Let $R$ be a ring with unity. Let's define skew polynomial over it. Let $\alpha \in Aut(R)$, then $R[x, \alpha]$ is a polynomial ring with standard summation and multiplication: $\sum a_i x^i * \sum b_j x^j =$ $\sum a_i \alpha ^{i}(b_j) x^{i+j}$. Let's consider $\mathbb{C}[x,$ $\alpha]$, where $\alpha: z \mapsto \bar{z}, z \in \mathbb{C}$. Prove that $Z(\mathbb{C}[x, \alpha]) \cong \mathbb{R}[x^2]$, where Z(R) means center of ring R. Prove that $\mathbb{C}[x, \alpha]/(x^2+1)$ $\cong \mathbb{H}$ ($\mathbb{H}$ is quaternions field).
I've somehow dealt with first isomorphism. Let $a_i = x+iy, b_i = s+it$. Than we have that $(x+iy)(s+(-1)^{j}it) = (s+it)(x+(-1)^{k}iy)$. So if $j$ and $k$ are even than this equation holds for every polynomial. This means that all $x^k$ in such polynomials have even k, so it is exactly $\mathbb{R}[x^2]$. But what should I do with second isomorphism? I only found out that $(x-i)(x+i) = x-2ix-1, (x+i)(x-i)=x^2+2ix-1$, so $i$ isn't a root of $x^2-1$. How to prove that it's irreducible? As i understood i have to prove that this is division ring and has dimension = 4. Then using Frobenius theorem i have isomorphism.
It would be very straightforward to argue this way: consider the set $\{1, i, x, ix\}\in \mathbb C[x;\alpha]/(x^2+1)$. Clearly they are $\mathbb R$-linearly independent. For if they weren't, they would have a linear combination lying in $(x^2+1)$, but obviously everything in that ideal has degree $2$ or more, and the linear combination only has degree $1$.
Furthermore it's also clear they generate $\mathbb C[x;\alpha]/(x^2+1)$ as an $\mathbb R$-module. (They span all polynomials of degree less than $2$.)
Notice that with $x=j$ and $ix=k$, the elements $i,j,k$ satisfy all the defining relations they are supposed to for the quaternion algebra. That's enough to identify it as $\mathbb H$.