I'm currently reading through Gallian, and I'm struggling on Example 16 in Chapter 14 (Ideals and Factor Rings). It follows like this:
The ideal $\langle x^2 + 1 \rangle$ is maximal in $\mathbb R[x]$. To see this, assume that $A$ is an ideal of $\mathbb R[x]$ that properly contains $\langle x^2 + 1 \rangle$. We will prove that $A = \mathbb R[x]$ by showing that $A$ contains some nonzero real number $c$. (This is the constant polynomial $h(x) = c$ for all $x$.) Then $1 = c(1/c) \in A$ and therefore, by Exercise 19, $A = \mathbb R[x]$.
Exercise 19 was asking if $A$ is an ideal of a ring $R$ and $1 \in A$, then $A = R$, which I had no problem with proving. However, its past this that I start to get confused:
To this end, let $f(x) \in A$ but $f(x) \not\in \langle x^2 + 1 \rangle$. Then $$f(x) = q(x)(x^2 + 1) + r(x)$$ where $0 \leq \deg(r(x)) < 2$ and $r(x) \neq 0$. Then $r(x) = ax + b$ where $a, b \neq 0$, and $$ax + b + r(x) = f(x) - q(x)(x^2 + 1) \in A$$
I'm assuming that this is a typo since he seemed to just pull out that $2r(x) = f(x) - q(x)(x^2 + 1)$. I want to say that this is supposed to be $ax + b = r(x) = f(x) - q(x)(x^2 + 1)$, unless I'm missing something. But then he suddenly claims this:
$$a^2x^2 - b^2 = (ax + b)(ax - b) \in A \text{ and } a^2(x^2 + 1) \in A$$ So $$0 \neq a^2 + b^2 = (a^2x^2 + a^2) - (a^2x^2 - b^2) \in A$$
My two issues are this: why do we need to do this in the first place? How does this relate to $\langle x^2 + 1 \rangle$ being a maximal ideal? As far as definitions go, I only know that maximal ideals are ideals $A$ such that if $B$ is an ideal of some commutative ring $R$, then $B = A$ or $B = R$ (I actually sort of struggle with this definition as well, so some elucidation may be helpful).
I'm also struggling where $(ax - b)(ax + b)$ came from. Is this because $x^2 = -1$? But then we only get $a^2x^2 - b^2 = -a^2 - b^2$, and I have no idea where this fact will be used to show that $\langle x^2 + 1 \rangle$ is a maximal ideal. I appreciate the help!