$ is divisible by $p$ hence $a\in (p)$?

Xam
  • 6,119
emmett
  • 787

2 Answers2

14

It's trivial to show that primes are irreducible. So, assume that $a$ is an irreducible in a UFD (Unique Factorization Domain) $R$ and that $a \mid bc$ in $R$. We must show that $a \mid b$ or $a \mid c$. Since $a\mid bc$, there is an element $d$ in $R$ such that $bc=ad$. Now replace $b,c$ and $d$ by their factorizations as a product of irreducibles and use uniqueness.

qwr
  • 10,716
W.Leywon
  • 388
  • what if $d$ is non invertible? –  Nov 25 '17 at 16:43
  • The uniqueness of the factorization implies that. There is no need to use invertibility to cancel out the elements,just simply compare the factors of both sides of the equation,they shall be equal,one-to-one. – W.Leywon Nov 30 '17 at 08:59
  • 1
    @user370967 We may assume $b, c$ are non-invertible. For otherwise, $a|b$ or $a|c$ are trivial. Under such assumption, $d$ is indeed non-invertible. We will show why. First, we can uniquely decompose $b$ and $c$ as product of irreducible elements, whose total number is at least $2$. On the RHS, we know $a$ is irreducible. If $d$ is invertible (i.e. an unit), then RHS only has $1$ irreducible element, which contradicts the definition of UFD. Therefore, $d$ is indeed non-invertible and we may decompose $d$ as well. At last, by comparing both sides of $bc=ad$, we show that $a|b$ or $a|c$. – Sam Wong Dec 21 '22 at 09:55
  • 1
    @user370967 We have also tacitly assumed that $b, c, d$ are all non-zero. For otherwise, it is quite trivial. – Sam Wong Dec 21 '22 at 09:57
  • @SamWong thank you for clarification. It helped me. But I have a question: we can compare both side of equality only when each element is irreducible. $d^{-1}$ is not irreducible because it is a unit. – user264745 Nov 29 '23 at 21:07
  • Ohh… I got answer. If $d$ is invertible, then $bcd^{-1}=a$. Since $a$ is irreducible, we have $b$ or $cd^{-1}$ is unit, by definition. Since $b$ is not unit, $cd^{-1}$ is unit. So $\exists r\in R$ such that $cd^{-1}r=c(d^{-1}r)=1_R$. That is, $c$ is a unit. Thus we reach contradiction. Hence $d$ is not invertible. Don’t delete your comment. – user264745 Nov 29 '23 at 21:16
8

The proof given above is probably the standard one to show that a factorial domain is an AP-domain. But there is another proof using the following application: for an irreducible $p\in D$, let's define $e_p\colon D\setminus \{0\}\rightarrow \Bbb{N}$ given by $a\mapsto e_p(a)=\#$ of times that $p$ or its associates appear in the irreducible factorization of $a$.

We notice that because $D$ is factorial the application given above it's well defined. Moreover, if $a\in D^{\times}$, then $e_p(a)=0$ for every irreducible $p$, and if $a\in D\setminus{D^{\times}_0}$, then $e_p(a)=0$ iff $p\not\mid a$. Equivalently, $e_p(a)>0$ iff $p\mid a$.

We have the following:

Lemma: Let $D$ be a factorial domain and $a,b\in D\setminus \{0\}$. Then $$e_p(ab)=e_p(a)+e_p(b)$$ for every irreducible $p\in D$.

Proof: As $D$ is factorial, we can write $a=p^{e_p(a)}\ldots $ and $b=p^{e_p(b)}\ldots $, then $$ab=(p^{e_p(a)}\ldots)(p^{e_p(b)}\ldots)=p^{e_p(a)+e_p(b)}\ldots $$ Hence, $e_p(ab)=e_p(a)+e_p(b)$.

Now we're going to prove that if $D$ is factorial, then $D$ is an AP-domain. Let $p$ be an irreducible element in $D$ and let $a,b\in D$ such that $p\mid ab$. If $ab=0$, then $a=0$ or $b=0$, so in this case clearly $p\mid a$ or $p\mid b$. If $ab\neq 0$, since $p\mid ab$ we have $e_p(ab)>0$, so by the above lemma we find $$e_p(ab)=e_p(a)+e_p(b)>0.$$ Therefore we deduce that necessarily $e_p(a)>0$ or $e_p(b)>0$, i.e., $p\mid a$ or $p\mid b$. Hence, $p$ is prime.

As a remark, this kind of ideas applied to $\Bbb{Z}$ can be found in the first pages of the book "A Classical Introduction to Modern Number Theory" by K. Ireland and M. Rosen.

Community
  • 1
Xam
  • 6,119