R be a simple ring that is finite-dimensional over its center k, show that R is isomorphic to a matrix
algebra over its center k iff R has a nonzero left ideal A with (dimkA)2 ≤ dimkR
1
Expert's answer
2012-10-31T08:46:11-0400
If R ∼ Mn(k), we can take A to be R · E11 for which (dimkA)2 = n2= dimkR. Conversely, suppose R has a nonzero left ideal A with(dimkA)2≤ dimkR. Then (nd)2= (dimkV )2≤ (dimkA)2≤ dimkR= n2d. This implies that d = 1, so D =k and R ∼ Mn(k).
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