$R$ is an integral domain. Let $x\neq0$ and $f\in{R[t]}$ of minimal degree such that $f(x)=x^n+r_1x^{n-1}+\dots+r_n=0$. Then $r_n\neq0$.
Why is this true? I know that integral domain means if we have $ab=0$ then $a=0$ or $b=0$.
We assumed $x\neq0$ so if we assume $r_n=0$, we have $$ f(x)=x^n+r_1x^{n-1}+\dots+xr_{n-1}=0 $$ and we can write $$ x(x^{n-1}+r_1x^{n-2}+\dots+r_{n-1})=0 $$ so we know $$ x^{n-1}+r_1x^{n-2}+\dots+r_{n-1}=0 $$ but I can't see why that would lead to a contradiction. Help please.
This leads to a contradiction since you found a monic polynomial $g(t)=t^{n-1}+r_1t^{n-2}+\dots+r_{n-1}$ of degree $n-1<n=\deg f$ such that $g(x)=0$.