While doing my Algebra HW, I "proved" that every unique factorization domain (UFD) is a principal ideal domain (PID). I know that this is not true, however I fail to see where exactly is the mistake. Could someone please pinpoint whats wrong ?
Firstly, to avoid confusion, as a UFD I denote such commutative ring $R$ that there is a subset $\mathbb{P} \subset R:$ All elements of $\mathbb{P}$ are prime, each prime element $p \in R$ has an associated element $p' \in \mathbb{P}: (p) = (p')$ and each element $x \in R\setminus ( \{0\} \cup R^*) $ can be written as $x = \varepsilon \cdot p_1 ^{\nu_1 } \cdot … \cdot p_m ^{\nu_m}$ , where $\varepsilon \in R^*$ invertable, $p_1, …,p_m \in \mathbb{P}$ and $\nu_1, …, \nu_m \in \mathbb{N} \cup \{ 0\} $. Moreover, such factorization is unique up to permutations.
Firstly I would like to prove the following statement:
Claim 1: $R$ is a UFD and $a, b \in R\setminus \{0\} : a \mid b $, and $b$ is factorized as $b= \varepsilon _1 p_1^{\nu_1}\cdot …\cdot p_m^{\nu_m} , p_1,.., p_m \in \mathbb{P}$ are primes and $\varepsilon \in R^*$ – an invertable element, then $a$ is factorized as $a = \varepsilon _2 p_1^{\nu'_1}\cdot ….\cdot p_m^{\nu_m' }$, where $0 \leq \nu_k ' \leq \nu_k , \forall k \in\{1,…, m\} $
Proof: Let $a \mid b \Leftrightarrow \exists c\in R\setminus \{0\} : b = a\cdot c$. W.l.o.g assume $c \notin R^* $ (otherwise $q$ is factorized as $q := q \cdot p_1^0 \cdot… \cdot p_m^0$). Then, if $a = \varepsilon_1 \cdot q_1 ^{l_1} \cdot …\cdot q_ n^{l_m}, c = \varepsilon _2 h_1 ^{j_1}\cdot …\cdot h_k^{j_k}$, $b = a\cdot c = \varepsilon_1 \varepsilon _2 q_1 ^{l_1} \cdot …\cdot q_ n^{l_m}h_1 ^{j_1}\cdot …\cdot h_k^{j_k}$ would be a factorization of $b$. But factorizations are unique up to permutations, so $q_1 ,.., q_n, h_1,.., h_k \in \{p_1, .., p_n\}$ so (w.l.o.g) $n=m , q_i = p_i, i = 1,.., n$ and $k = m , h_1 = p_1 ,…, h_m = p_m$. Morever, $p_i^{l_i + j_i } = p_i^{\nu_i}, l_i, j_i \geq 0 \Rightarrow l_i \leq \nu _i $ $\blacksquare$
Now, the "proof" itself:
Claim 2: Each UFD is PID
Proof: Assume for contradiction, i.e there is $I \subset R, I \neq \emptyset: \forall x \in R: I \neq (x) $. Consider $\mathcal{P} := \{ P \subset I : P \text{ is a principal ideal}\}$. The partial order on $\mathcal{P}$ is given by $"\subset"$ relation. I will show that $\mathcal{P}$ has a maximal element – that is, a principal ideal $P\subset I: \forall P' \subset I$ principal ideals : $P'\subset P$. For that I wold like to use Zorn's lemma – then, all I need to show is that every chain $\{ P_{\alpha }\} _{\alpha \in A } \subset \mathcal{P}$ is bounded from above: i.e there is an element $P \in \mathcal{P}: \forall \alpha \in A: P_{\alpha} \subset P$.
Firstly, $\{ P_{\alpha}\}_{\alpha \in A}$ is a chain means that $A$ is a totaly ordered set ($\forall \alpha_1, \alpha _2\in A: \alpha_1 < \alpha_2 \lor \alpha_1 =\alpha_2 \lor \alpha_2 > \alpha_1$) and $\alpha_1 \leq \alpha_2 \Rightarrow P_{\alpha_1}\subset P_{\alpha_2}$. Since $\{P_\alpha\}_{\alpha \in A}$ are principal ideals, we can associate this chian with another chain $\{ d_\alpha \}_{\alpha \in A}: \forall \alpha \in A: P_{\alpha} = (d_{\alpha})$. Since $(d_{\alpha_1}) \subset (d_{\alpha_2}) \Leftrightarrow d_{\alpha_2} \mid d_{\alpha_1}$, the relations translates to $\alpha_1 \leq \alpha_2 \Rightarrow d_{\alpha_2} \mid d_{\alpha_1} $.
