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Vector spaces seem to have a very similar definition to fields. Are vector spaces fields?


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How do you define multiplication of vectors? – user7530 Sep 6 '11 at 21:30
@user7530: Aha, thanks. – gareth Sep 6 '11 at 21:33
However, note that any field is a vector space over itself, and more generally if $F\subseteq K$ are both fields, then $K$ is a vector space over $F$ (if you "forget" how to multiply elements of $K$, and only remember how to multiply when at least one factor is in $F$). – Arturo Magidin Sep 6 '11 at 23:54
And sometimes vector spaces can be equipped with a multiplication that makes them fields, e.g. $\mathbb{R^2} \to \mathbb{C}$ or $\mathbb{F}_p^n \to \mathbb{F_{p^n}}$. – user7530 Sep 7 '11 at 0:25
up vote 4 down vote accepted

No. Vector spaces are not fields. Look at the definition of the vector space. First you consider a group $(V,+)$ which is commutative. Then you consider a field $\Bbb{K}$ whose operations are compatible to those of $V$, i.e.

$$ (\alpha+\beta)x=\alpha x+\beta x,\ \forall x \in V,\alpha,\beta \in \Bbb{K}$$

and so on. Notice that the scalar multiplication does not take place between two elements of $V$, but between ONE element of $\Bbb{K}$ and ONE element of $V$.

As a quick example, you could look at $\Bbb{R}^3=\{ (x,y,z) : x,y,z \in \Bbb{R}\}$ with addition $(a,b,c)+(d,e,f)=(a+d,b+e,c+f)$ and scalar multiplication $\alpha(a,b,c)=(\alpha a,\alpha b,\alpha c)$. This is a vector space over $\Bbb{R}$ (you can check all the axioms). But $\Bbb{R}^3$ is not a field: the product $(a,b,c)\cdot(d,e,f)$ is not defined. The product is defined between one scalar and one vector.

I hope this makes things a little bit clear. :)

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Thanks, Beni. This has indeed made things a lot clearer. :) – gareth Sep 6 '11 at 21:39

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