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I am confused how the operation $\cdot : F\times V\rightarrow V$ is defined for a vector space over a field. Is $\cdot : F\times V\rightarrow V$ simultaneously defined as $\cdot : F\times F\rightarrow F$ in regards to the field, or should there be symbols distinguishing the axioms below? Any help would be greatly appreciated. If I was to guess the operations are simultaneously defined, but I could be wrong.

Definition 1: A vector space over a field F is defined to be a set $V$. This set is said to have two operations $+: V\times V\rightarrow V$ and $\cdot: F\times V\rightarrow V$. Also, $V$ is defined to have the following conditions:

$(C1)$ $V, +$ forms a commutative group with an identity element $=0$,

$(C2)$ For all $a, b\in F$, for all $v\in V$, $(a⋅ b)⋅ v=a⋅ (b⋅ v)$,

$(C3)$ For all $v\in V$, $1⋅v=v$,

$(C4)$ For all $a\in F$, for all $u, v\in V$, $a⋅(u+v)=a⋅u+a⋅v$,

$(C5)$ For all $a, b\in F$, for all $v\in F$, $(a+b)⋅v=a⋅v+b⋅v$.


Definition 2: A $\textbf{field}$, $F$, is a set with two operations, $+$ and $\cdot$ which satisfies the following conditions:

$(C1)$ $F, +$ forms a commutative group with an identity element $= 0$,

$(C2)$ $F^{\neq 0}, \cdot$ forms a commutative group with an identity element $= 1$,

$(C3)$ For every $a, b, c\in F$, $a\cdot (b+c)=a\cdot b+a\cdot c$.

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1 Answer 1

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These are two distinct operations, for which we use the same symbol. It is usually easy to distinguish between them: if we see $\lambda.v$, then

  • if $\lambda\in F$ and $v\in V$, then we are talking about the function from $F\times V$ into $V$;
  • if $\lambda,v\in F$, we are talking about the product defined in $F$.

Of course, if it turns out that $V=F$, there is space for ambiguity here, but in this case both operations are usually the same.

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  • $\begingroup$ That makes sense. Thank you! $\endgroup$
    – W. G.
    Commented Mar 11, 2018 at 15:20
  • $\begingroup$ @W.G. If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ Commented Mar 11, 2018 at 15:22
  • $\begingroup$ Of course. You bet :) $\endgroup$
    – W. G.
    Commented Mar 11, 2018 at 15:24

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