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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.

https://dl.dropboxusercontent.com/u/17606191/fieldextensions.gif

Lammey
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2 Answers2

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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.

Daniel Fischer
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    Only just sat down and read through this, it is exactly the detailed answer I was looking for, thanks a lot! – Lammey Jan 05 '14 at 14:59
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    If I have a left C-vector space, does the term xi have any meaning? Or have we just not defined multiplication on the right yet? – Lammey Jan 05 '14 at 18:46
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    Formally, you need to define the meaning of an expression like $x\cdot i$ before you use it. Informally, when you have a left $\mathbb{C}$-vector space without an explicit definition of $x\cdot i$, it is [usually, at least] taken to mean the ordinary scalar multiplication that is usually written with the scalar on the left. Note that it is possible to have spaces where both $i\cdot x$ and $x\cdot i$ have defined but different meanings. – Daniel Fischer Jan 05 '14 at 18:53
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    Hmm ok that makes sense to me, but I'm having trouble making sense of the right scalar multiplication in this case: https://dl.dropboxusercontent.com/u/17606191/proof.gif The author seems to be defining a transformation by multiplication on the right which is also itself yet to be defined. I'm sure I'm mistaken but I can't see how! Thanks again for the speedy reply by the way, it's well appreciated! (The part I'm speaking of is about 2/3 of the way down the page I linked) – Lammey Jan 05 '14 at 19:06
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    "On the other hand multiplication on the right by $i$ ..."? Don't forget that $D$ is a (division) ring, and $i$ is an element of $D$. For every $d\in D$, the map $M_d\colon x\mapsto x\cdot_D d$ is an $\mathbb{R}$-linear map $D\to D$. In fact, with the $\mathbb{C}$-vector space structure just previously defined, it is $\mathbb{C}$-linear because of the associativity, $M_d(i\odot x) = M_d(i\cdot_D x) = (i\cdot_D x)\cdot_D d = i\cdot_D(x\cdot_D d) = i\odot(x\cdot_D d) = i\odot M_d(x)$. The author defines $T = M_i$. – Daniel Fischer Jan 05 '14 at 19:18
  • So the division algebra D is also a division ring, and since the field the algebra is over (Complex field) is a subring of D, the left and right scalar multiplication can just be seen as ring multiplication by a subset of the elements of D? And in a division algebra the ring multiplication is defined to be the bilinear product associated with the algebra? Sorry I'm taking so long to reply it's just taking a while to sink in. – Lammey Jan 05 '14 at 19:59
  • I mean that a division algebra viewed as a ring has it's ring multiplication defined to be the bilinear product associated with the algebra. – Lammey Jan 05 '14 at 20:16
  • Yes, any algebra is in particular also a ring, forgetting about scalar multiplication, and the multiplication in that ring is the bilinear product of the algebra. A division algebra is of course a division ring. Since the considered field is a subring of $D$, we can use the multiplication of $D$ to define the scalar multiplication that makes $D$ a $C$-vector space. – Daniel Fischer Jan 05 '14 at 20:18
  • Ah I see! Ok thanks again for your time! – Lammey Jan 05 '14 at 20:29
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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.

Ulrik
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