> [!NOTE] Theorem (Bézout's Identity) > 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 $f,g\in F[x]$ such that they're both not equal to zero. Let $d=\gcd\{ f,g \}$ denote their [[Greatest Common Divisor of Polynomial Forms Over Integral Domain|gcd]]. Then $d$ is the smallest degree non-zero polynomial that can be written in the form $fx+gy$ where $x,y \in D[x].$ Proof. Follows from [[Ring of Polynomial Forms over Field is a Euclidean Domain]] and [[Bézout's Lemma for Euclidean Domains]]. > [!NOTE] Corollary > Let $F$ be a [[Field (Algebra)|field]]. Let $f,g \in F[x]$ be non-zero polynomials that are [[Relatively Prime Polynomials over Integral Domain|relatively prime]]. Then there exists $x,y\in F[x]$ such that $fx+gy=1.$ *Proof*. If $f,g$ are relatively prime then their gcd is a unit. Multiply both sides by multiplicative inverse. *Proof*. Let $I = \{ fh+gk \mid h, k \in F[x] \}$. Firstly, the zero polynomial is in $I$ because $0 = f(x) \times 0 + g(x) \times 0$. Next, if polynomials $j_{1} (x), j_{2}(x) \in I$ then $j_{1}(x) = f(x)h_{1}(x)+g(x)k_{1}(x)$ and $j_{2}(x) =f(x)h_{2}(x)+g(x)k_{2}(x)$. Then $j_{1}(x)+j_{2}(x)=f(x)[h_{1}(x)+h_{2}(x)] +g(x)[k_{1}(x)+k_{2}(x)]$and since $h_{1}(x)+h_{2}(x) \in F[x]$ and $k_{1}(x)+k_{2}(x) \in F[x]$ then $j_{1}(x)+j_{2}(x) \in I$ and $(I,+)$ is closed. Finally, if $j(x) \in F[c]$ then $j(x) = f(x)h(x)+g(x)k(x)$ for some $h(x), k(x) \in F[x]$. Then $-j(x)=f(x)(-h(x))+g(x)(-k(x)) \in I$. Thus $I$ is a subgroup of $(F[x],+)$ by [[Two-Step Subgroup Test|two-step subgroup test]]. Also for any $r(x) \in F[x]$, $j(x)r(x) = f(x)h(x)r(x) + g(x)k(x)r(x)$which shows $j(x)r(x) \in I \implies r(x)j(x)$, by commutativity of $F[x]$. Thus $I$ is an [[Ideal of Ring|ideal]] of $F[x].$ By [[Ring of Polynomial Forms over Field is a Principal Ideal Domain|ring of polynomial forms over field is PID]], there exists $j(x) \in F(x)$ such that $I=j(x)F[x]$. Since $f(x),g(x) \in I=j(x)F[x]$, there exists $k_{1}(x),h_{1}(x)\in F[x]$ such that $f(x)=j(x)k_{1}(x)$ and $g(x)=j(x)h_{1}(x)$. This means that $j(x) \mid f(x)$ and $j(x) \mid g(x)$ and since $f(x)$ and $g(x)$ are relatively prime $j(x)$ is a unit in $F[x]$ so $I=F[x]$ by [[Ideal with Unit is Whole Ring|ideal with unit is whole ring]]. Since $1 \in F[x]=I$, there exists $h(x),k(x) \in F[x]$ such that $f(x)h(x)+g(x)k(x)=1$. # Applications **Consequences**: [[Euclid's Lemma for Irreducible Polynomial Forms over Field]] and so $F[x]$ is a unique factorization domain.