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
 A: 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.
A: First it is worth to look into the elements of the field  $\mathbb R[x]/\langle x^2+1\rangle$.
Since $\mathbb R$ is a field $\mathbb R[x]$ is an Euclidean domain so the division algorithm holds.
hence any element of $\mathbb R[x]/\langle x^2+1\rangle$ is of the form $a+b\alpha $ where $\alpha $ satisfies $\alpha ^2+1=0$ and $a,b\in \mathbb R$.
Now consider the homomorphism $\phi:\mathbb R[x]/\langle x^2+1\rangle\to \mathbb C$ by $\phi(a+b\alpha)=a+bi$
A: The two answers so far have explained why $\ker f=0$ implies that the map is bijective. For the proof given, one also needs to check that the map is well-defined since elements of $\mathbb{R}[x]/\langle x^2+1\rangle$ are cosets (so representatives are not unique).
A better proof is as follows: 
Consider the map $\phi:\mathbb{R}[x]\to\mathbb{C}$ given by $\phi(g)=g(i)$. It is straightforward to check that $\phi$ is a surjective ring homomorphism. By the Fundamental Homomorphism Theorem for rings, there is an isomorphism
$$\overline{\phi}:\mathbb{R}[x]/\ker\phi\to\mathbb{C}.$$
It is now left to show that $\ker\phi=\langle x^2+1\rangle$. Well, since $i^2+1=0$ is follows that $\langle x^2+1\rangle\subset\ker\phi$. On the other hand, since $x^2+1$ is irreducible in $\mathbb{R}[x]$ it follows that $\langle x^2+1\rangle$ is a maximal ideal and $\ker\phi=\langle x^2+1\rangle$ as required.
A: If $f(g_1) = f(g_2)$ then, since $f$ is a homomorphism of rings, $f(g_1-g_2)=0$, so that $g_1-g_2\in\ker f$.  If the only member of $\ker f$ is $0$, then $g_1-g_2=0$ so $g_1=g_2$.
