Proposition I.16

We verify the vector space axioms.

1.

$(V∕W,+)$ is an abelian group

• (Associativity) Let $[w+c]=C,[w+d]=D$ and $[w+e]=E$ be classes in $V∕W$. Consider $(C+D)+E$. By definition the former is the class containing $(w+c)+(w+d)$. The class $(C+D)+E$ is therefore the class containing $(w+c+w+d)+(w+e)$ for our $(w+c+w+d)\in C+D$ and some $w+e\in E$. Similarly, $C+(D+E)$ contains $(w+c)+(w+d+w+e)$. But $(w+c+w+d)+(w+e)\sim (w+c)+(w+d+w+e)$ and so the classes $C+(D+E)=(C+D)+E$ are equal.
• (Existence of identity) Let $0:=W+0=W$. We then have, for all other classes $C\in V∕W$: $C+W=C$. To see why, pick $[c+w]=C$. We then have $[c+w]+[w]=[c+2w]=C+W$. But $(c+2w)-(c+w)=w\in W$, i.e., $c+2w\sim c+w$ and so $C+W=[c+2w]=[c+w]=C$.
• (Existence of inverses) Let $[w+c]=C\in V∕W$ be arbitrary. Then set $-C:=[w-c]$. We have $C+(-C)=[w+c]+[w-c]=[2w]=[w]=W$.
• (Commutativity) Let $[w+c]=C$ and $[w+d]=D$ be arbitrary. Then $C+D=[w+c]+[w+d]=[(w+c)+(w+d)]=[(w+d)+(w+c)]=[w+d]+[w+c]=D+C$.

2.

Scalar multiplication is defined
We have $1C=1[c]=[1c]=[c]=C$.

3.

Addition and scalar multiplication are related by

• $r(C+D)=\mathit{rC}+\mathit{rD}$
  $r(C+D)=r([w+c]+[w+d])=[r(w+c+w+d)]=[(\mathit{rw}+\mathit{rc})+(\mathit{rw}+\mathit{rd})]=[\mathit{rw}+\mathit{rc}]+[\mathit{rw}+\mathit{rd}]=r[w+c]+r[w+d]=\mathit{rC}+\mathit{rD}$
• $(r+s)C=\mathit{rC}+\mathit{sC}$
  Follows similarly from the analogous property for the vector space $V$.
• $(\mathit{rs})C=r(\mathit{sC})$
  Follows similarly from the analogous property for the vector space $V$.