Why can we use the division algorithm for $x-a$?

In Theorem 5.2.3 in these notes, it is said that

Since $x − a$ has leading coefficient $1$, which is a unit, we may use the Division Algorithm…

Why is this true? I thought that the Division Algorithm is only guaranteed to work in Euclidean domains not any integral domain.

Thanks.

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For polynomials over any coefficient ring, the high-school polynomial long division algorithm works to divide with remainder by any monic polynomial, i.e any polynomial $\rm\:f\:$ whose leading coefficient $\rm\:c =1\:$ (or a unit), since $\rm\:f\:$ monic implies that the leading term of $\rm\:f\:$ divides all higher degree monomials $\rm\:x^k,\ k\ge n = deg\ f,\:$ so the division algorithm works to kill all higher degree terms in the dividend, leaving a remainder of degree $\rm < n = deg\ f.$

The division algorithm generally fails if $\rm\:f\:$ is not monic, e.g. $\rm\: x = 2x\:q + r\:$ has no solution for $\rm\:r\in \mathbb Z,\ q\in \mathbb Z[x],\:$ since evaluating at $\rm\:x=0\:$ $\Rightarrow$ $\rm\:r=0,\:$ evaluating at $\rm\:x=1\:$ $\Rightarrow$ $\:2\:|\:1\:$ in $\mathbb Z\:$ $\Rightarrow\!\Leftarrow$ Notice that the same proof works in any coefficient ring $\rm\:R\:$ in which $2$ is not a unit (invertible). Conversely, if $2$ is a unit in $\rm\:R,$ say $\rm\:2u = 1\:$ for $\rm\:u\in R,\:$ then division is possible: $\rm\: x = 2x\cdot u + 0.$

However, we can generalize the division algorithm to the non-monic case as follows.

Theorem (nonmonic Polynomial Division Algorithm) $\ $ Let $\,0\neq F,G\in A[x]\,$ be polynomials over a commutative ring $A,$ with $\,a\,$ = lead coef of $\,F,\,$ and $\, i \ge \max\{0,\,1\!+\!\deg G\!-\!\deg F\}.\,$ Then
$\qquad\qquad \phantom{1^{1^{1^{1^{1^{1}}}}}}a^{i} G\, =\, Q F + R\ \ {\rm for\ some}\ \ Q,R\in A[x],\ \deg R < \deg F$

Proof $\,\ $ If $\ \deg G < \deg F\,$ then let $\, Q = 0 ,\ R = a^i G.\, $ Else we induct on $\,\deg G.\,$ Let $\, k = \deg F,\,$ so $\,\deg G = k+j\,$ for $\, j \geq 0.\,$ Splitting $\,G,F\,$ into $ $ lead $\color{#c00}+$ rest $ $ terms:

$\begin{array}{lrl}
\ G = b x^{k+j\!}\color{#c00} + G’,\ \deg G'<k\!+\!j\!\!\!\!\!\! &\Rightarrow\quad aG\!\!\!\! &=\,abx^{k+j}+aG’\\
\ F = a x^k\color{#c00}+F’,\quad \deg F’ \!< k & bx^jF\!\!\!\! &=\, abx^{k+j}+bx^jF’\\
\qquad\ a,b\neq 0\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! & aG-bx^jF\!\!\!\! &=\, aG’-bx^jF’\ \ {\rm has\ \ deg} < k\!+\!j \end{array}$

$\begin{array}{lrl}{\rm Therefore,\ \ by\ \ induction} &a^j(aG-bx^jF) =&\!\!\! Q F + R\ \ {\rm for}\ \ Q,R\in A[x], \ \deg R < k\\
&\Rightarrow\quad a^{j+1} G\, =&\!\!\! \bar QF + R\ \ {\rm for}\ \ \bar Q = Q\!+b(ax)^j\end{array}$

Remark $\ $ Alternatively, if localizations are known, we can divide by the monic $\,a^{-1} F\in A[a^{-1}][x]\,$ then pullback the result to $\,A[x].$

Or, as in the AC-method, we can conjugate to the monic case: scaling $\,F\,$ by $\,a^{k-1}\,$ for $\,k = {\rm deg} F,\,$ we can rewrite $\,F\,$ as a monic polynomial in $\,X = ax,\,$ and similarly we can scale $\,G\,$ by $\,a^i\,$ to make it a polynomial in $\,X.\,$ Then we divide $\,G(X)\,$ by the monic $\,F(X),\,$ and finally replace $\,X\,$ by $\,ax.$

That’s the point really. It isn’t guaranteed to work for EVERY polynomial in the ring $R[x]$ you are dealing with, but it will work for some polynomials, and the polynomial $x-a$ is never a problem. A polynomial of the form $ax +b$ with $a$ a non-unit of $R$ would cause a problem. To deal explicitly with $x-a$ and a polynomial $p(x) \in R[x]$, we can work by induction on the degree of $p(x).$ Suppose that this is $n > 1$ and we can write polynomials of degree $n-1$ in the expected form (note that when $p(x) = cx+d$ has degree $1,$ we have $cx+d = c(x-a) + (ac+d)$, which starts the induction). We can certainly write $p(x) = xq(x) + r$ for some polynomial $q(x) \in R[x]$ of degree $n-1$ and some $r \in R.$ By assumption, we may write $q(x) = (x-a)s(x) + t$ where $s(x)\in R[x]$ has degree $n-1$ and $t \in R$. For the sake of space I omit some steps, but you can then see that $p(x) = (x-a) [xs(x) + t ] + (at+r).$