Exercise 5, § 5

Let E be a vector space over a commutative field K and let $T(E)$ be the tensor algebra of E.

(a) Show that $T(E)$ contains no divisors of 0 and that the only invertible elements in $T(E)$ are the scalars $\neq 0$. If $\dim(E) > 1$, $T(E)$ is a non-commutative K-algebra.

(b) Let $u, v$ be two elements of $T(E)$. Show that, if there exist two elements $a, b$ of $T(E)$ such that $ua = vb \neq 0$, one of the elements $u, v$ is a right multiple of the other. (Consider first the case where $u$ and $v$ are homogeneous; reduce it to the case where E is finite-dimensional and write $u = e_1u_1 + \cdots + e_nu_n$, $v = e_1v_1 + \cdots + e_nv_n$, where $(e_k)$ is a basis of E. Conclude by arguing by induction on the smallest degree of the homogeneous components of $ua$.) Deduce that, if $\dim(E) > 1$, the ring $T(E)$ admits no field of left fractions (I, § 9, Exercise 15).

(c) Show that for two non-zero elements $u, v$ of $T(E)$ to be permutable, it is necessary and sufficient that there exist a vector $x \neq 0$ in E such that $u$ and $v$ belong to the subalgebra of $T(E)$ generated by $x$ (use (b)). Deduce that, if $\dim(E) > 1$, the centre of $T(E)$ is equal to K.