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I ask for Wedderburn's Theorem.

my question is if $R$ is a simple ring, then $R = M _{n}(D)$ by the Theorem of Wedderburn ?

and how can conclude that $ R=D $ or $R = M _{2}(D)$ by the following lemma: If L is a proper left ideal of R then L is both minimal and maximal.

reference:

Bergen, Jeffrey, I. N. Herstein, and C. Lanski. Derivations with invertible values. Canad. J. Math 35.2 (1983): 300-310.

rschwieb
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H.Elmir
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  • Why would you cite a paper without including the author's name?! Is it Bergen, Herstein and Lanski? – rschwieb Feb 07 '17 at 14:34
  • Unfortunately it seems to be very hard to find a view online, so that will make it a lot harder to use the reference. I think in this case it would be nice to have a photo of the part you are trying to figure out. – rschwieb Feb 07 '17 at 14:38
  • yes ,the authors are: Bergen, Jeffrey, I. N. Herstein, and C. Lanski. Derivations with invertible values. Canad. J. Math 35.2 (1983): 300-310. – H.Elmir Feb 07 '17 at 20:05

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if R is a simple ring. then $R = M _{n}(D)$ by the Theorem of Wedderburn ?

No. The theorem you are thinking of only applies to simple Artinian rings.

There are lots of simple rings that aren't Artinian, and they are not isomorphic to such matrix rings.

If you have any further doubts, check this question: why is a simple ring not semisimple?

Also related is: Definition of a simple ring .

Perhaps the context of the paper includes Artinianness (or maybe works with finite dimensional algebras, which are Artinian), but I can't access the paper. Further details would be helpful.

how can conclude that $ R=D $ or $R = M _{2}(D)$ by the following lemma: If L is a proper left ideal of R then L is both minimal and maximal.

I don't think this lemma is stated correctly or completely. The only proper left ideal of a division ring $D$ is the zero ideal, which isn't a minimal left ideal. (Of course if you allowed the zero ideal to be minimal, then $M_2(D)$ would only have one minimal left ideal, and it wouldn't be maximal.)

Even overlooking that point, the statement is apparently incorrect as written: $\mathbb R[x]/(x^2)$ has exactly one proper ideal $(x)/(x^2)$, and it is both minimal and maximal, and this ring isn't semisimple because it has nonzero nilpotent elements. So, something is amiss.

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rschwieb
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  • Maybe there are some situations where ithe statement is true, see here Theorem 9.4.7. Maybe the definition of simple ring comes into play? – Marc Bogaerts Feb 07 '17 at 12:48
  • @MarcBogaerts Including Artinianess in the definition is rare. Aside from the book mentioned in this post I have never seen it used by an author. Even in the pdf you're referencing, the author as much admits he is modifying the standard definition just for the exercises (cont'd) – rschwieb Feb 07 '17 at 14:28
  • (cont'd) Specifically this passage on page 13: Problems 1–5 are the key steps in showing that a ring R is simple if and only if R is Artinian and has no two-sided ideals except 0 and R. Thus if a simple ring is defined as one with no nontrivial two-sided ideals, then the addition of the Artinian condition gives our definition of simple ring; in particular, it forces the ring to be semisimple. The result that an Artinian ring with no nontrivial two-sided ideals is isomorphic to a matrix ring over a division ring (Theorem 9.4.7) is sometimes called the Wedderburn-Artin theorem. – rschwieb Feb 07 '17 at 14:29
  • @MarcBogaerts Understandably, some authors want simple rings to be a subclass of semisimple rings. Personally, I kind of wish "semisimple" meant "finite product of simple rings" and that Artinian semisimple rings were called "Wedderburn rings". But the inertia on established terminology is so great... – rschwieb Feb 07 '17 at 14:32
  • I admit I'm uneasy with non commutative rings, what is the commutative version of this situation for commutative rings? PS if; in my comment I start with @srch your name should appear but it doesn't – Marc Bogaerts Feb 07 '17 at 15:17
  • @MarcBogaerts The @ notification feature does not work when you are directing the notification at the owner of a post (because they will be notified regardless of whether or not you include @! ) For commutative rings, the simple rings coincide with fields. I'm not sure I'm addressing the question you have in mind though... – rschwieb Feb 07 '17 at 15:30