# 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 $\operatorname{Ker} \phi$?

Clearly $x^2+1$ is in the kernel as $i^2 +1= 0$. Now, a non-zero real polynomial of degree at most $1$ is not $0$ at $i$ as it is $ai+b$ with real $a,b$ at least one non-zero. Thus the kernel cannot contain any non-zero polynomials of degree at most $1$ and the polynomial $x^2+1$ of degree $2$ is thus of minimal degree.

(That is, if we exclude the $0$ polynomial, which is likely intended. Otherwise that one is of minimal degree, but that's boring.)

Hint: Can $i$ be a root of a linear polynomial in $\mathbb R[x]$ ?

$f(x)=x^2+1$ has no zeros in $\Bbb{R}$, and so is irreducible over $\Bbb{R}$. Hence $\langle x^2 + 1 \rangle$ is a maximal ideal in $\Bbb{R}[x]$, which means $\Bbb{R}[x]/\langle x^2 + 1 \rangle$ is a field, being an extension of $\Bbb{R}$ of degree $2$ in which $x^2+1$ has a root.

Let $\alpha=x+\langle x^2 + 1 \rangle$, then $\Bbb{R}(\alpha)\cong\Bbb{R}[x]/\langle x^2 + 1 \rangle$ with elements $a+b\alpha$, for $a,b\in\Bbb{R}$.

Now in $\Bbb{R}[x]/\langle x^2 + 1 \rangle$ we have: \begin{align} \alpha^2+1&=(x+\langle x^2 + 1 \rangle)^2+(1+\langle x^2 + 1 \rangle)\\ &=(x^2+1)+\langle x^2 + 1 \rangle=0 \end{align} and $\alpha$ is a zero of $x^2+1$. It follows $a+b\alpha\in\Bbb{R}(\alpha)$ is equivalent to $a+bi\in\Bbb{C}$ implying $\Bbb{R}(\alpha)\cong \Bbb{C}$ a required.

$⟨x^2+1⟩$ 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 $\mathbb R[x]/⟨x^2+1⟩$.