How do you prove the injective nature of an isomorphism from $\mathbb{R}[x]/\langle x^2+1\rangle$ and $\mathbb{C}$ I've been reviewing my notes for a course I'm taking and I am confused about my professor shows that $\mathbb{F}=\mathbb{R}[x]/\langle x^2+1\rangle$ is isomorphic to $\mathbb{C}$. I understand most of it but when it comes time to prove that the homomorphism is a bijection I get confused. Essentially, what he does is he shows that $f:\mathbb{F} \to \mathbb{C}: f(g)=g(i),g\in \mathbb{F}$ is a homomorphism and then shows the injective nature by showing that $\ker(f)={1_\mathbb{F} }$ and then he shows the bijective nature. 
I dont understand why this is a valid way of showing the injective nature of $f$. Could someone please explain it to me? Also, if you dont mind, would you be able to walk me through the logic of this proof?
Thanks

It's a famous theorem that $\varphi$ is a monomorphism/injective, if and only if, the kernel of $\varphi$ is trivial, i.e $\ker\varphi = \{0\}$. I'll prove it here for you. Which means however you or he must have typoed when writing $1$ as you can never have $\varphi(1)=0$, unit goes to unit.
Assume $\varphi$ is injective, then $\varphi(a)=0=\varphi(b)$ has that $a=b$ and the only element must be $0$ because any homomorphism take $0\to 0$.
Assume that $\ker\varphi=0$, then let's assume that $\varphi(a)=\varphi(b)$, this gives us that
$$0=\varphi(a)-\varphi(b)=\varphi(a-b)$$
This means that $a-b=0$ and in turn $a=b$ hence it is a monomorphism.
