- #1
peteryellow
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Let A be a ring.
If every finitely generated right ideal is genreated by an idempotent then Jac(A) =0.
Here Jac(A) means jacobson radical.
Proof: Let a be in Jac(A) and pick an idempotent element e such that Ae = Aa, thus a = ae and e=xa for x in A. Hence a =axa so a(1-xa)=0. Since a is in Jac(A) also xa is in Jac(A)
and 1-xa is a unit , hence a=0.
My question is that why is a = ae and e=xa , please help me with this.
Because Ae = Aa, so I will say that ae = a'a. and we have that e=xa, is it because we can pick e in A so ee=e =xa, is it so, please help.
If every finitely generated right ideal is genreated by an idempotent then Jac(A) =0.
Here Jac(A) means jacobson radical.
Proof: Let a be in Jac(A) and pick an idempotent element e such that Ae = Aa, thus a = ae and e=xa for x in A. Hence a =axa so a(1-xa)=0. Since a is in Jac(A) also xa is in Jac(A)
and 1-xa is a unit , hence a=0.
My question is that why is a = ae and e=xa , please help me with this.
Because Ae = Aa, so I will say that ae = a'a. and we have that e=xa, is it because we can pick e in A so ee=e =xa, is it so, please help.