# Unique factorization of an element in an UFD

By definition:

An integral domain $$R$$ is a unique factorization domain if the following conditions are satisfied:

1. Every element $$a \in R$$, $$a \neq 0$$ that is not a unit can be factored into a product $$a = c_1 \cdots c_n$$ where $$c_1,\dots,c_n \in R$$ are irreducible elements.
2. If $$c_1,\dots,c_n$$ and $$d_1,\dots,d_m$$ are two factorizations of the same element of $$R$$ into irreducibles, then $$n = m$$ and $$d_j$$ can be renumbered so that $$c_i$$ and $$d_i$$ are associates.

I need to prove that every element $$a \in R$$, $$a \neq 0$$ which is not a unit can be written uniquely as: $$\begin{equation} a = up_1^{e_1} \cdots p_s^{e_s} \end{equation}$$ where $$u \in R$$ is a unit, $$p_1,\dots,p_s \in R$$ are irreducible elements mutually not associate and $$e_1,\dots,e_s \in \mathbb{N} \setminus \{0\}$$. I think I need to start with an arbitrary factorization $$a = c_1 \cdots c_n$$, then use the following result, but honestly I don't know how to put it formally.

Let $$R$$ be an integral domain and let $$a,b \in R$$. If $$a$$ and $$b$$ are associate elements, then $$a,b \neq 0$$ and $$a = b \cdot u$$ for some unit $$u \in R$$.

## 2 Answers

The action you take is identical to the following situation where you consider words that are monomials in several variables: for example $$(\frac{3}{4}x)(5y)(x)(\frac{5}{3}x)(\frac{2}{5}z)(\frac{1}{4}y)$$ First group all associates: $$(\frac{3}{4}x)(x)(\frac{5}{3}x).(5y)(\frac{1}{4}y). (\frac{2}{5}z)$$ Then for each group of accociates extract a unique unit: $$(\frac{3}{4}\frac{5}{3})(x)(x)(x).(5\frac{1}{4})(y)(y).(\frac{2}{5})(z)$$ Then bring all units together and simplify them and exponentiate the rest: $$\frac{5}{8}x^3y^2z$$

• I understand the idea, but how can I write a formal proof of the above statement? Maybe using induction on the length of the factorization? – Lele99_DD Aug 26 '19 at 13:21
• Follow the three steps in the example: First note that you can regroup the associates and relabel them with the same label but a different unit, like in $u_{1,1}p_1.u_{1,2}p_1.u_{1,3}p_1$ and $u_{2,1}p_2.u_{2,2}p_2.u_{2,3}p_2$ etc... Then show that in each group you can put the units in the front to form just one unit per group as in $u_{1,1}u_{1,2}u_{1,3}.p_1.p_1.p_1$ . Then show that all the units of each group can be put in front as one uniuqe unit followed by equal identical elements that can be exponentiated. – Marc Bogaerts Aug 26 '19 at 13:40

By definition of unique factorization domain, the given element $$a$$ has a factorization in irreducibles $$a = c_1 \cdots c_n$$. We proceed by induction on the length of the factorization ($$n \in \mathbb{N}$$, $$n \geq 1$$). The base of induction ($$n = 1$$) is trivial. In fact, if $$a = c_1$$ is a factorization of $$a$$ in irreducibles, then $$a = up_1^{e_1}$$ where $$u := 1$$, $$p_1 := c_1$$ and $$e_1 := 1$$.

In the induction step, we assume $$n \geq 2$$ and we suppose that every nonzero non-unit element with a factorization of lenght $$n-1$$ can be written uniquely in the desired form. Let $$b := c_1 \cdots c_{n-1}$$. Then $$b = c_1 \cdots c_{n-1}$$ is a factorization of $$b$$ in irreducibles and by inductive hyphotesis we have that $$b = vp_1^{d_1} \cdots p_r^{d_r}$$ for some unit $$v \in R$$, for some irreducible elements $$p_1,\dots,p_r \in R$$ which are mutually not associate and for some $$d_1,\dots,d_r \in \mathbb{N} \setminus \{0\}$$. Thus $$a = vp_1^{d_1} \cdots p_r^{d_r}c_n$$. Now we have two possibilities:

• $$c_n \nsim p_i$$ for all $$1 \leq i \leq r$$. In this case, let $$u := v$$, $$s := r+1$$, $$p_s := c_n$$, $$e_i := d_i$$ for all $$1 \leq i \leq r$$ and $$e_s := 1$$ so that $$a = up_1^{e_1} \cdots p_s^{e_s}$$ (remember that a unique factorization domain is a commutative ring by definition).
• $$c_n \sim p_i$$ for some $$1 \leq i \leq r$$. In this case, by applying the result mentioned in the question we get a unit $$u_i \in R$$ such that $$c_n = p_iu_i$$. Now let $$u := vu_i$$, $$s := r$$, $$e_j := d_j$$ for all $$1 \leq j \leq s$$ with $$j \neq i$$ and $$e_i := d_i + 1$$ so that $$a = up_1^{e_1} \cdots p_s^{e_s}$$ (again, $$R$$ is a commutative ring).