Showing that $\Bbb R[x] / \langle x^2 + 1 \rangle$ is isomorphic to $\Bbb C$ question 
Show that $\Bbb R[x] / \langle x^2 + 1 \rangle$ is isomorphic to $\Bbb C$.
Let $\phi$ be the homomorphism from $\Bbb R[x]$ onto $\Bbb C$ given by
  $f(x) \rightarrow f(i)$ (that is, evaluate a polynomial in $\Bbb R[x]$
  at $i$).  Then $x^2 + 1 \in \operatorname{Ker} \phi$ and is clearly a
  polynomial of minimum degree in $\operatorname{Ker} \phi$.  Thus,
  $\operatorname{Ker} \phi = \langle x^2 + 1 \rangle$ and $\Bbb R[x] /\langle x^2 + 1 \rangle$ is isomorphic to $\Bbb C$.

My question is:  How is $x^2 + 1$ a polynomial of minimum degree in $\operatorname{Ker} \phi$?

$⟨x^2+1⟩$ is an irreducible polynomial over $\mathbb R$, then it is a maximal ideal.Because of that $\mathbb R[x]/⟨x2+1⟩$ is a field. All elements in $⟨x^2+1⟩$ given by $a+bx$. Now we could easily define an isomorphism between $\mathbb C$ and  $\mathbb R[x]/⟨x^2+1⟩$.
