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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 $u \in R$.

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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: $$ \frac{5}{8}x^3y^2z $$

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  • $\begingroup$ 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? $\endgroup$ – Lele99_DD Aug 26 '19 at 13:21
  • $\begingroup$ 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. $\endgroup$ – Marc Bogaerts Aug 26 '19 at 13:40
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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).
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