More clearly the question is:
Prove that R is a field by showing that $\mathbb{R}$ forms an isomorphisim to $\mathbb{C}$ as rings (explicitly find an isomorphism between them and prove it is an isomorphism).
Let $$R=\left\{\begin{bmatrix}a&b\\-b&a\end{bmatrix}:a,b\in\mathbb{R}\right\}\subseteq M_2(\mathbb{R}).$$
So I need to show that this is a bijective homomorphisim. The definition that is given is:
Definition 4.1.7: A homomorphism $\mathbb{\phi}$: → ′ is injective if $\mathbb{\phi}$(1) = $\mathbb{\phi}$(2) implies 1 = 2 for any 1,2 ∈ . It is surjective if for every ′ ∈ ′, there is a ∈ such that $\mathbb{\phi}$() = ′. It is bijective if it is both surjective and injective.
Any hints would be greatly appreciated. To be honest I am not even sure what the question is asking. If someone could break this down it would be great. How do I show all the properties of a homomorphisism from the matrix of reals to the matrix of the complex (if that is even what is being asked)? I am even more confused on showing it is injective. This is my first time encountering material like this before. It is very rough.
For rings $R_1,R_2$, what it means for $R_1$ to be isomorphic to $R_2$ is that they are "essentially the same" with respect to ring structure, and an isomorphism is the way by which you identify the elements of the two rings.
For instance, say we consider a subset $R\subseteq\mathbb R^2$ given by $R=\{(x,x)\mid x\in\mathbb R\}$ and we define ring structure on $R$ by $(x,x)+(y,y)=(x+y,x+y)$ and $(x,x)(y,y)=(xy,xy)$. Now, clearly, this ring is pretty silly: it is just the same as $\mathbb R$, where we have just written each number twice next to itself. The ring $R$ is hence "essentially the same as" $\mathbb R$, with the element $(x,x)\in R$ being "the same as" $x\in\mathbb R$, so $R$ is isomorphic to $\mathbb R$ with the isomorphism $\phi:R\to\mathbb R, (x,x)\mapsto x$.
What you are supposed to show should be that the ring $$M=\left\{\begin{bmatrix}a&b\\ -b&a\end{bmatrix}\mid a,b\in\mathbb R\right\}$$ under matrix addition and multiplication is isomorphic to $\mathbb C$. So we need to find a function $\phi$ such that $\phi(M_1M_2)=\phi(M_1)\phi(M_2)$ and $\phi(M_1+M_2)=\phi(M_1)+\phi(M_2)$ for all $M_1,M_2\in M$. To find a candidate function $\phi$, first multiply out $M_1M_2$ and try to look for a possible choice of $\phi$ that satisfies the first condition at least. Then, check whether or not it satisfies the second condition. If it does, you're nearly done: you need to check that $\phi$ is a bijection (i.e. that the elements of $M$ is in 1-1 correspondence via $\phi$ to the elements of $\mathbb C$) and that $\phi(1_M)=1_{\mathbb C}$ (i.e. that the identity elements of both rings are identified with each other). Once you do that, you're done with the problem.