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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.

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

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If you do not have enough theory at your disposal, you still could verify the claim "manually":

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 $\Bbb R[X]$.

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