The elements and operations of the field $C = \Bbb R[x] / \langle x^2 + 1 \rangle$ 
$$C = \Bbb R[x] / \langle x^2 + 1 \rangle = \{[a + b x_{x^2 + 1}]\}$$

I know $C$ is a field since it has complex roots $(x+i)(x-i)$ and is irreducible over the reals, also since deg is $2$. 
How would I find elements of this field? To find the sum and products of two elements: $[a + b x_{x^2 + 1}]$ and $[c + d x_{x^2 + 1}]$.
 A: Any element of $\Bbb R[x]$ has the form $$p(x) + \langle x^2 + 1 \rangle,$$
and by polynomial long division we can find a unique representative of this element of the form
$$a + b x$$
Since any two representatives of a given element differ by a polynomial of the form $(x^2 + 1) q(x)$, which if $q \neq 0$ has degree $\geq 2$, all of the elements represented by the linear polynomials $a + b x$ are distinct. Thus the elements (without duplication) are precisely $a + b x + \langle x^2 + 1 \rangle$, $a, b \in \Bbb R$. This field turns out to be the complex numbers, $\Bbb C$, and for convenience, we usually write the class $x + \langle x^2 + 1 \rangle$ as $i$, and hence write the elements as
$$a + b i .$$
If we add two elements, we get
$$(a + b i) + (c + d i) = (a + c) + (b + d) i .$$
If we multiply two elements, we get
$(a + b i) (c + d i) = (a c) + (a d + b c) i + (b d) i^2.$
Now, $i^2 = (x + \langle x^2 + 1 \rangle)^2 = x^2 + \langle x^2 + 1 \rangle$, which has linear representative $$x^2 - (x^2 + 1) = -1,$$ so that $i^2 = -1$ (by construction we identify $i$ with the root of the generating polynomial $x^2 + 1$ in the splitting field $C$). Thus,
$$(a + b i) (c + d i) = (a c - b d) + (a d + b c) i .$$
