"The simple submodules of R (as a module over R ) are exactly the minimal left ideals of R . So R=⨁ S i where each S i is a minimal left ideal...''
Then comes the part that I don't understand:
"...In particular the element 1∈R can be written as a finite sum,
1=x i 1 +⋯+x i n
where x i j ∈S i j
I don't see why 1 can be written as this finite sum?
Thanks for your answer to my question
I read this note from " A first course in non-commutative rings(second edition page 26_27)" written by T.Y.Lam
Agree with Peter. The expression R=⨁ Si needs to be well understood:
First ⨁ is coproduct defined in R-mod. Therefore a generic element of ⨁ Si is an array of elements si in Si all of them are zero up to a finitely many; see Foundations of Module and Ring Theory Robert Wisbauer, p. 68 for the case R noncommutative; the commutative case is folklore.
Then use the isomorphism (sum +) to bring the (infinite) array to R. Now since 1 is in R it follows that 1 is in the image of ⨁ Si which is a finite sum ...
Let me to change my question
If R be a semi-simple ring with an identity we can decompose R into finite direct sum of it's minimal left ideals
why this proposition is true?
I know if R be semi-simple we can decompose R into direct sum of it's minimal left ideals but I don't know why this direct sum must be finite
Peter :we say ring R is semi-simple if R as R-module is semi-simple module
Yaser, It is because R itself is finitely generated. It's a consequence of the so-called Finite Sum Lemma:
Let M be a finitely generated (left) R-module. Suppose that M is a sum (not direct, just sum) of Sj where j is in J and J is infinite. Then there exists a finite subset I into J such that M is the sum of the Sj where j is in I.
Proof.- Since M is finitely generated, choose a set of generators x1,...,xk. Express each xi as a sum of finitely many sij. It follows that only a finitely many of the Sj appears in the above business. Therefore M is sum of that finitely many Sj.
Of course your problem M=R as module which is obviously f.g. and Sj are the ideals
Yes peter we know if a ring R be finitely generated and semi-simple Then R must be both Noetherian and Artinian but if R not be finitely generated I don't know what happen
@Peter, @Yaser. R itself is a left R-module, which is of course finitely generated by {1}. Therefore the answer to the problem stated by Yaser is in the result I wrote some days ago by setting M=R
@Peter any 2 decomposition of R is equals
R=⨁Bi and R=⨁Cj imply that I=j and Bi = Cj for all i
I think this may help you: https://books.google.com.au/books?id=5yvzCAAAQBAJ&pg=PA155&lpg=PA155&dq=%22minimal+left+ideal%22+definition&source=bl&ots=Ic25spiqq1&sig=TwQ1HNTE0o0Dhihj2bhxyP4rvMg&hl=en&sa=X&ved=0ahUKEwi8sbfzhOrJAhXH6xoKHUjfDpIQ6AEIOTAG#v=onepage&q=%22minimal%20left%20ideal%22%20definition&f=false
In case it does help, please write your conclusions here.
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