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I am trying to do the following exercises in Hungerford.

If $R$ is semisimple left Artinian, then $R$ is its own quotient ring.

By the Wedderburn-Artin theorem we know that $R$ is isomorphic to sum of matrix ring over division ring, but how can I using this property to prove the above result, or maybe there are different approachs to this question? Thanks you!

T100
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    What do you mean by "its own quotient ring"? That can't possibly mean $R\simeq R/I$, which is either trivially true ($I=0$) or false (e.g. $R$ is a field). – freakish May 20 '24 at 06:08
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    @freakish It means ring of fractions, in this case, i.e. $RS^{-1}$ where $S^{-1}$ is the set of regular elements. The thing about Artinian rings is that the regular elements are already units.. – rschwieb May 20 '24 at 11:22
  • @rschwieb the ring of fractions, or more generally localization, is well defined over commutative rings only, no? – freakish May 20 '24 at 11:34
  • @freakish No. The construction is also given for noncommutative rings, although the axioms are slightly more involved to make the operations work. Given any (possibly noncommutative) ring, and a nonempty multiplicative subset which one can say is "right permutable" and "right reversible", one can carry out the construction that works for commutative rings. (Permutable/reversible are known as the Ore conditions) Using the subset of regular elements, one can attempt to form a right or left ring of fractions. – rschwieb May 20 '24 at 11:37
  • This is usually known as the "(left/right) classical ring of quotients." – rschwieb May 20 '24 at 11:42

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For an Artinian ring, an element that is not a zero divisor on a side is already a unit.

So, when ones goes to form either ring of quotients for the ring, nothing new is gained, because the set of regular elements is already the set of units.

rschwieb
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