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Why $\frac{R[x]}{<x^2-1>}$ is not a field, but if $Q[x]$ is there instead of $R[x]$ it is. How to check $<x^2-1>$ is maximal ideal or not in the easiest way possible?

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    $\begingroup$ The correct formulation of this problem uses something like $x^2-2$, not $x^2-1$. $\endgroup$ – GEdgar Jun 29 at 11:32
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Actually, none of them is a field, since in both cases, $[x+1],[x-1]\neq0$, but$$[x+1]\times[x-1]=[x^2-1]=0.$$

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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.

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