Doubts about adjunction of elements to a ring I've just solved the following exercise

Suppose we adjoin an element $\alpha$ to $\mathbb{R}$ satisfying the relation $\alpha^2 = 1$. Prove that the resulting ring is isomorphic to the product ring $\mathbb R\times \mathbb R$, and find the element of $\mathbb R\times \mathbb R$ which corresponds to $\alpha$.

Here is my solution:
The aforementioned ring is $\mathbb{R} [x]/(x^2-1)=\{a_0+a_1 x:a_0,a_1 \in \mathbb{R}\}$, where addition is defined as usual and the multiplication is
$$(a_0+a_1 x)(b_0+b_1 x)=(a_0b_0+a_1b_1)+(a_0b_1+a_1b_0)x.$$
Consider the map $\varphi:\mathbb{R} \times \mathbb{R} \to \mathbb{R} [x]/(x^2-1)$ defined by $$(a,b) \mapsto \left( \frac{a+b}{2} \right)+\left( \frac{a-b}{2} \right)x.$$
One can show that it is a ring homomorphism, and it has the inverse
$$a+bx \mapsto (a+b,a-b). $$
This shows that $\mathbb{R} [x]/(x^2-1) \cong \mathbb{R} \times \mathbb{R}$, and the "new" square root of $1$ corresponds to $(1,-1) \in \mathbb{R} \times \mathbb{R}$.

Assuming my solution is correct, I have a few questions:


*

*How do we know that $\alpha$ is not a previously known (i.e. real) square root of $-1$? To elaborate, In the quotient ring $\mathbb{R}[x]/(x-2)$ (which is the construction of adjoining the reals another "2"), the relation $x-2=0$ forces $x$ to be an element of the reals, and then $\mathbb{R}[x]/(x-2) \cong \mathbb{R}$. So, how do we know that $x^2-1=0$ doesn't force $x$ to be $1$ or $-1$ (and consequently $\mathbb{R} [x]/(x^2-1) \cong \mathbb{R}$)?

*In general what exactly is this "$x$" in the polynomial ring. Is it necessarily some unfamiliar quantity? (I guess not, because in $\mathbb{R}[x]/(x-2)$ it's a well known real number).
Thank you!
 A: $x$ is a formal symbol that satisfies the relation $x^2=1$. This is the only relation that $x$ satisfies because it's the only property we've given $x$. Notably, $x$ fails to satisfy $x+1=0$ or $x-1=0$ because we didn't say that it did. Don't think of $x$ as a variable, think of it as a formal symbol that satisfies the properties that we give it, like $i$.
A: The simplest way to handle quotient rings is to avoid them.
Consider $\psi : \mathbb R [x] \to \mathbb R \times \mathbb R$ given by $\psi(f)=(f(1),f(-1))$.
Then $\psi$ is a surjective ring homomorphism whose kernel is $(x^2-1)$.
Therefore, $\psi$ induces an isomorphism $\bar\psi: A=\mathbb R [x]/(x^2-1) \to \mathbb R \times \mathbb R$.
$\alpha$ corresponds to the image of $x$ under $\psi$, which is $(1,-1)$.
$\pm 1, \pm \alpha$ are elements of $A$ that satisfy $t^2=1$.
They are all different elements of $A$ because they have different images under $\bar\psi$:
$\bar\psi(1)=(1,1), \bar\psi(-1)=(-1,-1), \bar\psi(\alpha)=(1,-1), \bar\psi(-\alpha)=(-1,1)$
A: I think that your solution is wrong right from the start, because of an unjustified expression of the quotient ring R[T]/(T^2-1) (I prefer to write T instead of x, to stress that T is a formal symbol in a polynomial ring, and not to confuse it with an element of another ring, as you do). In general, if f(T) is a polynomial of degree n in K[T], where K is any field, the quotient ring K[T]/(f(T)), considered as a vector space over K (obvious laws), is of dimension n, with basis 1,x,...,x^n-1,iff f(T) is irreducible in K[T], and is a field (good exercise!). If not, one must decompose f(T) into a product of powers of different irreducibles. If f(T) is the product of two co-prime polynomials, then the celebrated "Chinese remainder theorem"  gives an isomorphism between K[T]/(f(T)) and K[T]/(g(T)) x K[T]/(h(T)):this is the case with your f(T)=T^2-1. There is no "new" square root y of 1, because if you write y=(a,b), y^2=(1,1) means a^2=b^2=1. To elaborate, consider the more instructive example f(T)=T^2+1, which is irreducible over R. Then R[T]/(T^2+1) is isomorphic to the field C.
