Let x not be quasinilpotent, so l i m n &#x2192;<!-- → -->

bedblogi38am 2022-05-11 Answered
Let x not be quasinilpotent, so l i m n | | x n | | 1 / n = λ 0. Let π ( x ) = x ¯ , where π ( x ) : A A / r a d ( A ) is the canonical quotient map. Suppose that | | x ¯ n | | 1 / n = 0. Then it's spectrum σ ( x ¯ ) = 0, so x ¯ is not invertible in A / r a d ( A ), and thus generates a proper ideal in A / r a d ( A ). So then π 1 ( x ¯ ) generates a proper ideal in A containing r a d ( A ).
From here, if r a d ( A ) is maximal I think I'd have a contradiction, but I don't know if that's true. If not, does anyone have another strategy I could try?
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Answers (1)

Bentyrchjurvk
Answered 2022-05-12 Author has 15 answers
Assume x A such that x + r a d ( A ) is nilpotent in A / r a d ( A ). This implies that for all ε > 0 there is an n N and r r a d ( A ) for which | x n r | ε n .
r is quasi-nilpotent so there exists M N s.t. for all m > M, | r m | ε n m
We then show by induction that there exists A > 0 such that | x n m | A 2 m ε n m for all m. We choose A 1 such that this is true for all m [ 0 , M ].
Then if m > M, we have
x n m r m = ( x n r ) ( x n ( m 1 ) + + r m 1 ) ,
so
| x n m | ε n ( A 2 m 1 ε n ( m 1 ) + + A ε n ( m 1 ) ) + ε n m A ε n m ( 1 + 1 + 2 m 1 ) = A 2 m ε n m .
So we can conclude that ε > 0 lim inf | x n | 1 / n < 2 ε, which means that x r a d ( A ).
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