Cosets of $\mathbb{R}[x] / \langle x²+1 \rangle$ I'm trying to understand this question and I've found this pdf which explains it a little better. My book has no intuition and don't even mention what it's doing, it just throw proofs at my face.
So, for this part:

In another example, the book actually computes the quotient $A/I$
  where $A = \mathbb{R}[x]$ and $I = A \cdot (x²+1)$. It uses the theorem
  above to say that $L = A/I$ is a field since $x²+1$ is irreducible
  over $K[x]$ (shouldn't it be $\mathbb{R}[x]$?). Then, it computes $L$
  as follows:
$$p(x) = q(x)\cdot (x²+1)+r(x)$$
  where $r(x) = bx+a$. It then takes $p(x) \bmod I$ which gives the following:
  $$
 \overline{p(x)}
 = \overline{q(x)\cdot (x²+1) + r(x)}
 = \overline{r(x)}
 = \overline{b\cdot x + a}
$$
  Then, the book simply says that $L = \{b\cdot x + a: a,b \in
> \mathbb{R}\}$. Why is it that $L$ is the $p(x) \bmod I$? What am I
  missing about quotient rings? Can somebody clarify it for me?

By the definition in the PDF, two polynomials are congruent mod $p(x)$ if their difference is a multiple of $p(x)$, which is reasonable. So we define:
\begin{align*}
 a(x) = b(x) \bmod p(x)
&\iff a(x)-b(x) = q(x)\cdot p(x) \\
&\implies a(x) = q(x)\cdot p(x)+b(x)
\end{align*}
for some $q(x)\in F[x]$  (our field)
I understood it and it made perfect sense. The problem is that in the book, you can see that it uses the division theorem, therefore limiting $r(x)$ to $1$ degree. Following the PDF's definition, $b(x)$ should be of any degree, rigth?
 A: Let me try to shed some light on what’s going on here. For this let us first construct the quotient ring $\mathbb{R}[x]/(x^2+1)$ and then look into how this relates to the explanations you’ve seen so far. Because I don’t know about your knowledge of ring theory and quotients I try to make everything as explicit as possible (the more knowledgable reader may excuse the lack of abstraction and elegance).
Construction of the quotient ring $\mathbb{R}[x]/(x^2+1)$
We start by introducing a equivalence relation $\sim$ on $\mathbb{R}[x]$ via
$$
 p(x) \sim q(x)
\iff \text{there exists $k(x) \in \mathbb{R}[x]$ with $p(x)-q(x) = k(x)(x^2+1)$}.
$$
It is easy to check that this is an equivalence relation. Instead of $p(x) \sim q(x)$ it is common to write
$$
 p(x) \equiv q(x) \mod x^2+1,
$$
but we will stick to $\sim$ for now. Because $\sim$ is an equivalence relation we can talk about equivalence classes. For every $p(x) \in \mathbb{R}[x]$ let $\overline{p(x)}$ denote the equivalence class of $p(x)$ with respect to $\sim$. It is also common to refer to $\overline{p(x)}$ as the residue class of $p(x)$. From the definition of $\sim$ it immediately follows that
\begin{align*}
 \overline{p(x)}
 &= \{p(x) + k(x)(x^2+1) \mid k(x) \in \mathbb{R}[x]\} \\
 &= p(x) + \{k(x)(x^2+1) \mid k(x) \in \mathbb{R}[x]\}
 = p(x) + (x^2+1),
\end{align*}
where $(x^2+1) = \{k(x) (x^2+1) \mid k(x) \in \mathbb{R}[x]\}$ is called the ideal generated by $(x^2 + 1)$. These equivalence classes are the cosets you are referring to in your question title. Let
$$
 \mathbb{R}[x]/(x^2+1) = \mathbb{R}[x]/{\sim}
$$
denote the set of equivalence clases.
We can now define a ring structure on $\mathbb{R}[x]/(x^2+1)$ via
$$
 \overline{p(x)} + \overline{q(x)}
 = \overline{p(x)+q(x)}
 \quad\text{and}\quad
 \overline{p(x)} \cdot \overline{q(x)}
 = \overline{p(x) q(x)}
$$
for all $p(x), q(x) \in \mathbb{R}[x]$. Notice that we define the operations on $\mathbb{R}[x]/(x^2+1)$ by using representants, so we need to check that they are well-defined. This is rather easy, so we will skip this part. (If you have seen the construction of quotient groups or quotient vector spaces before, this will look familiar, and the calculations are more or less the same.)
Notice that $\overline{0}$ is the neutral element with respect to the addition, that $-\overline{p(x)} = \overline{-p(x)}$ for all $p(x) \in \mathbb{R}[x]$ and that $\overline{1}$ is the neutral element with respect to the multiplication. Because the multiplication on $\mathbb{R}[x]$ is commutative the same goes for the multiplication on $\mathbb{R}[x]/(x^2+1)$.
With this we have now constructed the quotient ring $\mathbb{R}[x]/(x^2+1)$.
Choosing nice representants
We have constructed $\mathbb{R}[x]/(x^2+1)$ as a set of equivalence classes, but each equivalence class is represented by many different polynomials. For example $\overline{0} = \overline{x^2+1}$ and
$$
 \overline{x}
 = \overline{x^2+x+1}
 = \overline{x^4-x^2+x-2}.
$$
To better understand $\mathbb{R}[x]/(x^2+1)$ it is now natural to search for a nice system of representants of the equivalence classes.
This is where the division theorem comes in.

