> [!NOTE] **Theorem** (Ring of Univariate Polynomial Forms over Field form Unique Factorization Domain) > > Let $F$ be a [[Field (Algebra)|field]]. Let $F[x]$ denote the [[Ring of Polynomial Forms|ring of polynomial forms]] over $F$ in $x.$ Let $f\in F[x]$ be a non-constant polynomial. Then $f$ can be uniquely expressed as a product of [[Irreducible Polynomial|irreducible polynomials]] over $F$ up to multiplication by units: that is if $f=g_{1}\dots g_{m}=h_{1}\dots h_{n}$where $h_{i}$ and $g_{j}$ are irreducible, then $m=n$ and after renumbering the $h_{i}s we have $g_{1}(x)=a_{1}h_{1}(x), \;g_{2}(x)=a_{2} h_{2}(x),\; \dots \;, g_{m}(x) = a_{n}h_{n}(x) $where each $a_{i}$ is a [[Units of Ring of Polynomial Forms over Integral Domain|unit]] in $F[x].$ ***Proof of Existence of Complete Factorization**: For $n\in\mathbb{N}_{\geq 2},$ let $P(n)$ be the statement that every non-constant polynomial $f\in F[x]$ with $\deg(f)<n$ can be expressed as a product of irreducible polynomials over $F.$ Any degree one polynomial over $F$ is irreducible and so $P(2)$ is true. Suppose $P(k)$ is true for some $k\in \mathbb{N}_{\geq 2}$ and so every polynomial $f$ with $1\leq \deg(f)\leq k-1$ has a complete factorization. Let $g\in F[x]$ with $\deg(g)=k.$ If $g$ is irreducible then we are done. Otherwise, $g$ is a product of polynomials of smaller degree and by $p(k)$ they have complete factorisations. It follows that $g$ also has a complete factorisation and $p(k)\implies p(k+1).$ Therefore, by [[Induction Principle|mathematical induction]], every non-constant polynomial is a product of irreducible polynomials. ***Proof of Uniqueness Complete Factorization**:* Suppose. $p_{1}p_{2}\dots p_{s}=q_{1}q_{2}\dots q_{t}$where all the $ps and $qs are irreducible over $F.$ Then $p_{1}\mid q_{1}q_{2}\dots q_{t}$ and by [[Euclid's Lemma for Irreducible Polynomial Forms over Field|Euclid's lemma]], $p_{1}\mid q_{i}$ for some $i.$ So $q_{i}=kp_{1}$ for some $k\in F[x].$ Since $q_{i}$ is irreducible either $\deg(k)=0$ or $\deg(p_{1})=0$ but not both. $p_{1}$ is irreducible so $\deg(p_{1})>0$ which implies $\deg(k)=0$ i.e. $k$ is a unit in $F[x].$ Thus $q_{i}=u_{i}p_{1}$ for some $u_{i}\in F.$ Cancelling $p_{1}$ from both sides, we proceed similarly with $p_{2},$ and so on. After a finite number of steps, we determine that $s=t$ and that $q_{1},\dots,q_{t}$ are associates of $p_{1},\dots,p_{s}$ in some order. ***Proof***: Follows from [[Ring of Polynomial Forms over Field is a Euclidean Domain]] and [[Euclidean domain is UFD]]. ***Proof***: Follows from [[Ring of Polynomial Forms over Field is a Principal Ideal Domain]] and [[Principal Ideal Domain is Unique Factorization Domain]].