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Can anyone give me a relatively simple proof or Some reference for the following fact.(I know that there is a proof of this theorem in Gerard J. Murphy'book: "$C^*$-Algebras and Operator Theory", but I'm sure that there should be a simple proof of this.

Every hereditary C*-subalgebra of a simple $C^*$-algebra is also simple!

Maybe this is easy for someone, but it makes me confused for a long time. I am a novice!

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    $\begingroup$ You should unaccept the irrelevant answer and accept the relevant answer. $\endgroup$
    – Yemon Choi
    May 1, 2012 at 9:46
  • $\begingroup$ I agree with Captain Oates. $\endgroup$ Jan 21, 2014 at 4:44

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I don't know much about C* algebras, but I would guess that there shouldn't be a simple proof of the theorem you state. Here's an algebraic example which seems to be a counterexample to the theorem once the modifier C* is removed from the statement. Let A be the Weyl algebra, i.e. the algebra generated by x,y with the relation xy-yx=1. It isn't hard to show this algebra is simple and that it has global dimension 1. The subalgebra generated by x also has global dimension 1, but of course it is far from simple.

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    $\begingroup$ Peter, I don't think the hereditary that you use in algebra is the same as the hereditary that people use for $C^\ast$-algebras. The latter has nothing to do with homological dimension in the Tor-Ext sense, or probably in any other sense come to think of it. $\endgroup$
    – Yemon Choi
    Feb 20, 2012 at 1:31
  • $\begingroup$ Thank you for your interest in my question! Yes, the hereditary that people define in $C^*$-algebra is not the same as the hereditary that use for general algebra. $\endgroup$
    – Aviv
    Feb 20, 2012 at 2:25
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    $\begingroup$ Ah, that's unfortunate terminology, then. I guess this is evidence that the first phrase in my first sentence is correct :-) $\endgroup$ Feb 20, 2012 at 3:32
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Here is a direct argument which may not differ much in its essence from the one in the book that you mention:

Say $A$ is simple and $B$ is a closed hereditary subalgebra of $A$. This means that if $a\in A$ and $b_1,b_2\in B$ then $b_1ab_2\in B$. Let $x\in B$ be non-zero and let us show that it generates $B$ as a closed two sided ideal. Since $x$ generates $A$ as a closed two-sided ideal, the finite sums of elements of the form $axa'$, with $a,a'\in A$, form a dense subset in $A$. In particular, if $y\in B$ and $\epsilon>0$ then there exists an element of the form $$ \sum_{i=1}^{n} a_ixa'_i $$ within a distance $\epsilon$ of $y$. The problem with this is that the $a_{i}$s and $a'_i$s are not in $B$. They are only in $A$. This is fixed using that in a C*-algebra one always has that $|c^*|^{1/n}c|c|^{1/n}\to c$ for any $c$ (alternatively, you can use an approximate unit for $B$). Then for $k$ large enough the element $$ \sum_{i=1}^{n} (|y^*|^{1/k}a_i|x^*|^{1/k}) x (|x|^{1/k}a'_i |y|^{1/k} ) $$ is also within a distance of $\epsilon$ of $y$.

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  • $\begingroup$ Thank you very much Leonel! I have to say that this proof may be what I expect! I think your method is similar to the proof of the following fact: Let $A$ be a simple $C^*$-algebra and $q \in A$ a projection. Then $qAq$ is a simple $C^*$-algebra. I once tried to imitate the proof of the above easy result, But I didn't get it. Thank you again! $\endgroup$
    – Aviv
    Feb 21, 2012 at 1:25
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    $\begingroup$ You're welcome. The same argument with very little modification shows also that every closed two-sided ideal of the hereditary subalgebra is the intersection of an ideal of the algebra with the hereditary subalgebra. $\endgroup$ Feb 22, 2012 at 0:32
  • $\begingroup$ Yes! Let $B$ be a hereditary subalgebra of a simple $C^*$-algebra $A$, essentially, as in Gerard J. Murphy'book, and also your argument, let $I$ be any closed ideal of $B$ and $J$ be the closed ideal of $A$ generated by $I$, i.e., $J = \overline{(AIA)}$. Then, [J \cap B = \overline{(BJB)}=I.] then my question follows.(one direction is trivial.) $\endgroup$
    – Aviv
    Feb 22, 2012 at 1:46

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