# Prove that the quotient and remainder of division a and b in $\mathbb{Q}$[X] also belong to $\mathbb{Z}$[X].

Suppose a and b are polynomials with integer coefficients and b has a leading coefficient 1. Prove that the quotient and remainder of division a and b in $$\mathbb{Q}$$[X] also belong to $$\mathbb{Z}$$[X].

While i understand when b has an leading coefficient 1. It can atleast divide the highest degree of a as long as the degree of a is higer than b. Meaning the quotient q will be an integer. But for the remainder r i cannot fully comprehend that it also will be an integer. Any hints or assumptions I should take?

• Just look at how division is performed in $\mathbb{Q}[X]$ and you'll realize that you never go outside $\mathbb{Z}$ if both $a$ and $b$ are in $\mathbb{Z}[X]$ and $b$ is monic. – egreg Oct 14 '18 at 22:07

Let $$a = q\, b + r.\,$$ If $$\,q\,$$ has a coef $$\not\in\Bbb Z\,$$ write $$\,q = p + q',\,$$ for $$\, p\in\Bbb Z[X]\,$$ and $$\,q'\,$$ with lead coef $$\not\in\Bbb Z.\,$$ $$a = b (p+q') + r$$ $$\Rightarrow\, a - bp = bq' + r.\,$$ so comparing lead coefs using $$\,b$$ monic & $$\deg r < \deg b\,$$ yields $$\,q'$$ has lead coef $$\in\Bbb Z,\,$$ contradiction. Thus $$\,q\in\Bbb Z[x],\,$$ so $$\,r = a\!-\!qb\in\Bbb Z[X]$$.
Alternatively it is the special case $$c=1$$ of the nonmonic division algorithm, which implies that $$\ c^k a = q\,b + r\,$$ for $$\,q,r\in\Bbb Z[X],\ c$$ = lead coef of $$\,b.$$