$(x^2+1)\mathbb{R}[x]$ is a maximal ideal in $\mathbb{R}[x]$

Question

Prove that ideal $M:=(x^2+1)\mathbb{R}[x]$ is a maximal ideal in the ring $\mathbb{R}[x]$. Which field is isomorphic to $\mathbb{R}/M$?

Please, help me to solve this problem. I have an exam tomorrow and I have no idea what's going on in this task. Please, help!

Hint: this is a field extension of $\mathbb R$ of degree 2. – Ian Coley Feb 17 '15 at 20:27 — Ian Coley, Feb 17 '15 at 20:27
You should use "field" instead of "body" (as you know). – Dietrich Burde Feb 17 '15 at 20:35 — Dietrich Burde, Feb 17 '15 at 20:35
Ok. In Polish is body :) – Kasia Pisaniak Feb 17 '15 at 20:36 — Kasia Pisaniak, Feb 17 '15 at 20:36
exactly in Polish "field" - "body" - "ciało" :) – Kasia Pisaniak Feb 17 '15 at 20:43 — Kasia Pisaniak, Feb 17 '15 at 20:43
Probably Polish took the term from the French corps. – egreg Feb 17 '15 at 21:58 — egreg, Feb 17 '15 at 21:58

score 3 · Answer 1 · answered Feb 17 '15 at 20:28

3

Since $\mathbb{R}[x]/(x^2+1)\cong \mathbb{C}$ is a field, the ideal ${\frak{m}}=(x^2+1)$ is maximal.

answered Feb 17 '15 at 20:28

Dietrich Burde

140,055

What theorem apply? – Kasia Pisaniak Feb 17 '15 at 20:32
This theorem here. – Dietrich Burde Feb 17 '15 at 20:33

score 1 · Accepted Answer · answered Feb 17 '15 at 21:01

Let us consider the application$$\phi:\mathbb{R}\left[X\right]\ni P\longmapsto P\left(i\right)\in\mathbb{C}.$$ I let you check that this is a surjective homomorphism.

Now we want to determine its kernel. We have $\phi\left(X^{2}+1\right)=i^{2}+1=0$ whence $$\left(X^{2}+1\right)\mathbb{R}\left[X\right]\subset\ker\phi.$$ The difficult part is to show the converse. If $Q\in\ker\phi$, let us consider the euclidian division of $Q$ by $X^{2}+1$ : we write$$Q=\left(X^{2}+1\right)R+S$$ with $Q,S\in\mathbb{R}\left[X\right]$ and with $\deg S\leq\deg\left(X^{2}+1\right)-1=1$. Hence, $S=aX+b$ with $a,b\in\mathbb{R}$. Now, since $Q\in\ker\phi$ and since $\phi$ is a homomorphism, we have$$0=\phi\left(Q\right)=\phi\left(\left(X^{2}+1\right)R+S\right)=\phi\left(X^{2}+1\right)\phi\left(R\right)+\phi\left(S\right)=\phi\left(S\right)=ai+b$$ and thus $a=b=0$, i.e. $S=0$ and $Q\in\left(X^{2}+1\right)\mathbb{R}\left[X\right].$

By the factorization theorem, we get$$\mathbb{R}\left[X\right]/\left(X^{2}+1\right)\mathbb{R}\left[X\right]\simeq\mathbb{C}.$$ The fact that the ideal $\left(X^{2}+1\right)\mathbb{R}\left[X\right]$ is maximal comes form the fact that $\mathbb{C}$ is a field.

score 0 · Answer 3 · edited Feb 17 '15 at 20:41

0

$x^2+1$ is irreducible in $\mathbb{R}[X]$.

edited Feb 17 '15 at 20:41

Bernard

179,256

answered Feb 17 '15 at 20:40

marwalix

17,045

It proves only it's a prime ideal. – Bernard Feb 17 '15 at 20:42
1

But in a PID, all nonzero prime ideals are maximal. – lokodiz Feb 17 '15 at 20:45
@Bernard In a PID every prime ideal $\ne 0$ is maximal. Being Euclidean, $,F[x],$ is a PID. But I agree it should be explicitly mentioned in any proof. – Bill Dubuque Feb 17 '15 at 20:47
I think it has to be mentioned. – Bernard Feb 17 '15 at 20:48
What is a PID? I don't know. – Kasia Pisaniak Feb 17 '15 at 21:04
Principal ideal domain; every ideal is of the form $aR$ for some $a \in R$. – lokodiz Feb 17 '15 at 21:27

$(x^2+1)\mathbb{R}[x]$ is a maximal ideal in $\mathbb{R}[x]$

3 Answers3