Understanding the equivalence classes for $\Bbb R[x]/\langle x^2+1\rangle$

Question

I was reading my class notes on equivalence classes and came across the following:

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$(\Bbb R[x],+,.)$ is a commutative ring with identity which implies:

$(\Bbb{R}[x],+)$ is an Abelian group

$(\Bbb{R}[x],.)$ is a monoid.

Now, consider the polynomial $x^2+1 \in \Bbb R[x]$.

$x^2+1$ has no root in the field $\Bbb R$.

Let us define a relation $\rho$ on $\Bbb R[x]$ as follows:

$f(x)\rho g(x)$ iff $(x^2-1)|(f(x)-g(x))$

Exercise: Prove that $\rho$ is an equivalence relation on $\Bbb{R}[x]$ Then $\rho$ partitions the set $\Bbb {R}[x]$ into mutually disjoint equivalence classes.

General form of representation of equivalence classes:

$2x^2+4x+5 \in \Bbb{R}[x]$

$f(x) = (x^2+1)q(x)+ Ax + B$

For different choices of $A,B$:

$$2x^2+4x+5 = 2(x^2+1) + (4x+3)$$

When divided by $(x^2+1)$ we get an equivalence class of remainder $(ax+b)$ i.e.

$$(ax+b)+\langle x^2+1 \rangle $$ (where $\langle x^2+1 \rangle$ represents any multiple of $(x^2+1)$)

$ax+b$ and $a'x + b'$ should be of same class if $a-a'$ and $b-b'$ are divisible by $(x^2+1)$ but we can't get such $a',b'$.

$\therefore$ We don't get such things.

Let $\langle x^2+1\rangle = I$

If you add $2x + 3 + I$ and $4x+5+I$ you get $6x + 8 + I$

Similarly, if you add (modulo $4$) $7 + 4\Bbb Z$ and $9+4\Bbb Z$ you get $0+4\Bbb{Z}$.

Examples:

$(7+4z)(9+4z) = 63 + 4z \equiv 3 + 4z$

$(2x+3+I)(4x+5+I)= 8x^2 + 22x + 15 + I \equiv (22x + 7) + I$

Thus, $(22x+7) \in \Bbb R[x]/I$

This is a field under $+$ and $.$

All non-zero elements have an inverse and it is isomorphic to the complex field $\Bbb C$.

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Now, there are a few things here which I don't understand.

Questions:

1. What do they mean by:

   $ax + b$ and $a'x + b'$ should be of same class if $a-a'$ and $b-b'$ are divisible by $(x^2+1)$ but we can't get such $a',b'$?

   I don't understand what they mean by "we can't get" in this context. Also, I don't understand what they mean by "if $a-a'$ and $b-b'$ are divisible by
   $x^2+1$". Firstly, how can numbers ($a-a'$ and $b-b'$) be divisible by a polynomial expression in the first place? Secondly, if I understand this correctly whenever a polynomial with real coefficients is divided by $(x^2+1)$ the remainder will be of the form $\alpha x + \beta$ such that $\alpha,\beta \in \Bbb R$. That is there will be infinite different possible equivalence classes, since $\alpha$ and $\beta$ can acquire any value from the infinite number of reals.

2. They are saying that $\Bbb R[x]/\langle x^2+1\rangle$ is isomorphic to the complex field $\Bbb C$. I can't see how! For being a field it is necessary that an inverse has to exist for every non-zero element under the multiplication operation. But say the inverse of $2x+3$ will be $\frac{1} {2x+3}$ which is surely not in $\Bbb R[x]/\langle x^2+1\rangle$.

3. In the exercise given in the notes:"Prove that $\rho$ is...", the relation clearly doesn't look symmetric to me. If $(x^2-1)$ divides $(f(x)-g(x))$ how can we say that $f(x)-g(x)$ also divides $(x^2-1)$? Doesn't look right to me.

Answer 1

2)

$(2x+3)(2x-3)-4(x^2+1)=4x^2-9-4x^2-4=-13$

So in $\mathbb{R}[x]/(x^2+1)$

$$ \frac{1}{2x+3}=\frac{2x-3}{-13} $$

3) To see symmetry of $\rho$ you need to check that if $f \rho g$ then $g \rho f$

Suppose $f \rho g$ then $x^2+1$ divides $f(x)-g(x)=(-1)(g(x)-f(x))$.

So $(x^2+1)$ also divides $g(x)-f(x)$ and thus $g \rho f$

Answer 2

For (1): Here, the author is trying to show that every element of $\Bbb R[x]/(x^2+1)$ can be uniquely represented in the form $ax+b$. Existence of such a representation follows from polynomial long division. To prove uniqueness, they suppose you are given two polynomials $ax+b$ and $a'x+b'$ which are equal modulo $x^2+1$. This by definition means that $$(a-a')x+(b-b')=f(x)(x^2+1)\tag{$*$}$$ for some $f(x).$ I am not quite sure how they conclude $(x^2+1)$ divides $(a-a')$, but here is a different way to proceed. If $a\neq a'$ or $b\neq b'$, then the left side of $(*)$ would be a nonzero polynomial of degree $0$ or $1$, while the right hand side would be either the zero polynomial, or a polynomial of degree at least two. Either way, the degrees do not match up, so this is impossible, and we conclude $a=a'$ and $b=b'$.

By the way, it does make sense for $x^2+1$ to divide $a-a'$, since $a-a'$ is a number, which is a constant polynomial. In particular, $x^2+1$ divides $a-a'$ if and only if $a-a'=0$.