# Polynomial ring/Field/One variable/Euclidean division/Fact/Proof

Proof

We prove the statement about the existence by induction over the degree of ${\displaystyle {}P}$. If the degree of ${\displaystyle {}T}$ is larger than the degree of ${\displaystyle {}P}$, then ${\displaystyle {}Q=0}$ and ${\displaystyle {}R=P}$ is a solution.

Suppose that ${\displaystyle {}\operatorname {deg} \,(P)=0}$. By the remark just made also ${\displaystyle {}\operatorname {deg} \,(T)=0}$ holds, so ${\displaystyle {}T}$ is a constant polynomial, and therefore (since ${\displaystyle {}T\neq 0}$ and ${\displaystyle {}K}$ is a field) ${\displaystyle {}Q=P/T}$ and ${\displaystyle {}R=0}$ is a solution.

So suppose now that ${\displaystyle {}\operatorname {deg} \,(P)=n}$ and that the statement for smaller degrees is already proven. We write ${\displaystyle {}P=a_{n}X^{n}+\cdots +a_{1}X+a_{0}}$ and ${\displaystyle {}T=b_{k}X^{k}+\cdots +b_{1}X+b_{0}}$ with ${\displaystyle {}a_{n},b_{k}\neq 0,\,k\leq n}$. Then setting ${\displaystyle {}H={\frac {a_{n}}{b_{k}}}X^{n-k}}$ we have the relation

{\displaystyle {}{\begin{aligned}P'&:=P-TH\\&=0X^{n}+{\left(a_{n-1}-{\frac {a_{n}}{b_{k}}}b_{k-1}\right)}X^{n-1}+\cdots +{\left(a_{n-k}-{\frac {a_{n}}{b_{k}}}b_{0}\right)}X^{n-k}+a_{n-k-1}X^{n-k-1}+\cdots +a_{0}.\end{aligned}}}

The degree of this polynomial ${\displaystyle {}P'}$ is smaller than ${\displaystyle {}n}$ and we can apply the induction hypothesis to it. That means there exist ${\displaystyle {}Q'}$ and ${\displaystyle {}R'}$ such that

${\displaystyle P'=TQ'+R'{\text{ and with }}\operatorname {deg} \,(R')<\operatorname {deg} \,(T){\text{ or }}R'=0.}$

From this we get altogether

${\displaystyle {}P=P'+TH=TQ'+TH+R'=T(Q'+H)+R'\,,}$

so that ${\displaystyle {}Q=Q'+H}$ and ${\displaystyle {}R=R'}$ is a solution.

To prove uniqueness, let ${\displaystyle {}P=TQ+R=TQ'+R'}$, both fulfilling the stated conditions. Then ${\displaystyle {}T(Q-Q')=R'-R}$. Since the degree of the difference ${\displaystyle {}R'-R}$ is smaller than ${\displaystyle {}\operatorname {deg} \,(T)}$, this implies ${\displaystyle {}R=R'}$ and so ${\displaystyle {}Q=Q'}$.