Which one is Field

Why $$\frac{R[x]}{}$$ is not a field, but if $$Q[x]$$ is there instead of $$R[x]$$ it is. How to check $$$$ is maximal ideal or not in the easiest way possible?

• The correct formulation of this problem uses something like $x^2-2$, not $x^2-1$. – GEdgar Jun 29 at 11:32

Actually, none of them is a field, since in both cases, $$[x+1],[x-1]\neq0$$, but$$[x+1]\times[x-1]=[x^2-1]=0.$$
In both cases $$x^2-1$$ factorizes as $$(x-1)(x+1)$$, so the ideal it generates can't be maximal both if you are on $$\mathbb{Q}$$ or $$\mathbb{R}$$.
Notice that in both cases the class of $$x-1$$ is a zero-divisor, so the quotient can't be a field.
Maybe you are thinking about the ideal generated by $$x^2+1$$ which is maximal on the real but not on the complex.