Clarifying operations identifying $\mathbb{R}[x]/\left$ with $\mathbb{C}$ On page 281 of "A First Course in Abstract Algebra" by John B. Fraleigh, we have:

Let $F = \mathbb{R}$, and let $f(x) = x^2 + 1$, which is well known to have no zeros in $\mathbb{R}$, and thus is irreducible over $\mathbb{R}$. Then $\left<x^2 + 1\right>$ is a maximal ideal in $\mathbb{R}[x]$, so $\mathbb{R}[x]/\left<x^2 + 1\right>$ is a field. Identifying $r\in\mathbb{R}$ with $r + \left<x^2 + 1\right>$, we can view $\mathbb{R}$ as a subfield of $E = \mathbb{R}[x]/\left<x^2 + 1\right>$.

I understand the bit about generating a field from the quotient of R over the maximal ideal. But my question is: why does identifying "$r\in\mathbb{R}$" as described above allow us to view $\mathbb{R}$ as a subfield of $E$? 
The text goes on to say:

Let $\alpha = x + \left<x^2 + 1\right>$. 
  Computing in $\mathbb{R}[x]/\left<x^2 + 1\right>$, we find
  $$
  \alpha^2 + 1 = (x + \left<x^2 + 1\right>)^2 + (1 + \left<x^2 + 1\right>)
  = (x^2 + 1) + (x^2 + 1) = 0.
$$
  Thus $\alpha$ is a zero of $x^2 + 1$.

I don't understand how we arrive at: $(x + \left<x^2 + 1\right>)^2 + (1 + \left<x^2 + 1\right>)$. It seems to me that $\alpha^2 + 1$ should simply be equal to the first term: $(x + \left<x^2 + 1\right>)^2 + 1$. I don't understand why we add the ideal again.
I also don't understand our choice of $\alpha$. There is something about factoring  polynomials that I am missing.

It should be $\overline{1}$ in ${\bf{R}}[x]/\left<x^{2}+1\right>$, so $\overline{1}=1+\left<x^{2}+1\right>$.
