Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots.
Thoughts so far:
I feel like I might be missing something here. If $ R $ is an integral domain, then so is $ R[X] $. The factor theorem gives a correspondence between roots and factors. Clearly (is this clear?) a polynomial of degree $ d $ can't have more than $ d $ irreducible factors.
Best Answer
It is only mildly clear; you have to show that, when $R$ (and hence $R[x]$) is an integral domain, the degree function does in fact satisfy $$\text{deg}(fg)=\text{deg}(f)+\text{deg}(g).$$ Then you can argue that (by the factor theorem), if $a_1,\ldots,a_n$ were roots of $f$, then $f=(x-a_1)\cdots(x-a_n)g$ for some $g\in R[x]$, so that $\deg(f)=n+\deg(g)$ and $\deg(g)\geq0$, hence $\deg(f)\geq n$.
When $R$ is not an integral domain, we might have $\text{deg}(fg)<\text{deg}(f)+\text{deg}(g)$ (do you see why?)