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

Proof

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

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

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

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

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

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

From this we get altogether

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

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

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

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