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

So suppose now that $\mathrm{deg}(P)=n$ 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 $a_n,b_k\ne 0,k\le n$. Then setting $H=\frac{a_n}{b_k}{X}^{n-k}$ we have the relation

$\begin{array}{cc}{P}^{\prime }& :=P-TH\\ & =0X^n+(a_{n-1}-\frac{a_n}{b_k}{b}_{k-1})X^{n-1}+\cdots +(a_{n-k}-\frac{a_n}{b_k}{b}_0)X^{n-k}+a_{n-k-1}{X}^{n-k-1}+\cdots +a_0.\end{array}$

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

$P^{\prime }=T{Q}^{\prime }+R^{\prime }\text{ and with }\mathrm{deg}(R^{\prime })<\mathrm{deg}(T)\text{ or }{R}^{\prime }=0.$

From this we get altogether

$P=P^{\prime }+TH=T{Q}^{\prime }+TH+R^{\prime }=T(Q^{\prime }+H)+R^{\prime },$

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

To prove uniqueness, let $P=TQ+R=T{Q}^{\prime }+R^{\prime }$, both fulfilling the stated conditions. Then $T(Q-Q^{\prime })=R^{\prime }-R$. Since the degree of the difference $R^{\prime }-R$ is smaller than $\mathrm{deg}(T)$, this implies $R=R^{\prime }$ and so $Q=Q^{\prime }$.

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