Primordial element (algebra)

Let ${\displaystyle V}$ be a vector space over a field ${\displaystyle \mathbb {F} }$ and let ${\displaystyle \left(e_{i}\right)_{i\in I}}$ be an ${\displaystyle I}$-indexed basis of vectors for ${\displaystyle V.}$ By the definition of a basis, every vector ${\displaystyle v\in V}$ can be expressed uniquely as

$${\displaystyle v=\sum _{i\in I}a_{i}(v)e_{i}}$$

for some ${\displaystyle I}$-indexed family of scalars ${\displaystyle \left(a_{i}\right)_{i\in I}}$ where all but finitely many ${\displaystyle a_{i}}$ are zero. Let

$${\displaystyle I(v)=\left\{i\in I:a_{i}(v)\neq 0\right\}}$$

denote the set of all indices for which the expression of ${\displaystyle v}$ has a nonzero coefficient. Given a subspace ${\displaystyle W}$ of ${\displaystyle V,}$ a nonzero vector ${\displaystyle p\in W}$ is said to be primordial if it has both of the following two properties:[1]

1. ${\displaystyle I(p)}$ is minimal among the sets ${\displaystyle I(w),}$ where ${\displaystyle 0\neq w\in W,}$ and
2. ${\displaystyle a_{i}(p)=1}$ for some index ${\displaystyle i.}$