Principal Ideal Domains and Unique Factorization Domains

Principal Ideal Domains

Let R be an integral domain. Then R is a PID (Principal Ideal Domain) if every $I \subseteq R$ ideal is principal. A GCD always exists in a PID (unique up to unit factors).

Proposition: let R be a PID, $I \subseteq R$ prime. Then I is maximal.

Proof: $I = (a)$ for some $a \in R$. By Krull’s theorem there is $m \in R$ such that $(a) \subseteq (m)$ and (m) is maximal. if $(a) \subseteq (m)$ then a = xm for some $x \in R$. Since (a) is prime we need x or m in (a) If $m \in (a)$ then $(m) \subseteq (a)$ and $(m) = (a)$ so (a) is maximal. If $x \in (a)$ then x = au for some $u \in R$. So a=xm=aum=1,which means $um = 1$ and hence $m \in R^*$ so $(m) = R$ which would mean (m) is not maximal which is a contradiction.

Proposition: Let R be a ring. The following are equivalent: 1) R is a field 2) $R[x]$ is Euclidean 3) $R[x]$ is a PID.

Proof: $1 \rightarrow 2$ was proved in last post, $2 \rightarrow 3$ follows because Euclidean $\rightarrow$ PID. $3 \rightarrow 1$: $R[x] \leq R[x]$ is integral $\rightarrow$ R is integral. $R[x]/(x) \cong R$ by the first isomorphism theorem so (x) is prime and maximal (since $R[x]$ is a PID), hence $R[x]/(x) \cong R$ is a field.

Let R be an integral domain and $a,b \in R$. 1) $r \in R \backslash (R^* \cup \{0\})$ is irreducible if whenever r = ab, we have $a \in R^*$ or $b \in R^*$. 2) if r is not irreducible we say it is reducible. 3) $r \in R \backslash \{0\}$ is prime if (r) is prime.

Lemma: if $r \in R$ is irreducible and $u \in R^*$ then ur is irreducible.

Proof: ur = ab, $r = u^{-1}ab = (u^{-1}a)b$. r irreducible implies $u^{-1}ab = (u^{-1}a)b$ if $b \not\in R^*$.

$r, s \in R$ are associated if (r) = (s).

Proposition: let R be an integral domain and $r \in R$. Then r prime $\rightarrow$ r irreducible.

Proof: Assume r = ab for $a,b \in R$. THen ab = $r \in (r)$ prime so we must have $a \in (r)$ or $b \in (r)$. Assume without loss of generality that $a \in (r)$. Then $a = ru$ for some $u \in R$. Thus $r = ub \rightarrow 1 = rb \rightarrow b \in R^*$ so r is irreducible. (Note: in general the converse is not true)

Proposition: if R is a PID then prime = irreducible

Proof: $\rightarrow$ was proved above. Let $r \in R$ be irreducible. Let $I \subseteq R$ be an ideal containing (r). Then $r \in (r) \subseteq m$ so r = mu for sme $u \in R$. Since r is irreducible either m is a unit or u is a unit. If m is a unit then I = R so (r) is maximal, which means (r) is prime so r is prime If u is a unit then $I = (r)$.

Unique Factorization Domains

Let R be an integral domain. R is a UFD (Unique Factorization Domain) if 1) for all $r \in R \backslash (R^* \cup \{0\})$ there exists $p_0 ... p_{n-1}$ irreducible elements in R such that $r = \prod_{i=0}^{n-1}p_i$. 2) The decomposition is unique: if $p_i, q_j$ are irreducible and $\prod_{i=0}^n p_i = \prod_{i=0}^m q_j$ then n=m and there exists $\sigma \in S_n$ such that $p_i$ is associated to $q_{\sigma(i)}$. Note that the $p_i$ might be non distinct.

Examples: $\mathbb{Z}$ is a UFD, any PID is a UFD. If R is a UFD then $R[x]$ is a UFD. $\mathbb{Z}[x]$ is a UFD that is not a PID.

