Notes - Rings and Modules HT24, Factorisation

Every element in $R \setminus \{ 0 \}$ is either a unit, or can be rewitten as a product of irreducible elements.

Every element in $R \setminus \{0\}$ is either a unit, or can be written as a product of irreducible elements (i.e. it is a factorisation domain) unique up to reordering and multiplication by units.

$\mathbb Z[\sqrt{-5}]$, since $6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$ and $2, 3, (1 + \sqrt{-5}), (1 - \sqrt{-5})$ are irreducible.

$(1) \implies (2)$:

Suppose $R$ is a UFD and $p$ is irreducible. We aim to show that $p$ is prime (point 2 is immediate, since UFDs imply that $p$ already can be written as a product of irreducibles). Suppose $p \mid ab$ for some $a, b \in R$. We want to show $p \mid a$ or $p \mid b$.

If $a$ or $b$ zero, then $p \mid a$ or $p \mid b$ immediately. Since $p \mid ab$, $ab = pd$ for some $d$. If either $a$ or $b$ is a unit, then by pre- or post- multiplying by its inverse, we see $p \mid a$ or $p \mid b$ immediately also. Otherwise, write $a, b, d$ as a product of irreducibles, say

\[\begin{aligned} a &= a_1\cdots a_k \\\\ b &= b_1\cdots b_\ell \\\\ d &= d_1\cdots d_s \end{aligned}\]

Then we have

\[(pd_1 \cdots d_s) = (a_1 \cdots a_k) (b_1 \cdots b_\ell)\]

Since the factorisation into irreducibles in a UFD is unique up to reordering and units, because $p$ is irreducible it must be the case that $p = a _ i$ or $p = b _ j$ up to units for some $i$ or $j$. Then $p \mid a$ or $p \mid b$ as required.

$(2) \implies (1)$:

Now suppose that every irreducible element is prime and that every non-zero non-unit $a \in R$ can be written as a product of irreducibles.

Fix some $a \in R$. We want to show that it has a factorisation into irreducibles unique up to reordering and units, so we proceed by inducting on $M(a)$, the minimum number of irreducibles in the factorisation of $a$ into irreducible elements.

By the assumptions, we have that $a$ can be written as a product of irreducibles. Suppose we have two products:

\[a = p_1 \cdots p_{M(a)}\]

and

\[a = q_1 \cdots q_k\]

are two factorisations of $a$, and $k \ge M$. Since $p _ 1$ is prime, and since $p _ 1 \mid q _ 1 \cdots q _ k$, there must exist $j$ with $p _ 1 \vert q _ j$. We can assume by reordering that $j = 1$. Since $q _ 1$ is irreducible, we have

\[q_1 = u_1 p_1\]

where $u _ 1$ is a unit (otherwise $p \vert q _ 1$ implies that $q _ 1$ would be reducible). Hence we see that

\[\begin{aligned} &p_1p_2 \cdots p_{M(a)} = (q_1) q_2 \cdots q_k \\\\ \implies& p_1 p_2\cdots p_{M(a)} = (u_1 p_1) q_2\cdots q_k \\\\ \implies& p_1 p_2\cdots p_{M(a)} = p_1 u_1q_2\cdots q_k &&(1\star) \\\\ \implies& p_2 \cdots p_{M(a)} = u_1 q_2 \cdots q_k &&(2\star) \end{aligned}\]

where the brackets are just for clarity showing what was replaced, $(1\star)$ is just reordering, and $(2\star)$ follows from the fact that since $R$ is an integral domain, we have the cancelation law (if $a\ne 0$, then $ab = ac \implies b = c$).

Since $p _ 2 \cdots p _ {M(a)} = u _ 1 q _ 2 \cdots q _ k$ is two factorisations into irreducibles with $M(a) - 1$ elements, by induction on $M(a)$, we are done (I’m not sure this is fully correct).

$(2) \implies (3)$

Since prime elements are irreducible by assumption, this is immediate.

$(3) \implies (2)$

We need to check that irreducible elements are prime. If $a \in R$ is irreducible and $a = p _ 1 p _ 2 \cdots p _ k$ as a product of primes, then by the definition of irreducibility, $k = 1$, so $a$ is prime.

Since we are in a PID, and all irreducible elements are prime in a PID, it suffices to just show every element has some factorisation into irreducibles and not worry about uniqueness (since if every irreducible element is prime and if every non-zero non-unit $a \in R$ can be written as a product of irreducibles, then $R$ is a UFD).

• Suppose $\exists a _ 1 \in R$ such that $a _ 1$ is not a product of irreducibles.
• In particular, $a _ 1$ itself cannot be irreducible, so $\exists b, c$ that are not units s.t. $a _ 1 = bc$
• At least one of $b, c$ can’t be written as a product of irreducible elements (if both could be, then $a _ 1$ would be a product of irreducibles). Denote this element $a _ 2$.
• Repeating the process for $a _ 2$, we get a sequenece $a _ 1, a _ 2, \cdots$ of elements that are not products of irreducibles.
• But then $\langle a _ 1 \rangle \subsetneq \langle a _ 2 \rangle \subsetneq \cdots$, because the fact $a _ 1 = b a _ 2$ (wlog) implies $\langle a _ 1 \rangle \subseteq \langle a _ 2 \rangle$, and the containment is strict since if $\langle a _ 1 \rangle = \langle a _ 2 \rangle$, then $a _ 1 = u a _ 2$ for some unit and since $a _ 1 = bc$ would imply $b$ were a unit, but it was assumed not to be.
• This is a contradiction, because we have an ascending chain of ideals that never stabilises, but any PID is Noetherian and so must stabilise.

(This proof actually shows that any ring satisfying the ACCP (ascending chain condition on principal ideals) and that where any irreducible element is prime is a UFD).

It is also a UFD.