> [!NOTE] Theorem (Ideals of Ring of Polynomials over a Field) > Let $F$ be [[Field (Algebra)|field]]. Let $F[x]$ be the [[Ring of Polynomial Forms|ring of polynomial forms]] over $F$ in $x.$ Then $F[x]$ is a [[Principal Ideal Domain|principal ideal domain]]: that is for any [[Ideal of Ring|ideal]] of $F[x],$ denote $I,$ $I=\{ fg \mid g\in F[x] \}$ for some $f\in F[x].$ **Notation**: we may write $\{ fg \mid g\in F[x] \}$ as $fF[x]$ (just like we write $n\mathbb{Z}$ instead of $\{ nm \mid m\in \mathbb{Z} \}$ which are the [[Ideals of Integers|ideals of the integers]]). **Proof**: If $I=\{ 0 \}$ then the result is true with $f=0$. Now suppose $I \neq \{ 0 \}.$ Let $f \in I$ have the smallest possible degree among the non-zero polynomials in $I$ which exists by [[Well-Ordering Principle|well-ordering]]. Since $I$ is an ideal, by definition we have $\{ fg \mid g(x) \in F[x] \} \subset I.$ Now suppose $h \in I$. By [[Division with Remainder Theorem for Ring of Polynomial Forms over Fields|division with remainder]], there exists $q,r \in F[x]$ such that $h=fq+r$ and $\deg(r)<\deg(f).$ But then $r=h-fq \in I$ since $h,fq\in I$ and $(I,+)$ is a group and is closed. Since $r$ has degree less than that $f,$ by minimality of the degree of $f$ it must be the zero polynomial. Thus $h=fq\in \{ fg \mid \in F[x] \}$ and $I \subset \{ fg \mid g\in F[x] \}.$ Therefore $I=\{ fg \mid g\in F[x] \}.$ *Proof*. We have that [[Ring of Polynomial Forms over Field is a Euclidean Domain]]. We also have [[Euclidean Domain is Principal Ideal Domain]]. # Applications **Consequences:** We have [[Bézout's Identity for Ring of Polynomial Forms Over Field]] so in particular if two polynomials over a field $f,g$ are relatively prime then there there exists polynomials over the same field $x,y$ such that $fx+gy=1.$