Are $\Bbb Q[x]/<x^2-1>$ and $\Bbb Q[x]$ isomorphic to any subring of $\Bbb R$?
We know that $x^2-1$ is not irreducible so the quotient ring $\Bbb Q[x]/<x^2-1>$ is not a field. But I am wondering whether it is possible to identify that by some subring of $\Bbb R.$ What is confusing about this is that we are like adjoining a square root of $1$ but we know that the square root of $1$ already exists in the field of $\Bbb Q.$ Also, how about $\Bbb Q[x]$? Is $\Bbb R$ an extension of $\Bbb Q[x]?$
First of all, $\Bbb Q[x] \cong \Bbb Q[\pi]$ (since $\pi$ is transcendental) and $\Bbb Q[\pi]$ is a subring of $\Bbb R$. So $\Bbb Q[x]$ is isomorphic to a subring of $\Bbb R$.
As for $\Bbb Q[x] / (x^2 - 1)$. Since $\Bbb R$ is zero divisor free, any subring of $\Bbb R$ will also be zero divisor free. But $\Bbb Q[x] / (x^2 - 1)$ isn't (since $(x-1)(x+1) = 0$), so is can't be isomorphic to any subring of $\Bbb R$.