Theorem: Let $p(x), q(x) \in \mathbb{R}[x]$ with $q(x) \neq 0$. Then there exist unique $k(x), r(x) \in \mathbb{R}[x]$ such that $p(x) = k(x)q(x) + r(x)$ and $\deg(r(x)) < \deg(q(x))$.

Now take some $p(x) \in \mathbb{R}[x]$ and let $k(x), r(x) \in \mathbb{R}[x]$ with $p(x) = k(x) (x^2+1) + r(x)$ and $\deg r(x) < \deg(x^2+1) = 2$. Because $\overline{x^2+1} = 0$ it follows that
$$
 \overline{p(x)}
 = \overline{k(x) (x^2+1) + r(x)}
 = \overline{k(x)} \cdot \underbrace{\overline{x^2+1}}_{=\overline{0}} + \overline{r(x)}
 = \overline{r(x)}.
$$
So $r(x)$ is a representant of the equivalence class $p(x)$. Because $\deg(r(x)) < 2$ we can write $r(x) = b x + a$ with $b,a \in \mathbb{R}$. Therefore
$$
 \overline{p(x)}
 = \overline{r(x)}
 = \overline{bx+a}.
$$
With this we have shown that every equivalence class $z \in \mathbb{R}[x]/(x^2+1)$ can be written as $z = \overline{bx+a}$, i.e. every equivalence class can be represented by a linear polynomial.
The nice thing about this is that this representant is actually unique: Suppose that we have $b_1, b_2, a_1, a_2 \in \mathbb{R}$ with $\overline{b_1 x + a_1} = \overline{b_2 x + a_2}$. This means that $(b_1 x + a_1) \sim (b_2 x + a_2)$, so there exists some $k(x) \in \mathbb{R}[x]$ with
$$
 (b_1 - b_2)x + (a_1 - a_2)
 = (b_1 x + a_1) - (b_2 x + a_2)
 = k(x) (x^2+1).
$$
The polynomial of the left side is at most of degree $1$ and $\deg(k(x)(x^2+1)) = \deg(k(x))+2$, so this is only possible if $k(x) = 0$. But then the right hand side of the above equality is $0$, so $b_1 x + a_1 = b_2 x + a_2$. (Here we basically proved the uniqueness of $k(x)$ and $r(x)$ in the division theorem.)
Putting our above abservations together we find that for every equivalence class $z \in \mathbb{R}[x]/(x^2+1)$ there exists unique $b,a \in \mathbb{R}$ with $z = \overline{bx+a}$. Therefore
$$
 \mathbb{R}[x]/(x^2+1)
 = \{\overline{bx+a} \mid a,b \in \mathbb{R}\},
$$
and the representation of $z \in \mathbb{R}[x]/(x^2+1)$ as $z = \overline{bx+a}$ is unique (i.e. $b$ and $a$ are unique).
The quotient ring $\mathbb{R}[x]/(x^2+1)$ is a field
We have now constructed $\mathbb{R}[x]/(x^2+1)$ as a commutative ring with $1$. But so far we haven’t used that $x^2+1$ is irreducible. This is because the constructions so far does not really depend on $x^2+1$ and can be made with every polynomial. That $x^2+1$ is irreducible now comes into play to show that $\mathbb{R}[x]/(x^2+1)$ is already a field. To see how this works we will use the following theorem:

Theorem: Let $p(x),q(x) \in \mathbb{R}[x]$. Then there exist some polynomials $k(x),l(x) \in \mathbb{R}[x]$ with $\mathrm{gcd}(p(x),q(x)) = k(x)p(x)+l(x)q(x)$, where $\mathrm{gcd}(p(x),q(x))$ denotes the greatest common denominator of $p(x)$ and $q(x)$.

