Corollary

Let $K$ be a commutative field and $M$ a vector space over $K$. For every element $z \in T(M)$, there exists a smallest vector space $N$ of $M$ such that $z \in T(N)$ and $N$ is of finite rank over $K$.

It is understood in this statement that for every vector subspace $P$ of $M$, $T(P)$ is canonically identified with a subalgebra of $T(M)$. Let $z \in T(M)$; $z$ can be expressed as a linear combination of elements each of which is a finite product of elements of $M = T^1(M)$; all the elements of $M$ which occur in these products generate a vector subspace $Q$ of finite rank and $z \in T(Q)$.

Let $\mathscr{F}$ be the (non-empty) set of vector subspaces $P$ of finite rank such that $z \in T(P)$. Every decreasing sequence of elements of $\mathscr{F}$ is stationary, since they are vector spaces of finite rank. Hence $\mathscr{F}$ has a minimal element $N$ (Set Theory, III, § 6, no. 5). It remains to verify that every $P \in \mathscr{F}$ contains $N$; now, $z \in T(P) \cap T(N) = T(P \cap N)$ (Proposition 4); in view of the definition of $N$, this implies $N \cap P = N$, that is $P \supset N$.

The subspace $N$ of $M$ is said to be associated with $z$.