Problem
Prove that the ideal <$x^2+1$> is maximal in $R[x]$
Attempt
Assume $A$ is an ideal of $R[x]$.
My attempt is to show that $A$ contains some non-zero real number $c$. Then $(1/c) c \in A$ and therefore A=R[x].
Let $f(x) \in A $ ,but $f(x) \notin $ <$x^2+1$>. Then $f(x)=q(x)(x^2+1)+r(x)$, where $r(x) \neq 0$ and degree of $r(x)$ is less than 2. It follows that $r(x)=ax+b$,where a and b are not both zero .
$ax+b= r(x)-q(x)(x^2+1) \in A$.
If $(ax-b)(ax+b) \in A$ ,then
$a^2x^2-b^2 \in A$ and $a^2x^2+x^2 \in A$. From these two we get $0\neq a^2+b^2 \in A$
Doubt
Are there any shorter proof for this? I am trying to attempt another proof by using this theorem .
Theorem
Let $R$ be a commutative ring with unity and let $A$ be an ideal of $R$. Then $R/A$ is a field if and only if $A$ is maximal