Prove that $\langle x^2+1\rangle$ is a maximal ideal of $\Bbb R[x]$ I'm trying to prove that $\langle x^2+1\rangle$ is a maximal ideal of $\Bbb R[x]$, but I have trouble. Let $I=\langle x^2+1\rangle$ and $R=\Bbb R[x]$. First, $I\neq R$. Second, let $J$ be any ideal of $R$ such that $I\subseteq J$. Suppose $J\neq I$, I want to claim that $J=R$.
Let $f(x)\in J\setminus I$. By division algorithm, we know that there exists $q(x),~r(x)\in R$ such that $f(x)=(x^2+1)q(x)+r(x)$ and $0\leq \deg r(x)\leq 1$. We let $r(x)=ax+b$.
I totally don't understand the rest part of my teacher's solution. Why did he consider such weird polynomial $a^2x^2-b^2$? And does there exist other solutions in this final part?
 A: Another way to see that $\langle x^2+1\rangle$ is maximal is by noting that $\Bbb R[x]/\langle x^2+1\rangle$ is a field, namely $\Bbb C$. Even if you don't want to go that way, it at least motivates whwre the use of $-ax+b$ together with $ax+b$ is motivated from: This corresponds to the complex conjugate $\overline {ai+b}=-ai+b$ of $ai+b$, which is often useful. Basically, we want to show that $ax+b$ has a multiplicative inverse modulo $\langle x^2+1\rangle$. Knowing from $\Bbb C$ that $\frac1z=\frac{\overline z}{z\overline z}$ where $z\overline z$ is always real, we arrive at the calculation as shown when translating this all back to $\Bbb R[x]$.
A: The basic idea is to show that $J=\mathbb{R}[x]$ by producing unity in $J$. Because once unity is in the ideal $J$ then by the absorption property everything in $\mathbb{R}[x]$ will be in $J$.
So your teacher basically wanted to show the existence of $1$ in $J$. The way he/she did this is as follows:


*

*Started out with something $f(x) \in J \setminus I$. 

*Using ideal properties got $r(x) \in J$ ($\because$ both $f$ and $x^2+1$ are in $J$).

*Using the polynomial $a^2x^2-b^2$ he/she produced a non-zero constant $a^2+b^2 \in J$.

*Then by absorption he/she produced $1 \in J$.

A: The strategy here is to show that there is some scalar (i.e., some member of $\mathbb{R}\setminus \{0\}$) in your $I$ (your teacher's $J$); then, since the scalar is a unit (as $\mathbb{R}$ is a field), we show that $1 \in I$, thus $(1) = R \subset i$.
We know that the linear factor $ax + b \in I$. We want to somehow get rid of the linear part $ax$. The best way to do this is to take a multiple of $ax+b$ with no linear factor: that's where $(ax+b)(ax-b) = a^2x^2 - b^2 \in I$ comes in. Now we have a quadratic factor, but that doesn't bother us: we apply the division algorithm again, explicitly this time, and the remainder is a scalar.
A: This is hugely clear if you know that $\mathbb R[x]$ is a PID. Say $M\subset I$, where $I$ is an ideal. There exists a polynomial $d$ so $I=\langle d\rangle$. So $d$ must divide $x^2+1$; hence $d$ is a (non-zero) scalar, hence $I=\mathbb R[x]$.
A: It is obvious $x\notin M$ so, $M\neq R[x]$ now suppose $M$ is not maximal then there exists ideal $U=<a_nx^n+\cdots +a_0>$ such that $M\subset U\subset R[x]$. Because $M\subset U$ then $x^2+1 \in U$ and then $n$ should be in $\{0,1,2\}$. If $n$ is $2$ then easily you can check $U$ is equal to $M$ or $R[x]$. If $n$ is $0$ then $U=R[x]$ and if $n$ is $1$ because of $x^2+1 \in U$ you can check $x$ should be in $U$ and this means that $U=R[x]$.
