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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?enter image description here

Eric
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  • I think your prof is relying on advanced information: $\mathbb{R}[x]/\langle x^2+1\rangle$ is isomorphic to $\mathbb{C}$. Multiplying $ax+b$ and $ax-b$ is multiplying a complex number with its complex conjugate, to get a real. – ziggurism Jan 12 '18 at 14:18
  • Looks like you wrote down all the symbols but none of the language, but I realize this may be your teacher's fault (if they are in the habit of only writing symbols while perhaps narrating details.) In mathematical writing, it is really imperative (for students and teachers) to include language in proofs and notes. Using only bare symbols leads to precisely the problem you find yourself in. – rschwieb Jan 12 '18 at 14:23
  • @ziggurism It looks rather like a brute-force proof that takes great pains to avoid the quotient. (Although most of us would probably prefer to prove it using the quotient.) – rschwieb Jan 12 '18 at 14:24
  • @rschwieb Yes, I agree.. – Eric Jan 12 '18 at 14:48

5 Answers5

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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:

  1. Started out with something $f(x) \in J \setminus I$.
  2. Using ideal properties got $r(x) \in J$ ($\because$ both $f$ and $x^2+1$ are in $J$).
  3. Using the polynomial $a^2x^2-b^2$ he/she produced a non-zero constant $a^2+b^2 \in J$.
  4. Then by absorption he/she produced $1 \in J$.
Anurag A
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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 $\Bbb R[x]$.

Hagen von Eitzen
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  • I don't know if I'd say that's what motivates using $-ax+b$. I think there's plenty motivation to use $-ax+b$ without noticing the connection to $\mathbb{C}$. Regardless, it's a good connection to notice. – BallBoy Jan 12 '18 at 14:47
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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.

BallBoy
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  • Thanks! This answers the all unnatural steps in the solution. Also thanks to the other people that have pointed out the idea of using the isomorphism or PID. I think I'll use these solution in the practice, since they seems more elegant and straightforward. – Eric Jan 12 '18 at 14:45
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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 $I=\mathbb R[x]$.

David C. Ullrich
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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 $U=R[x]$.

GhD
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