Vector space over a field

Let $F$ be a field (Field), with additive identity $0_F$, multiplicative identity $1_F$, and the field axioms as stated there. A vector space over $F$, also called an $F$-vector space, consists of

• a set $V$, whose elements are called vectors;
• a binary operation $+:V\times V\to V$ on $V$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation), the vector addition;
• an element $0_V\in V$, the zero vector;
• a map $\cdot :F\times V\to V$, the scalar multiplication, written $\lambda v:=\cdot (\lambda ,v)$;

subject to the following axioms, in which $u,v\in V$ and $\lambda ,\mu \in F$ are arbitrary.

• (V1) $(V,+,0_V)$ is an abelian group (Group and abelian group): $+$ is associative and commutative, $0_V$ is a two-sided identity for $+$ (Left identity, right identity, and two-sided identity for a binary operation), and every $v\in V$ has an additive inverse (Left inverse, right inverse, and invertible element of a monoid).
• (V2) $\lambda (u+v)=\lambda u+\lambda v$.
• (V3) $(\lambda +\mu )v=\lambda v+\mu v$.
• (V4) $(\lambda \mu )v=\lambda (\mu v)$.
• (V5) $1_{F}v=v$.

The elements of $F$ are called scalars. When several vector spaces are in play we write $0_V$ for the zero of $V$, and we write $-v$ for the additive inverse of $v$ and $u-v:=u+(-v)$.

The notation $0_V$ and $-v$ is legitimate. Axiom (V1) asserts only that some two-sided identity and some additive inverses exist. That there is at most one two-sided identity for $+$ is A left identity and a right identity for the same binary operation are equal; hence there is at most one two-sided identity, and that an invertible element of a monoid has exactly one inverse is In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided; both are proved before Group and abelian group and are inherited here with the group structure. So $0_V$ and $-v$ denote well-defined elements, and nothing below re-derives them.

What (V1) buys, and why it is not restated. Associativity, commutativity, the identity law $v+0_V=v=0_V+v$, the inverse law $v+(-v)=0_V$, cancellation (Cancellation in a group: $gx=gy$ or $xg=yg$ forces $x=y$; equivalently left and right translation by $g$ are bijections of $G$, so $gx=h$ and $xg=h$ each have exactly one solution) and the inverse identities (In a group $e^{-1}=e$, $(g^{-1}{)}^{-1}=g$ and $(gh{)}^{-1}=h^{-1}{g}^{-1}$, the order of the last product being essential) are facts about abelian groups. They are quoted from the group page wherever they are used and are never proved again for vectors.

• Scalar multiplication is not a binary operation on a set. It is a map $F\times V\to V$ with arguments from two different sets, so Binary operation on a set; associativity, commutativity, and a subset closed under the operation, which is about a map $S\times S\to S$, does not apply to it and is never cited for it. The definition above cites that item for the vector addition only. In particular "closed under scalar multiplication" below always means $\lambda v\in W$ for $\lambda \in F$ and $v\in W$, which is not an instance of the closure condition defined there.

• (V5) is an axiom, not a consequence of (V2)–(V4). Take any abelian group $(V,+,0_V)$ and define $\lambda v:=0_V$ for every $\lambda$ and $v$. Then (V2), (V3) and (V4) all hold, both sides of each being $0_V$, while (V5) fails as soon as $V\mathrm{\ne }\{0_V\}$. So (V5) has to be imposed, and it is what ties the scalar action to the identity of $F$.

• Two structures, one set. A vector space is data: the set $V$, the addition, the zero, and the scalar multiplication, over a fixed field $F$. The same set may carry vector-space structures over different fields, and the field is part of the statement of every result below. A field is a vector space over itself, and over any subfield $K\subseteq F$ every $F$-vector space is a $K$-vector space by restricting the scalars is the first place where that matters.

• The field is the published one. No field axiom is restated here; $F$ is a field in the sense of Field, whose axiom (A) already says that $(F,+)$ is an abelian group and whose axiom (M) says the same of $(F\setminus \{0_F\},\cdot )$. In particular every field is a vector space over itself, which is A field is a vector space over itself, and over any subfield $K\subseteq F$ every $F$-vector space is a $K$-vector space by restricting the scalars.