1
$\begingroup$

Let $A$ be an associative unital n-dimensional algebra over field $F$.

Show that if $ab=1$ for some $a,b \in A$ then $a=b^{-1}$

  • 1
    $L$ is surjective, in fact a linear operator on the vector space $A$ of dimension $n$ is surjective iff it's injective, which is a consequence of the dimention formula $dim(im(L))+dim(Ker(L))=n$2012-08-11
  • 2
    OP may be talking about $\mathrm{End}_A(A)\cong A^{op}$, not $\mathrm{End}_k(A)$ (because of the comment about considering $A$ as a left module over itself). What confuses me is that $\mathrm{End}_A(A)=A^{op}$ (because $a\mapsto\phi(a)=a\phi(1_A)$), and $A^{op}$ is not always isomorphic to $A$, so I wonder what $A\to\mathrm{End}_A(A)$ is supposed to be. Otherwise $\mathrm{End}_k(A)\cong M_n(k)$ has $\dim=n^2$ when $\dim_k A=n$, so $A\to\mathrm{End}_k(A)$ is certainly not surjective.2012-08-11

2 Answers 2