# Is there a ring which doesn't satisfy Division Algorithm for polynomials?

Division Algorithm: If $f(x)$ is a polynomial of degree $n$ and $g(x)$ is a polynomial of degree $m≧0$, then there exist unique polynomials $q(x)$ and $r(x)$ such that $f(x)=q(x)g(x)+r(x)$ where the degree of $r(x)$ is less than $m$.

I know this holds for any field $F$, but what about ring $R$? Since this holds for $\mathbb{Z}$, i cannot think of any counterexample in my mind.

The uniqueness part of the statement is false for any ring $R$ with zero-divisors (e.g. $\mathbb{Z}/4\mathbb{Z}$).
If $a,b\in R$ are non-zero and $ab=0$, then $$a=0\cdot bx+a$$ as well as $$a=a\cdot bx+a$$ (non-zero elements of $R$ are polynomials of degree 0.)
Here's the best you can do in general. Let $R$ be a commutative ring with unity and $p(x) \in R[x]$ monic. Then any polynomial $f(x) \in R[x]$ can be written as
$$f(x)=q(x)p(x)+r(x)$$
where $\deg r(x) < \deg p(x)$. The standard argument used for the Euclidean algorithm goes through in this instance. The key point being that we can always match coefficients because $p(x)$ is monic.
This is not true for $\mathbb Z[X]$. Try to divide $x^2+1$ by $2x$...