Remainder theorem states that for a $c$ in $F$ and $f(x)$ in $F(x)$. When we divide $f(x)$ by $x - c$ then the remainder is $f(c)$.
the Root Factor Theorem states that for $c$ in $F$ is a root of $f(x)$ in $F(x)$ if $f(c) = 0 $
Now my question is to show that the two above statements would be valid even when the coefficients of $f(x)$ lie in an integral domain $R$. But the following may not hold.
Given nonzero polynomials $f(x), g(x)$ in $F(x)$, let $$S = \{h(x) \in F(x) : h(x) = a(x)f(x)+b(x)g(x),\; a(x),b(x) \in F(x)\}$$ then there is a polynomial $d(x)$ in $S$ of smallest degree, and every $h(x)$ in $S$ is divisible by $d(x)$. (the hint given )
Please let me know if you have a way of proving this, I don't know how to approach it. What I tried to do is mimic the division algorithm.
Thank you,
Samir
The thing to notice is that in an integral domain, if the divisor polynomial is monic, then the division algorithm works in just the same way it does for a field (and by the same proof). Therefore your two facts above are valid in any integral domain. Your statement "Given nonzero polynomials..." is false, though: Look at the polynomials $2$ and $x$ in $\mathbb Z[x]$. The lowest-degree linear combinations of these two polynomials is are the even constants, but $x$ is not a multiple of any of them.