Why are the quaternions not an algebra over the complex numbers? I just began to study about algebras over rings and quickly came across the fact that the quaternions are not an algebra over the complex numbers. I would prefer an answer as elementary as possible.
 A: Because the center of $\mathbb H$ is $\mathbb R$.

Why?
Suppose $a+bi+cj+dk\in Z(\mathbb H)$, the center of $\mathbb H$. Then $i(a+bi+cj+dk)=-b+ai-dj+ck$ and $(a+bi+cj+dk)i=-b+ai+dj-ck$ have to be equal, so $c=d=0$. Now, $(a+bi)j=aj+bk$ and $j(a+bi)=aj-bk$ so $b=0$. Thus, $Z(\mathbb H)\subseteq\mathbb R$, and the opposite inclusion is obvious.

Why does this suffice?
If $\mathbb H$ were a $\mathbb C$-algebra, then since $\mathbb C$ is commutative, we need $\mathbb C\subseteq Z(\mathbb H)$.
In general, if $A$ is a commutative ring, for any $A$-algebra $B$ we need $A\subseteq Z(B)$.
A: If it were a $\Bbb C$ algebra, it would have to be dimension $2$ and contain a copy of $\Bbb C$. 
Taking any $x$ not in the copy of $\Bbb C$ situated in $\Bbb H$, the span of $\Bbb C$ and $x$ is the whole ring. But products between elements of this span commute with each other, and that means the span is a commutative ring. This contradicts the fact the quaternions aren't commutative.
At another level, the Artin-Wedderburn theorem says that the only possible simple Artinian $\Bbb C$-algebras are the square matrix rings over $\Bbb C$, but none of them have dimension $2$. ($\Bbb H$ is a simple Artinian ring.)
A: For an algebra we have the following axiom for scalar multiplication
$$\alpha x \centerdot \beta y = (\alpha \beta) x \centerdot y.$$
This does not hold for the quaternions with complex scalars. Check for $\alpha = \beta = i$ and $x=y=j$.