We can directly apply the theorem to our aid: Let $z \in \mathbb{R}[x]/(x^2+1)$  with $z \neq \overline{0}$. Then $z = \overline{bx+a}$ for unique $b,a \in \mathbb{R}$; because $z \neq \overline{0}$ we have $bx+a \neq 0$. Because $x^2+1$ is irreducible (!) and $\deg(bx+a) < 2 = \deg(x^2+1)$ it follows that $\mathrm{gcd}(bx+a,x^2+1) = 1$ (otherwise $bx+a$ would divide $x^2+1$, contrary to the irreducibility of $x^2+1$). So by the above theorem there exist $k(x),l(x) \in \mathbb{R}[x]$ with $k(x)(bx+a) + l(x)(x^2+1) = 1$. Now observe that
$$
 \overline{k(x)} \cdot z
 = \overline{k(x)} \cdot \overline{bx+a} + \underbrace{\overline{l(x)(x^2+1)}}_{=\overline{0}}
 = \overline{k(x)(bx+a)+l(x)(x^2+1)}
 = \overline{1}.
$$
So $z$ is invertible in $\mathbb{R}[x]/(x^2+1)$. As every non-zero element $z \in \mathbb{R}[x]/(x^2+1)$ is invertible it follows that $\mathbb{R}[x]/(x^2+1)$ is a field.
Epilogue
If we compare the above construction with the one given in the book and the pdf (as far as I’ve read it) it becomes pretty clear what happens: In your book the division theorem is used to represent each equivalence class $z = \mathbb{R}[x]/(x^2+1)$ as $z = \overline{bx + a}$ with unique $b,a \in \mathbb{R}$. But of course $z$ could also be represented by polynomials of higher degree, as it is an equivalence class of inifintely many polynomials (of various degrees), which is what happens in the pdf.
We can also see where we need that $x^2+1$ is irreducible: The irreducibility of $x^2+1$ is equivalent to every non-zero polynomial $p \in \mathbb{R}[x]$ with $\deg(p) < \deg(x^2+1) = 2$ having $\mathrm{gcd}(p,x^2+1) = 1$. This is what we needed to construct multiplicative inverses.
If we were to use the construction above with some reducible polynomial $q(x)$ instead of $x^2+1$ then there would be some divisor $p(x)$ of $q(x)$ with $\deg(p(x)) \geq 1$. Then every combination $k(x)p(x) + l(x)q(x)$ with $k(x),l(x) \in \mathbb{R}[x]$ is divisable by $p(x)$ as well, and thus cannot be $1$.
If on the other hand $q(x) \in \mathbb{R}[x]$ is irreducible then we can replace $x^2+1$ by $q(x)$ in the above construction and get a field $\mathbb{R}[x]/(q(x))$. The main difference is that every $z \in \mathbb{R}[x]/(q(x))$ can then be uniquely represented as $z = \overline{a_{d-1} x^{d-1} + \dotsb + a_1 x + a_0}$ where $d = \deg(q(x))$ and $a_{d-1}, \dotsc, a_0 \in \mathbb{R}$, .
I hope this helps trying to understand what’s going on here.
A: Any coset representative of $\langle x^2+1\rangle$ is represented by a polynomial $\overline{p(x)}$ and any power of $x$ in that polynomial whose degree is greater than one can be reduced to a linear by substituting $-1$ for $x^2$, because $\overline{x^2}=\overline{-1}$.  So for example $\overline{x^3}$ is the same as $\overline{-x}$.  So every polynomial of degree one gives a distinct coset.
Does that help at all?
The space itself $\Bbb R[x]/\langle x^2+1\rangle$ is actually isomorphic to $\Bbb C$, because it's basically adding an element to $\Bbb R$ whose square is $-1$.
