Show that $\langle x^{2}+1\rangle$ is a maximal ideal of $R[x]$

Question

Let $I=\langle x^{2}+1\rangle.$

Consider A be an ideal containing $I$ properly . How can I prove that A contains some nonzero number? After that I can do it

Does $,R = \Bbb R?\ \ $ – Bill Dubuque Aug 01 '17 at 16:28 — Bill Dubuque, Aug 01 '17 at 16:28

score 3 · Answer 1 · answered Aug 01 '17 at 16:26

You're supposing $\langle x^2+1\rangle\subsetneq A\subsetneq \mathbb{R}[x]$.

Pick an $f\in A\setminus I$ and consider the remainder of $f$ divided by $x^2+1$. What form must it take? What can you do with that element and $x^2+1$? (Remeber, you're aiming to get a constant.)

lhf · Answer 2 · 2017-08-01T16:31:40.117

3

Hint for an alternative solution: Consider the ring homomorphism $\mathbb R[x] \to \mathbb C$ given by $f(x) \mapsto f(i)$. Prove that it is surjective and find its kernel.

edited Aug 01 '17 at 16:31

answered Aug 01 '17 at 16:31

lhf

216,483

score 2 · Answer 3 · answered Aug 01 '17 at 16:25

2

If it was not a maximal ideal, $(x^2+1)\subset (p)$, you would have $x^2+1=p(x)q(x)$ where $p,q$ are not constant, and their degree is $1$. This implies that $x^2+1$ has a root, contradiction.

answered Aug 01 '17 at 16:25

Tsemo Aristide

87,475

1

This doesn't answer the question asked ("how can I prove $A$ contains....") – Bill Dubuque Aug 01 '17 at 16:30

score 1 · Answer 4 · answered Aug 01 '17 at 16:41

1

Hint $\ p= x^2\!+1\,$ is irreducible so $\ f\not\in (p)\,\Rightarrow\, p\nmid f\,\Rightarrow\, 1 = \gcd(f,p) = a f + b p\in (f,p) $

answered Aug 01 '17 at 16:41

Bill Dubuque

272,048

I presume that $,R = \Bbb R,$ (or any field where $x^2+1$ is irreducible). Otherwise it has a proper factor $f$ so $(x^2+1)\subsetneq (f)\subsetneq (1)\ $ so it is not maximal. – Bill Dubuque Aug 01 '17 at 16:53

Show that $\langle x^{2}+1\rangle$ is a maximal ideal of $R[x]$

4 Answers4

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