Name for algebraic structure like a field that's forgotten its multiplicative identity I know a torsor is often called a "group that's forgotten its identity". Is there a name for a structure that's like a field that's forgotten its multiplicative identity?
Consider a field F and a vector space V on F equipped with a division operation that divides any vector by a non-zero vector to yield a scalar of F. This satisfies the obvious multiply/divide identities with the vector scaling function. There's an isomorphism between the vectors and the scalars for each non-zero vector that you pick as the "unit", but there's no canonical isomorphism.
I'm using this for physical quantities such as length, where one can add and scale lengths, but there is no canonical "unit length".
 A: If you are forgetting just the multiplicative identity, the structure is a $1$-dimensional vector space, i.e. an abelian group $L$ equipped with an action $\alpha\colon F\otimes_{\mathbb Z} L\to L$ (this just means that $\alpha\colon F\times L\to L$ is bilinear for inteeger scalars) so that $\alpha(\mu(x,y),p)=\alpha(x,\alpha(y,p))$, $\alpha(1,p)=p$, and such that for some non-zero $p\in L$, $x\mapsto \alpha(x,p)$ is a bijection. 
If you want to forget the additive identity as well, the resulting structure is an affine line $A$, which consists of an action $a\colon L\times A\to A$ so that $a(p_1+p_2,q)=a(p_1,a(p_2,q))$ and $a(0,q)=q$, such that for some non-zero $q$, $p\mapsto a(p,q)$ is a bijection.
A: One could say that homogeneous coordinates "forget length." Since the coordinates are given by $\Bbb R^{n+1}\setminus\{0\}$ under the equivalence $v\sim w\iff \exists \lambda\in \Bbb R(w=\lambda v)$, equivalence classes contain elements of all lengths, so they forgot the lengths they began with.
Projective geometry uses homogeoneous coordinates. In the context of projective geometry, distance is not invariant.

Just as $\Bbb R^n$ acts faithfully and transitively on $\Bbb R^n$ by addition, making every point "as good as the origin," the scaling action of $\Bbb R$ on $X=\{\lambda v\mid \lambda\in \Bbb R\setminus\{0\}\}$ for a nonzero vector $v$ is transitive. This could be viewed as "the action of $\Bbb R$ making all vectors in $X$ as good as one another."
A: The closest algebraic structure to what you asked foris a ring. Rings dont necessarily have a multiplicative identity. Notice that in that case there cant be inverses and is not a field, so you can't define a vector space over it.
