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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.

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    $IJ=-JI\neq JI$. Therefore $I$ is not in the center of $\mathbb H$. This calculation doesn't a priori rule out some different embedding of $\mathbb C$ into $\mathbb H$, but you've written $i$ as if it were just something lying outside of the span of $1,I,J,K$, and you can't because the span is all that we define the quaternions to be.2011-12-28
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    Wait, but how is I=i. Like my understanding is that I,J and K aren't complex number.2011-12-28
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    Setting $I=i$ would be one way to embed $\mathbb C$ into $\mathbb H$, but it wouldn't be an embedding into the center. Your definitions of algebra might be different, but when I think of associative algebras over a field, I require the field be contained in the center of the algebra. This is in contrast to Lie algebras where the algebra is just a vector space over the field. However, Alex has shown below that even this is impossible.2011-12-28
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    @simplicity: What is, for example $iI$ for you? If it is a quaternion, then it has to be one of $a+bI+cJ+dK$ with $a,b,c,d$ real. If you allow the scalars $a,b,c,d$ to have complex values, you avoid that problem, but then your algebra is no longer the quaternions. Instead you get something bigger (and less interesting).2011-12-28

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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.

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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.