polynomial with a root but no linear factor I have proven that if $R$ is UFD, $f(x) \in R[x]$, then $f(x)$ has a root $r\in R$ if and only if $f$ has a linear factor over $R$. 
Is this true for rings in general, or domains in general? I have been trying to construct an example of a polynomial in $\mathbb{Z}[\sqrt{-5}]$ with a root but no linear factor, but I cannot seem to find an example. 
 A: The implication $f$ has a root $\implies$ $f$ has a linear factor is true over any unitary ring. 
Proof Since $1$ is an unit in $R$, we can do long division by $x-a$ in $R[X]$. Then 
$$f(x)=(x-a)Q(x)+R(x)$$
with $R$ a constant polynomial. Plugging in $a$ we get $R=0$, therefore $x-a$ is a factor of $f(x)$.
P.S. The problem in non UFD case is what do we mean by a factor. For example in $\mathbb Z/6\mathbb Z[X]$ we have
$$X^2-5X=(X-2)(X-3)=X(X-5)$$
Can we call all these factors of $f$?
A: The first claim that $f(x)$ has a root if and only if $f$ has a linear factor is not true for UFDs. Just consider the product of linear factors of the form $ax+b$ such that $a\neq 0$ is not a unit in $R$ (of course this does not exist if $R$ is a field). Jyrki already gave an example. 
However, the division algorithm for polynomials works in $R[X]$ for any ring $R$, provided that the leading coefficient of the divisor is a unit. So if there is a root $r$, then we can divide by $(x-r)$. Hence there is no polynomial over a ring having a root but no linear factor.
