Let $\mathbb{F}$ be a field and $R=\mathbb{F}[x]$, the polynomial ring over $\mathbb{F}$. Is the ideal $(x^2-1)$ maximal in $R$? Let $\mathbb{F}$ be a field and $R=\mathbb{F}[x]$, the polynomial ring over $\mathbb{F}$. Is the ideal $(x^2-1)$ maximal in $R$? Does the answer depend upon $\mathbb{F}$?
I think of this isomorphism $\mathbb{F}[x]/(x^2-1) \cong \mathbb{F}[i]$ where $\mathbb{F}[i]=\lbrace a+bi:a,b \in \mathbb{F} \rbrace$. Since $\mathbb{F}[i]$ is a field (which I am not quite sure about it), $(x^2-1)$ is maximal. 
Can anyone explain to me whether $\mathbb{F}[i]$ is a field or not.
 A: Hint:
$$x^2-1=(x-1)(x+1)\implies x-1+\langle x^2-1\rangle \;\;\text{is a divisor of zero in}\;\;\Bbb F[x]/\langle x^2-1\rangle\ldots$$
A: Hint: If so, the quotient structure is a field. But the quotient structure has zero divisors. 
We can also solve the problem "from the definition," without quoting the above result. Prove that $(x-1)$ is an ideal that properly contains $(x^2-1)$. 
A: If $p(x)$ is irreducible over a field $F$, then the ideal generated by $p(x)$ is maximal. If $p(x)$ reducible, the ideal  is not maximal. Furthermore, if $F$ is a commutative ring with unity, $F/J$ is a field if and only if $J$ is a maximal ideal. So to prove $F[x]/(x^2 - 1)$ is not a field, show that $x^2 -1$ is reducible over $F$.
A: Whilst $(x^2 - 1)$ is never maximal, the same is not true of $(x^2 + 1)$:
If $\mathbb{F} = \mathbb{C}$ then $(x + i)(x - i) = (x^2 + 1)$ so $\mathbb{C}[x]/(x^2 + 1)$ is not even an integral domain.
However, if $\mathbb{F} = \mathbb{R}$ then the map $f: \mathbb{R}[x] \rightarrow \mathbb{C}$ given by $f(x) = i$, $f(r) = r$ for $r \in \mathbb{R}$ is a surjective ring homomorphism with kernel $(x^2 + 1)$, and so $\mathbb{R}[x]/(x^2 + 1)$ is a field, so $(x^2 + 1)$ is a maximal ideal.
(In general, $(x^2 + 1)$ is maximal in $\mathbb{F}[x]$ iff there does not exist solutions to $x^2 + 1$ in $\mathbb{F}$.)
I'm posting this because the fact that the OP talks about $\mathbb{F}[x]/(x^2-1) \cong \mathbb{F}[i]$ suggests he might have meant $(x^2+1)$ in the first place.
A: Hint $\rm\,\ (x^2\!-\!1)\:$ is not maximal since  $\rm (1)\supsetneq (x\!-\!1)\supsetneq (x^2\!-\!1)\ $ by $\rm\ 1\mid x\!-\!1\mid x^2\!-\!1\:$ all properly.
Remark $\ $ For principal ideals: $ $ contains $\equiv$ divides, i.e. $\rm\: (a)\supseteq (b)\!\iff\! a\mid b.\:$ Thus, having no proper containing ideal (maximal) is equivalent to having no proper divisor (irreducible). 
A: All you need to check is whether $\mathbb F[x]/\langle x^2-1\rangle$ is a field. Note that, 
$$x+1,x-1\notin \langle x^2-1\rangle$$
$($$\mathbb F$ is an integral domain $\implies$ $\deg f(x)g(x)=\deg f(x)$$+\deg g(x)$$~\forall~f(x),g(x)\in\mathbb F[x]-\{0\}).$ 
Consequently $(x+1)+\langle x^2-1\rangle,(x-1)+\langle x^2-1\rangle$ are non-zero elements of $\mathbb F[x]/\langle x^2-1\rangle.$ 
But $((x+1)+\langle x^2-1\rangle)((x-1)+\langle x^2-1\rangle)=(x^2-1)+\langle x^2-1\rangle=\langle x^2-1\rangle$ implies that $\mathbb F[x]$ has divisors of zero.
Thus $\langle x^2-1\rangle$ is not maximal in $R.$
For your second question you should simply note that we've just used the property of an integral domain.
