Show that $R\left [ x \right ]/\left \langle x^{2}+1 \right \rangle$ is a field.

Recall: Theorem: Let R be a commutative ring R with unity. Let I be a proper Ideal of a ring R. Then, R/I is a field IFF I is a maximal Ideal.

Recall: Theorem: Let I be a proper Ideal of a commutative ring R. I is a Maximal Ideal If whenever B is an Ideal with $I \subseteq B\subseteq R$, then, either $a \in I$ or $b \in I$

I am aware that showing I is a maximal Ideal is a sufficient condition to proving the Quotient ring is a field. Any hints to get the ball rolling?

Thanks in advance.

  • 4
    $\begingroup$ The last “theorem” you mention is meaningless. $\endgroup$ – egreg May 29 '16 at 13:39


$\mathbf R[x]$ is a P.I.D.. In a P.I.D., the ideal generated by an irreducible element is maximal.

  • $\begingroup$ Is it possible to avoid using Principle Ideal Domain? This exercise precedes P.I.D $\endgroup$ – Mathematicing May 29 '16 at 13:45
  • $\begingroup$ Do you know the Euclidean algorithm and Bézout's theorem for polynomials? $\endgroup$ – Bernard May 29 '16 at 13:48

Suppose $I$ is an ideal of $\mathbb{R}[x]$ properly containing $\langle x^2+1\rangle$ and let $f(x)\in I\setminus\langle x^2+1\rangle$. Then $$ f(x)=q(x)(x^2+1)+ax+b $$ for some $q(x)\in\mathbb{R}[x]$, $a,b\in\mathbb{R}$. Since $f(x)\notin\langle x^2+1\rangle$, you have $ax+b\ne0$, that is, not both $a$ and $b$ are zero.

If $a=0$, then $b\ne0$ and so $I=\mathbb{R}[x]$.

Suppose $a\ne0$. Then also $x+a^{-1}b\in I$; set $a^{-1}b=c$. Then $$ (x+c)(x-c)=x^2-c^2=(x^2+1)-(1+c^2)\in I $$ and therefore $-(1+c^2)\in I$, forcing $I=\mathbb{R}[x]$.

More easily: consider the ring homomorphism $\varphi\colon\mathbb{R}[x]\to\mathbb{C}$ defined by $\varphi(x)=i$. What is the kernel of $\varphi$?


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