Now I would like to show that $P := \bigcup\limits_{\alpha \in A }P_{\alpha}$ is an upper boundary of the chain
1. $P$ is an ideal: Let $x \in P$. Then, $\exists \alpha \in A: x \in P_{\alpha}$. Then, $\forall r \in R: rx \in P_{\alpha}$ since $P_{\alpha }$ is an ideal $\Rightarrow rx \in P$. Let $x_1 , x_2 \in P$. Then $\exists \alpha_1, \alpha_2 \in A : x_1 \in P_{\alpha_1 }, x_2 \in P_{\alpha_2 }$. W.l.o.g $\alpha_1 \leq \alpha_2 \Rightarrow P_{\alpha_1} \subset P_{\alpha_2} \Rightarrow x_1, x_2 \in P_{\alpha_2} \Rightarrow x_1 + x_2 \in P_{\alpha_2} \Rightarrow x_1 +x_2 \in P$. Also its clear that $P \subset I$
2. $P$ is a principal ideal: If $|A|< \mathbb{N}$ the statement is obvious. Otherwise consider $\alpha \in A $. Then, there are two options: Either $P_\alpha = P$ (and $\forall \alpha' \in A : \alpha ' \leq a $), and we are done (since $P_{\alpha}$ is a principal ideal), or there is $\alpha ' \in A : \alpha ' > \alpha$ and $P_{\alpha } \subset P_{\alpha'}$ is a proper subset. However, per note above, $(d_\alpha) = P_{\alpha} \subset P_{\alpha '} = (d_{\alpha'})$ implies $d_{\alpha'} \mid d_{\alpha}$. We can consider $d_{\alpha }, d_{\alpha'} \neq 0$ since if $\{ P_{\alpha}\} = \{ \{0\} \}$, it is trivialy bounded from above. So, per claim 1, if $p_{\alpha} = \varepsilon p_1^{\nu_1} \cdot … \cdot p_n^{\nu_n}$ is a factorization of $p_{\alpha}$, then $p_{\alpha'} = \varepsilon' p_1^{\nu_1'} \cdot … \cdot p_n^{\nu_n'}$, $0 \leq \nu_1 ' \leq \nu_1 ,…., 0 \leq \nu_n' \leq \nu _n$ is a factorization of $p_{\alpha'}$. Moreover, $(d_\alpha) $ is a proper subset $(d_{\alpha'})$, i.e $p_{\alpha'}$ and $p_{\alpha}$ are not associated, so there must be at least one $k \in \{1 ,…, n\} : \nu_k' < \nu_k$. Now repeat the considerations for $P_{\alpha'}$: Either $P_{\alpha'} = P$ and the statement is true, or there is a proper overset $P_{\alpha''} \supset P_{\alpha'}$. For $d_{\alpha''}$ the factorization is given by $p_{\alpha'} = \varepsilon' p_1^{\nu_1''} \cdot … \cdot p_n^{\nu_n''}$, where $\nu_k '' \leq \nu _k', \forall k \in\{ 1,…, n\}$ and for at least 1 $k \in \{1,…,n\}: \nu_k '' < \nu_k' $. Then consider $P_{\alpha''}$ and so on. However this process cant go on forever – indeed, if it continues for at most $\nu_1 + \nu _2 + … + \nu_n :=N$ steps we would get an ideal $P_{\alpha^{(N)}} = (d_{\alpha^{(N)}})$, where $d_{\alpha^{(N)}}$ can be factorized as $d_{\alpha^{(N)}} = \varepsilon^{(N)}p_1^0 \cdot … \cdot p_n^0 \in R^* \Rightarrow (d_{\alpha^{(N)}}) = (1) = R$, however we assumed that $P_{\alpha^{(N)}} \subset I$ where $I$ is not a principal ideal, so a contradiction. This means that the process breaks down after finitely many steps, and we would get that $P = P_{\alpha^{(K)}} = (d_{\alpha^{(K)}}) $, so its a principal ideal
Now we have shown that $P \subset I$ and $P$ is a principal ideal $\Rightarrow P \in \mathcal{P}$ and $\forall \alpha \in A : P_{\alpha } \subset P$, so its an upper boundary. Lemma of Zorn now yields that there is such $P \in \mathcal{P}: \forall P' \in \mathcal{P}: P' \subset P$. But this gives us a contradiction: On one hand, $P \subset I$ is a proper subset, since we assumed that $I$ is not a principal ideal. On the other hand $\forall x \in I: x \in (x) \subset I$, where $(x)$ is a principal ideal, so $(x) \subset P \Rightarrow x \in P \Rightarrow I \subset P$. $\blacksquare$
I fail to see whats exactly worng here – I guess that would be either the way I use Zorn's lemma (which is still a new concept to me), or the part where is show that $P$ is a principal ideal. Would be very grateful if someone could say which part is wrong.
Best Answer
Zorn’s Lemma says that maximal elements exist, not that there is a unique maximal element (supremum). So you obtain a principal ideal $P\subseteq I$ with the property that if $P’$ is another principal ideal and $P\subseteq P’\subseteq I$, then $P=P’$. What you do not get is that every principal ideal is contained in $P$. The poset may have many distinct maximal ideals.
Here’s a situation where one can see this quite easily. Take an algebraically closed field $k$ and consider homogeneous ideals in $k[x,y]$ which are contained in $(x,y)$. This has many distinct maximal principal ideals, of the form $(ax-by)$ for $a,b\in k$, but $(x,y)$ is itself not principal.