# Show that the division algorithm holds for the polynomial ring with integer coefficients when the divisor is of the form g(x) = x-a?

Let $f(x)\in \Bbb Z$ and let $g(x)$ be of the form $g(x) = x-a$. Show that the division algorithm holds in this case.

I can see why this holds; it's because $g(x)$ has coefficient one for x so that it divides more or less nicely into $f(x)$, leaving a constant remainder.

I'm looking at Gallian's proof for the division algorithm to get an idea. I think this is what we do.

We proceed by induction. Suppose that $n = deg f(x) \geq deg g(x)$.

Let $f_1(x) = f(x) - (a_mx^{m-1})g(x)$ (this follows from the divisional algorithm: divide $f(x)$ by $g(x)$ and call the remainder $f_1(x)$, write $f_1(x)$ as the difference between f(x) and the product of g(x) and the quotient, $a_mx^{m-1}$).

Then $f_1(x) = 0$ or $deg f_1(x) < deg f(x)$.

Following Gallian, he applies the induction hypothesis but I have no clue where he provided the hypothesis in his proof. I'm a bit stuck.

Can somebody please help?

Thank you.

## 1 Answer

You already know that it works over $\mathbb Q$, so you get:

$$f(x) = h(x)\cdot (x-a) + c$$

with $h(x) = \sum_{i=0}^n a_ix^i$ and $a_i, c \in \mathbb Q$.

Now you can compare coefficients (starting with $x^{n+1}$, going downwards until $x^0$) in order to deduce $a_i,c \in \mathbb Z$.

More generally - with the same argument - you can show that division with remainder works in an integral domain whenever the leading coefficient of the denominator is a unit.