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Wedderburn-Artin's theorem:
A ring is left semisimple ring iff it is finite product of $M_{n_i}(D_i)$ for some division ring $D_i$.
Due to this theorem,we know that
if a ring is left semisimple,then it is also right semisimple.
I wonder how to prove this result without using the theorem.the result is equivalent to the following:
if $l.gl.dim(R)=0$,then $r.gl.dim(R)=0$.
is there a direct proof?
Thank you in advance!
I'm not sure there is anything more straightforward that proves symmetry of semisimplicity.
You could appeal to the fact that a semisimple ring is quasi-Frobenius, and therefore the lattice of right ideals is anti-isomorphic to the lattice of left ideals, with meets, joins and complements preserved, but that doesn't seem simpler.
I'm not sure this is really more direct than using Wedderburn-Artin, but a condition on a ring $R$ that is clearly left-right symmetric and is equivalent to semisimplicity is:
There is a decomposition $1=e_1+\dots+e_n$ into orthogonal idempotents such that, for every $i$, $e_iRe_i$ is a skew field and, for every $i$ and $j$, either $e_i$ and $e_j$ are conjugate or $e_iRe_j=0$.
