# Division algorithm in A[x] (A NOT a field!)

1. Jul 19, 2010

### sidm

is it possible? I'm reading a proof where A is a local ring and f is a monic polynomial, v another polynomial in A[x] then the author says there is q,r with v=qf+r and degf<r.

I thought the algorithm required divison of coefficients?! Maybe it's true we can always do this with f monic?

thanks

2. Jul 19, 2010

### rasmhop

I'm assuming you mean to require deg r < deg f (not the other way around).

No division of coefficients is required as long as f is monic (you don't need that A is local either). If f has degree larger than v you can just let r=f, q=0. Thus you can assume $deg(f) \leq deg(v)$. Let (v,f) be a pair of polynomials for which no such division exists, and for which the degree of v is minimal.

If deg v = 0, then deg f = 0 and since f is monic f=1 so we can let q=v, r=0.

If deg v > 0 let:
$$h(x) = v(x) - Ax^{\deg(v) - \deg(f)}f(x)$$
where A is the coefficient of the term of highest degree in v(x). Then deg h < deg v and therefore we can divide h by f to give it polynomials q(x) and r(x) with deg r < deg f such that:
$$h(x) = q(x)f(x) + r(x)$$
so
$$v(x) = (q(x)+Ax^{\deg(v) - \deg(f)})f(x) + r(x)$$
which shows that division is possible.

3. Jul 20, 2010

### sidm

ahhh, this is a good fact to know, much appreciated!