Proposition: let R be a PID, then R is a UFD.

Proof: Let $r \in R \backslash (R^* \cup \{0\})$ if r is irreducible we’re done. Assume r is not irreducible. Then $r = r_1\cdot s_1$ where $r_1, s_1 \not\in R^* \cup \{0\}$ such that $r_i = r_{i+1}\cdot s_{i+1}$. This process stops if $r_{i0}$ is irreducible. Assume the process does not stop. Then $r_i = r_{i+1}\cdot s_{i+1} \rightarrow (r_i) \subseteq (r_{i+1})$. Therefore $s_{i+1} \not\in R^*, (r_i) \subset (r_{i+1})$. $I = \bigcup_{i \in \mathbb{Z}_{>0}}r_i$ is an ideal since an increasing union of ideals is always an ideal. Since R is a PID I = (a) for some a. Then there exists i such that $a \in (r_i). (r_{i+1}) \subseteq I = (a) \subseteq (r_i)$ which is a contradiction. Hence there exists i0 such that $r_{i0}$ is irreducible and $r = r_{i0} \cdot \prod_{i < i0}s_i x_i$. By the same proof applied to $x_i$ we get that $x_1 = p_2 \cdot x_2$. We go on by induction (we stop if $x_i$ is a unit). $(x_1) \subset (x_2) \subset (x_3) \subset ... \subset (x_i) \subset ...$. Because we have no infinite increasing chains of ideals there exists i0 such that $x_{i0} \in R^*$ and $r = x_{i0} \cdot \prod_{i < i0}p_i = (x_{i0} \cdot p_{i0}) \cdot \prod_{i < i0}p_i$ so r is a product of irreducibles. Uniqueness: we’ll proceed by indcution on n: for n = 1 $p_0 = \prod_{j=0}^{m-1}q_j = q_0 \cdot (\prod_{j>0}q_j)$. If $\prod q_j$ is a unit then $(q_j) \cap R^*$ is nonempty so $q_i$ is a unit which is not possible because an irreducivle element cannot be a unit, hence m=1 and $p_0 = q_o$. Assume case n holds $\prod_{i=0}^n p_i = \prod_{j=0}^{m-1}q_j$. Then $(\prod_{i=0}^{n-1}p_i)p_n= \prod_{j=0}^{m-1}q_j$ hence $p_n \vert \prod_{j=0}^{m-1} q_j$. Since R is a PID there exists $j_0$ such that $q_{j_0} \in (p_n)$ therefore $q_{j0} = p_n \cdot u$ where u is a unit. Then $(\prod_{i<n}p_i) \cdot p_n = q_{j0}\cdot \prod_{j \neq j0}q_j = p_n \cdot u \prod_{j \neq j0}q_j$ so $\prod_{i < n} p_i = (u \cdot q_{j1}) \cdot \prod_{j \neq jo \cdot j1}q_j$ and by induction we are done.

Fundamental theorem of arithmetic: $\mathbb{Z}$ is a UFD

Proof: $\mathbb{Z}$ is euclidean $\rightarrow \mathbb{Z}$ is a PID $\rightarrow \mathbb{Z}$ is a UFD.

Proposition: let R be a UFD, then prime $\Leftrightarrow$ irreducible

Proof: prime $\rightarrow$ irreducible follows from the ring being integral. Irreducible $\rightarrow$ prime: assume $p, a, b \in R$ are such that $p \vert ab$ so there exists a c in R such that cp =ab. Let $a = \prod p_i, b = \prod q_j, c = \prod r_p$ $p \cdot \prod r_p = \prod p_i \prod q_j$. By uniqueness we might assume that p and $p_0$ are associated so $p_0 = up, u \in R^*$. Therefore $a = p_0 \cdot \prod_{i > 0} p_i = p \cdot (u \cdot \prod_{i >0} p_i)$ so p divides a and we’re done.