How is $\mathbb{H}$ a $\mathbb{R}$-Algebra, but not a $\mathbb{C}$-algebra? It says that $\mathbb{C}$ is not in the center of $\mathbb{H}$. Definition of $\mathbb{K}$-algebra for a ring if $Z(R)=K$.
However, you can do this $(a+bI+cJ+dK)(e+fi)=(e+fi)(a+bI+cJ+dK)$. So I don't see how  $\mathbb{H}$ is not a $\mathbb{C}$-algebra.
Unless assuming I=i. Which, is weird.
 A: For a ring $R$ to be a $F$-algebra it suffices that $F\subseteq Z(R)$. For example extension fields of $F$ are surely all $F$-algebras. IOW the field $F$ does not have to be equal to the center.
Aaron's comment explains what is wrong, if we identify $i=I$. But if we don't do such an identification (or identify the complex number $i$ with any of the uncountably quaternions $u$ with the property $u^2=-1$), then how do you view $\mathbf{C}$ as a subset of $\mathbf{H}$ in the first place? Remember that for (Hamilton's) quaternions the coefficients $a,b,c,d$ in $a+bI+cJ+dK$ are all real. How do you multiply such a thing with the complex number $i$? What's your definition?
It is certainly possible to keep Hamilton's rules for multiplying the quantities $1,I,J,K$, and allow $a,b,c,d$ to range over all the complex numbers. In that way we get a 4-dimensional algebra over $\mathbf{C}$ with basis $\{1,I,J,K\}$. This is a common process known as extension of scalars. The resulting $\mathbf{C}$-algebra is then an 8-dimensional vector space over the reals. Thus it is no longer the Hamilton's quaternions. That $\mathbf{C}$-algebra can also be constructed as the tensor product $\mathbf{C}\otimes_{\mathbf{R}}\mathbf{H}$. I leave it as an exercise to prove that this algebra is isomorphic to the algebra of 2x2 complex matrices, so it is nothing to get very excited about. For example it is not a division algebra. In a more technical language this fact is often expressed with the phrase "$\mathbf{C}$ is a splitting field of $\mathbf{H}$". Look up Brauer groups for more information.
A: We wish to show that the center of $\mathbb H$ is just $\mathbb R$.  Let $z=a+bI+cJ+dK$.  If $z$ were in the center, we could subtract off the real part to get something else in the center, so we might as well assume that $a=0$ (just to simplify computations).
We can compute the commutator $[z,I]=zI-Iz$, and similarly $[z,J],[z,K]$ by first computing the commutators of $I,J,K$ pairwise with each other, and then extending by linearity.
Since these basis elements anti-commute, we have $[I,J]=2k,[J,K]=2I,[K,I]=2J$.  Therefore $[z,I]=-2cK+2dJ$.  Thus, for $z$ to be in the center we must have $c=d=0$.  Taking the commutator with either $J$ or $K$ shows that $b=0$.  Thus, the center is just $\mathbb R$, and so we cannot have a copy of $\mathbb C$ in the center.
