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.
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7What is your definition of "algebra over a ring"? The probably big problem is that $w\alpha\neq \alpha w$ for $w$ complex and $\alpha$ a quaternion. – Thomas Andrews Dec 28 '15 at 19:25
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Possible duplicate of Quaternions as a counterexample to the Gelfand–Mazur theorem – Noah Schweber Dec 28 '15 at 20:36
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2The usual definition of an algebra $A$ over a commutative ring $R$ is: “$A$ is a ring and there is a ring homomorphism $R\to Z(A)$”, where $Z(A)$ is the center of $A$. – egreg Dec 28 '15 at 20:55
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1This question seems different from the suggested dupe. Over there, the OP was oblivious that the quaternions were not a complex algebra and needed help clearing up the resulting "contradiction" with Gelfand-Mazur. Here the poster is aware it's not a complex algebra and is looking for explanations why it is not the case. – rschwieb Dec 28 '15 at 22:05
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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.)
rschwieb
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How can I see that $\mathbb{C}Q_8$ is a two dimensional algebra over $\mathbb{C}$? Can we define this algebra in two different ways, one in which complex i interacts with the quaternion i and one in which they behave separately? – AgentSmith May 15 '18 at 18:15
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The group algebra $\mathbb C[Q_8]$ where $Q_8$ is the quaternion group? By definition, that has dimension $8$ over $\mathbb C$. If you mean you want to show that $\mathbb H$ is two dimensional over the subring $\mathbb C$, then it's well known that you can represent $\mathbb H$ inside of $M_2(\mathbb C)$ as the set of matrices $\begin{bmatrix}a&b\-\bar{b}&\bar a \end{bmatrix}$. See this – rschwieb May 15 '18 at 20:06
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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 $A\subseteq Z(B)$.
Kenta S
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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 $x=y=j$.
