The following argument appears in Cohn's Algebra, Vol. 2, page 65.
Consider any field extension $F/k$. Given elements $\alpha_1, \dots,\alpha_n$, of $F$, we write $k[\alpha_1, \dots,\alpha_n]$ for the subring generated by $\alpha_1, \dots,\alpha_n$, over $k$ and $k(\alpha_1, \dots,\alpha_n)$ for the subfield generated by $\alpha_1, \dots,\alpha_n$ over $k$. For a finite extension we actually have equality: $$k(\alpha_1, \dots,\alpha_n) = k[\alpha_1, \dots,\alpha_n], \tag{7}$$ for the right-hand side is a finite-dimensional $k$-algebra without zerodivisors, so the multiplication by any non-zero element $c$ is a linear transformation which is injective and hence invertible (cf. also Vol. 1, p. 359). Thus any non-zero element has an inverse and the right-hand side of $(7)$ is a field, necessarily equal to the left-hand side.
I understand that $A = k[\alpha_1,\dots,\alpha_n]$ is a $k$-algebra, and if $c\in A$, then $T_c = c(\cdot) \in L(A)$ is an injective linear transformation on $A$. Thus, if $A$ is finite dimensional, then $T_c$ has an inverse in $L(A)$, say $T_c^{-1}$. But I think Cohn is claiming that $(*)$ $T_c^{-1} = T_b$ for some $b \in A$, hence $bc(\cdot)=I$, so $bc=1$ and $b$ is an inverse of $c$ in $A$. I don't see how $(*)$ follows. Am I right about the intended argument, and if so, how do we justify $(*)$?
