# If we take $R$ to be the field of real numbers. Show that $R[x]/\langle x^2 + 1\rangle$ is a field.

I know that $\langle x^2+1\rangle$ is the ideal generated by $x^2 +1$ in $R[x]$ and I know that the set of real numbers does construct a field. I also know that a field is an integral domain (commutative ring with unity that has no zero divisors) in which every non zero element has a multiplicative inverse. I am having trouble connecting all these and forming my thoughts in a way in which to show this.

• This is also known as the complex numbers $\mathbb{C}$ where $x$ is the imaginary unit $i$. – Quang Hoang Apr 5 '17 at 3:50

More generally if $R$ is a commutative ring with $1$ and $M$ is an ideal of $R$, then the quotient $R/M$ is a field if and only if $M$ is maximal.
So you have to show that $\left<x^2 + 1\right>$ is a maximal ideal of $\mathbb{R}[x]$. This amounts to showing that $x^2 + 1$ is not divisible by any polynomial of positive degree in $\mathbb{R}[x]$.
First show that the elements of $\Bbb R[X]/\langle X^2+1\rangle$ are of the form $a+bX+\langle X^2+1\rangle$; and that different $a,b$ lead to different elements. Thus let us represent the elements by the pair $(a,b)$. Next, verify that the addition in $\Bbb R[X]/\langle X^2+1\rangle$, expressed interms of such pairs is given by $(a,b)+(c,d)=(a+c,b+d)$, and that multiplication is given by $(a,b)\cdot(c,d)=(ac-bd,ad+bc)$. Finally, verify all field axioms. Actually, in the last step you need only verify the existence of multiplicative inverses, for everything else follows from the validity in $\Bbb R[X]$.