Identify a polynomial $f$ giving an isomorphism $\mathbb{R}[x]/(f)\to\mathbb{C}$ I am trying to find a polynomial $f$ that gives an isomorphism of  $\mathbb{R}[x]/(f)\to\mathbb{C}$. My guess is to find an element that looks like $i$ as $\mathbb{C}$ are all of the complex numbers. I know that an isomorphism is a bijective homomorphism so I would need to find a polynomial $f$ that is both a one-to-one and onto homomorphism. Any feedback would be appreciated.
 A: You want an isomorphism $\phi:\mathbb{R}[x]/(f)\to\mathbb{C}$, and it's important not to confuse $f$ and $\phi$. For example, you remarked that you need $f$ to be one-to-one and onto $\mathbb{C}$, but actually what you want is for $\phi$ to be one-to-one and onto $\mathbb{C}$. $f$ is a polynomial with real coefficients, and $\phi$ is a homomorphism of rings.
To better understand the difference between $f$ and $\phi$, and to help figure out what polynomial to pick for $f$, it will help to understand where the map $\phi$ comes from.
The way we will construct $\phi$ is by first finding a ring homomorphism $\Phi:\mathbb{R}[x]\to\mathbb{C}$. Once we have $\Phi$, the first isomorphism theorem will give us a homomorphism $\phi:\mathbb{R}[x]/\text{ker }\Phi\to\text{Im }\Phi$.
Note that since $\mathbb{R}[x]$ is a P.I.D. and $\text{ker }\Phi$ is an ideal of $\mathbb{R}[x]$, it follows that $\text{ker }\Phi=(f)$, for some polynomial $f\in\mathbb{R}[x]$. So if we can find a surjective ring homomorphism $\Phi:\mathbb{R}[x]\to\mathbb{C}$, then the first isomorphism theorem will give us an isomorphism $\phi:\mathbb{R}[x]/(f)\to\mathbb{C}$, where $f$ is the polynomial that generates $\ker\Phi$.
Let $\alpha\in\mathbb{C}$, and define $E_{\alpha}:\mathbb{R}[x]\to\mathbb{C}$ by letting
$$E_{\alpha}\left(a_nx^n+\ldots+a_1x+a_0\right)=a_n\alpha^n+\ldots+a_1\alpha+a_0.$$
for each polynomial $a_nx^n+\ldots+a_1x+a_0\in\mathbb{R}[x]$. $E_{\alpha}$ is the evaluation map, which sends each polynomial $p(x)\mapsto p(\alpha)$
Exercise 1: Show that $E_{\alpha}$ is a ring homomorphism.
Exercise 2: Show that if $\alpha=i$, then $E_{\alpha}$ is surjective.
So we'll let $\alpha=i$, and let $\Phi=E_{\alpha}$. This gives us an isomorphism $\phi:\mathbb{R}[x]/(f)\to\mathbb{C}$, where $(f)=\text{ker }\Phi$. It remains to figure out what $f$ must be.
Earlier it was mentioned that $f$ generates $\text{ker }\Phi$. What does it mean for a polynomial $p(x)$ to be in $\text{ker }\Phi$?
$\begin{align*}
\text{Note that }p(x)\in\text{ker }\Phi &\text{ iff }\Phi(p(x))=0 \\
&\text{ iff }E_i(p(x))=0 \\
&\text{ iff }p(i)=0 \\
&\text{ iff }i\text{ is a root of }p(x) \\
\end{align*}$
So $\text{ker }\Phi$ is the set of polynomials with $i$ as a root.
And $\text{ker }\Phi=(f)$, so there is one polynomial that divides all the polynomials that have $i$ as a root.
Here's the trick for finding $f$: pick a polynomial that has $i$ as a root, for example $x^2+1$. Since $x^2+1\in\text{ker }\Phi=(f)$, it follows that $f$ divides $x^2+1$. Hence $f$ is degree at most two. If $f$ was degree one, then $f$ wouldn't have real coefficients. So $f$ is degree two. Hence $f(x)=x^2+1$.
Our map $\phi:\mathbb{R}[x]/(x^2+1)\to\mathbb{C}$ is an isomorphism.
