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Let $A$ be a Banach algebra, $I$ be a closed two-sided ideal in $A$ and $J$ be a closed two-sided ideal in $I$. When $J$ is a closed two-sided ideal in $A$? (Except modes $I$ be a complemented ideal in $A$, or $A$ be a $C^*$-algebra.)

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    $\begingroup$ At the risk of stating the obvious: it's certainly closed, so the question is whether it is a two-sided ideal. A sufficient condition would be $J \subseteq IJI$. $\endgroup$ – Robert Israel Jan 10 '17 at 0:17
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    $\begingroup$ Further to @RobertIsrael's comment: the condition in his comment is always satisfied if J has a bounded approximate identity, by Cohen's factorization theorem. $\endgroup$ – Yemon Choi Jan 10 '17 at 0:25
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    $\begingroup$ The condition is of course not necessary: you can take $J=z^2A(D)$ and $I=zA(D)$ where $A(D)$ denotes the disc algebra $\endgroup$ – Yemon Choi Jan 10 '17 at 0:32
  • $\begingroup$ Thanks, can you give me another conditions for $A$ or $I,J$ that give us transitivity of ideals? $\endgroup$ – Albert harold Jan 10 '17 at 6:54
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    $\begingroup$ I think it is up to you: after all, this is part of doing proper research. One does not merely go shopping for convenient criteria that reduce one's problem to a trivial corollary of someone's big result. I think that @RobertIsrael's nice observation, together with the example in my previous comment and the counterexample in my answer, provide a more than adequate starting point for you to do your own investigation. $\endgroup$ – Yemon Choi Jan 10 '17 at 15:46
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Here is a simple example to show that "being an ideal in" is not transitive, even for finite-dimensional commutative algebras.

Take $A$ to be the algebra generated by $1$ and an element $z$ satisfying $z^3\neq 0$, $z^4=0$. (E.g. take $z$ to be a suitable upper triangular matrix.) Take $I=z^2A={\rm lin}(z^2,z^3)$ and $J={\rm lin}(z^2)$. Then $J$ is an ideal in $I$ but $z^3\in zJ \setminus J$, so $J$ is not an ideal in $A$.

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