Is it possible to construct a quasi-vectorial space without an identity element?
Solution 1:
In an old sci.math post by Dave Rusin, he discusses dropping sundry axioms form the usual set of axioms of vector spaces. What is below is taken from there.
So, let's recall what axioms we are dealing with: normally, a vector space $\mathbf{V}$ over a field $\mathbb{F}$ is defined to be a set, together with operations $+\colon\mathbf{V}\times\mathbf{V}\to\mathbf{V}$ and $\cdot\colon \mathbb{F}\times\mathbf{V}\to\mathbf{V}$, written in their usual infix notation, which satisfies the following conditions:
- For all $x,y\in\mathbf{V}$, $x+y=y+x$.
- For all $x,y,z\in\mathbf{V}$, $(x+y)+z = x+(y+z)$.
- There exists a vector $\mathbf{0}\in\mathbf{V}$ such that for all $x\in\mathbf{V}$, $x+\mathbf{0}=x$.
- For each $x\in\mathbf{V}$ there exists $y\in\mathbf{V}$ such that $x+y=\mathbf{0}$.
- For all $x\in\mathbf{V}$, $\alpha,\beta\in \mathbb{F}$, $\alpha(\beta x) = (\alpha\beta)x$.
- For all $x\in\mathbf{V}$, $1x = x$.
- For all $x,y\in\mathbf{V}$, $\alpha\in\mathbb{F}$, $\alpha(x+y) = \alpha x + \alpha y$.
- For all $x\in\mathbf{V}$, $\alpha,\beta\in\mathbb{F}$, $(\alpha+\beta)x = \alpha x + \beta x$.
You cannot just drop 3, because that would make 4 unintelligible. A way around it is to replace 4 with another statement which, in the presence of all other axioms, is equivalent to 4; namely:
4'. For all $x,y,w\in\mathbf{V}$, if $y + x = w + x$, then $y=w$.
That is, we have right cancellation. (This is important; if you set up cancellation on the "other side" as the identity, you run into some difficulties below)
Note that if you have 1-8, then you get 4'. And if you have 1-3, 4', and 5-8, then you get 4: first, note that since $\mathbf{0}+0x = 0x = (0+0)x = 0x + 0x$, then 4' implies that $0x = \mathbf{0}$. Then given any $x\in\mathbf{V}$, we have $x + (-1)x = (1+(-1))x = 0x = \mathbf{0}$, so 4 holds in this case. That is, 1, 2, 3, 4', 5, 6, 7, and 8, are an alternative way of defining vector spaces, with the added advantage that now you can drop any of the eight and the remaining statements are still intelligible.
If you take 1, 2, 3, 4', 5, 6, 7, and 8, then you can construct objects that are not vector spaces and satisfy any seven of these and not the eighth.
All but 1: Take $\mathbf{V}=\mathbb{R}$, and define scalar multiplication by $\alpha x = x$, and $x+y = x$ for all $x$ and $y$.
All but 2: Take $\mathbf{V}=\mathbb{R}^2$, define scalar multiplication the usual way, and $$x+y = \left\{\begin{array}{ll} (0,0) &\mbox{if $x=(0,0)$ or $y=(0,0)$;}\\ |\cos(\theta)|(x+y) & \mbox{if $x\neq(0,0)\neq y$, and $\theta$ is the angle from $x$ to $y$.} \end{array}\right.$$
All but 3: Take $\mathbf{V}=\emptyset$, with the empty addition and multiplication!
All but 4 or 4': Take $\mathbf{V}=\mathbb{R}\cup\{\bigcirc\}$. Define addition so that it is the usual addition in $\mathbb{R}$, and $r+\bigcirc=\bigcirc+r = r$ for all $r\in\mathbb{R}$, and $\bigcirc+\bigcirc=\bigcirc$. Scalar multiplication is regular multiplication on $\mathbb{R}$, and $r\bigcirc = \bigcirc$ for all reals $r$.
All but 5: Take $\mathbf{V}=\mathbb{R}$, and let $\sigma\colon\mathbb{R}\to\mathbb{Q}$ be any additive homomorphism of abelian groups. Define addition as usual, and scalar multiplication by $r x = \sigma(r)\cdot x$, where the multiplication on the right hand side is the usual real multiplication.
All but 6: Take $\mathbf{V}=\mathbb{R}$ with the usual addition, but zero multiplication: $\alpha x = 0$ for all $x$ and all $\alpha$.
All but 7: Take $\mathbf{V}=\mathbb{C}^2$, addition defined the usual way, and scalar multiplication given by: $$\alpha(x,y) = \left\{\begin{array}{ll} (\alpha x,\alpha y) & \mbox{if $x\neq 0$,}\\ (0, \overline{\alpha}y) & \mbox{if $x= 0$.} \end{array}\right.$$ where $\overline{\alpha}$ is complex conjugation.
All but 8: Take $\mathbf{V}=\mathbb{R}$ with usual addition, and scalar multiplication $r x = r^2\cdot x$, where the multiplication on the right hand side is the usual multiplication of real numbers.
Okay, but "All but 3" was almost cheating. What if we require that $\mathbf{V}$ be nonempty? Then you cannot have a structure that satisfies 1, 2, 4', 5, 6, 7, and 8, and does not satisfy 3:
Suppose $\mathbf{V}$ satisfies 1, 2, 4', 5, 6, 7, and 8, and is nonempty. Let $x\in V$. Then for all $y\in V$, we have: \begin{align*} (0x + y) + 0x &= 0x + (0x+y) &\quad&\mbox{(by 1)}\\\ &= (0x+0x) + y &\quad&\mbox{(by 2)}\\\ &= (0+0)x + y &&\mbox{(by 8)}\\\ &= 0x + y\\\ &= y + 0x &&\mbox{(by 1)} \end{align*} By 4', this means that $0x+y = y$, so $\mathbf{0}=0x$ shows that 3 is satisfied.