Left Vector Space
In this proof I'm going through (it's slightly too advanced for me, hence the difficulties I'm running in to) I don't know what the author means by "view D as a left C-vector space." Neither does google apparently. Could anyone clarify? Thanks for any replies.
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$\begingroup$It means here that the operation of scalar mutiplication of $\mathbf{C}$ on $\mathbf{D}$ shall be the multiplication with the element of $\mathbf{C}$ on the left,
$$\odot \colon \mathbf{C}\times \mathbf{D}\to \mathbf{D};\quad z \odot m = z\cdot m,$$
where $\cdot \colon \mathbf{D}\times \mathbf{D}\to\mathbf{D}$ is the multiplication on $\mathbf{D}$, and not on the right, $z\otimes m = m\cdot z$.
For non-commutative rings $R$, there is a real difference between left and right $R$-modules, for left $R$-modules, the scalar multiplication $\odot \colon R\times M \to M$ satisfies
$$(rs)\odot m = r\odot (s\odot m),$$
and for right $R$-modules, the scalar multiplication $\widehat{\otimes} \colon R\times M \to M$ satisfies
$$(rs)\widehat{\otimes} m = s\widehat{\otimes} (r \widehat{\otimes} m),$$
which is somewhat awkward, whence it's usually written $\otimes \colon M\times R\to M$, which satisfies the nicer
$$m\otimes (rs) = (m\otimes r)\otimes s.$$
If $R$ is a skew field (division ring), then left resp. right $R$-modules are also called left resp. right $R$-vector spaces.
If $R$ is commutative, left and right $R$-modules are same (every left $R$-module is a right $R$-module with the same operation and vice versa), so there is no algebraic distinction between left and right $R$-modules.
But here, where we have a division ring $\mathbf{D}$, and a field $\mathbf{C}\subset \mathbf{D}$, there are two "natural" choices how to define the operation of scalar multiplication to make $\mathbf{D}$ a $\mathbf{C}$-vector space, by multiplying on the left, or on the right [the two operations are the same if and only if $\mathbf{C}$ is contained in the centre of $\mathbf{D}$]. Saying "left $\mathbf{C}$-vector space" tells you which one is chosen, multiplication on the left.
$\endgroup$ 9 $\begingroup$Generally, if $F$ and $E$ are fields with $F \subseteq E$, then one can view $E$ as a vector space over $F$ by defining scalar multiplication to be the usual multiplication in the field, i.e. for $k \in F$ and $\alpha \in E$, the product $k \cdot \alpha$ is just the product with respect to the field multiplication.
So, since $\mathbb{C}$ is a subfield of $D$ in your example, we can view $D$ as a vector space over $\mathbb{C}$, i.e. as an $\mathbb{C}$-vector space.
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