> [!NOTE] Theorem (Euclid's Lemma for Ring of Polynomial Forms over Field) > Let $F$ be a [[Field (Algebra)|field]]. Let $F[x]$ be the [[Ring of Polynomial Forms|ring of polynomial forms]] over $F$ in $x.$ Let $f$ be an [[Irreducible Polynomial|irreducible polynomial]] over $F$ in $x.$ Let $g,h\in F[x]$ such that $f$ [[Divisibility in Ring of Polynomial Forms|divides]] $gh,$ denoted $f \mid gh.$ Then $f\mid g$ or $f\mid h.$ *Proof*. Let $f\mid gh.$ Suppose $f \not \mid g.$ If for some $j\in F[x],$ $j\mid f$ and $j\mid g$ then $f=jk_{1}$ and $g=jl_{1}$ for some $k_{1},l_{1}\in F[x].$ Since $f$ is irreducible either $j$ or $k_{1}$ is a unit. If the latter $k_{1}=\alpha$ where $0\neq \alpha\in F$ then $j=\alpha^{-1}f.$ But then $g=jl_{1}=\alpha^{-1}fl_{1}$ and $f\mid g,$ a contradiction. So $j$ is a unit. Thus $f$ and $g$ are [[Relatively Prime Polynomials over Integral Domain|relatively prime]]. Therefore by [[Bézout's Identity for Ring of Polynomial Forms Over Field|Bézout's identity]], there exists polynomials $x,y\in F[x]$ such that $fx+gy=1$Then $h=hfx+hgy$Since $f\mid gh,$ we have $f\mid hfx+hgy=h.$ # Applications **Consequences**: [[Ring of Polynomial Forms over Field is Unique Factorisation Domain]].