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From one of the talks I attended long back, I vaguely seem to remember the following fact:

Let $A$ and $B$ be $C^{\ast}$-algebras. If either $A$ is exact or $B$ is nuclear then every closed ideal of $A\otimes_{min}B$ is of the form $A \otimes_{min} J$ for some ideal $J$ of $B$.

Is this fact true or it needs some additional hypothesis? I tried to find references but could not find one. Any references?

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    $\begingroup$ Nik beat me to it. I honestly think that someone who has looked up the definitions of nuclear and exact Cstar algebras should think more carefully before asking a question for which $A= {\mathbb C}^2$ is a counterexample. It is essential to know and work with actual examples of Cstar algebras before launching into highbrow results about general Cstar algebra theory $\endgroup$
    – Yemon Choi
    Jul 25 '20 at 16:07
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    $\begingroup$ That said, I have a feeling that when A is simple the result you are thinking of might follow from some old results of Zacharias (Proc AMS 2001) which restrict the possible intermediate subalgebras between $A \otimes 1$ and $A\otimes B$ $\endgroup$
    – Yemon Choi
    Jul 25 '20 at 16:10
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    $\begingroup$ I’m voting to close this question because the OP should have put more thought/work into it before asking. $\endgroup$
    – Yemon Choi
    Jul 25 '20 at 16:11
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    $\begingroup$ I think this is a thread with quite a lot of interesting activity. Absolutely against closing. I wish there were fewer condescending comments though. $\endgroup$ Jul 29 '20 at 11:42
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    $\begingroup$ @MatthiasLudewig The comments are a reaction to a pattern of behaviour by various users. While they may seem harsh, they are intended to encourage said users to think more carefully about the mathematics they are starting to learn. There is a regrettable tendency in abstract functional analysis for beginners to only ever push around definitions in a soft way, which will lead after 5 years to people only engaging with their research specialism in a superficial way that does not reflect their own earlier potential. I speak from refereeing experience. $\endgroup$
    – Yemon Choi
    Jul 29 '20 at 12:50
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When A is simple, it's proved in: https://doi.org/10.1002/mana.201700009

As per the suggestion (below) by leo monsaingeon, here are the details reproduced from above paper:

Statement: Let $A$ and $B$ be $C^*$-algebras where $A$ is topologically simple. If either $A$ is exact or $B$ is nuclear, then every closed ideal of the $C^*$-algebra $A \otimes^\min B$ is a product ideal of the form $ A \otimes^\min J$ for some closed ideal $J$ in $B$.

Proof: Let $I$ be a non-zero closed ideal in $A \otimes^\min B$. Consider the collection $$ \mathcal{F} := \{ J \subseteq B: J \ \mathrm{is\ a\ closed\ ideal\ in}\ B\ \mathrm{and}\ A \otimes^\min J \subseteq I\}.$$ By Proposition 4.5 of [ASS], $I$ contains a non-zero elementary tensor, say, $a \otimes b$. If $K$ and $J$ are the non-zero closed ideals in $A$ and $B$ generated by $a$ and $b$, respectively, then by simplicity of $A$, we have $K = A$ and $A \otimes^\min J \subseteq I$. In particular, $\mathcal{F} \neq \emptyset$.

Note that, by injectivity of $\otimes^\min$ and the fact that a finite sum of closed ideals is closed in a $C^*$-algebra, it is easily seen that $$A \otimes^\min (\sum_i J_i) = \sum_i (A \otimes^\min J_i)$$ for any finite collection of closed ideals $\{J_i\}$ in $B$. So, with respect to the partial order given by set inclusion, every chain $\{ J_i : i \in \Lambda\}$ in $\mathcal{F}$ has an upper bound, namely, the closure of the ideal $\{\sum_{\mathrm{finite}} x_i : x_i \in J_i\}$ in $\mathcal{F}$, implying thereby that there exists a maximal element, say $J$, in $\mathcal{F}$.

We show that $A \otimes^\min J = I$. Consider the map $$\mathrm{Id} \otimes^\min \pi : A \otimes^\min B \rightarrow A \otimes^\min (B / J).$$ If $A$ is exact, then by the definition of exactness, its kernel is $A \otimes^\min J$; and, if $B$ is nuclear, then so are $J$ and $B/J$ and it is known (see Blackadar's book or Guichardet's paper on tensor products) that the sequence $$ 0 \rightarrow A \otimes^\max J \rightarrow A \otimes^\max B \rightarrow A \otimes^\max (B/J) \rightarrow 0$$ is always exact and, therefore, we obtain $$\ker(\mathrm{Id}\otimes^\min \pi) = \ker(\mathrm{Id}\otimes^\max \pi) = A \otimes^\max J = A \otimes^\max J.$$ Since $\mathrm{Id} \otimes^\min \pi$ is a surjective $*$-homomorphism, $$\widetilde{I}:=(\mathrm{Id} \otimes^\min \pi)(I)$$ is a closed ideal in $A \otimes^\min (B / J)$. It is now sufficient to show that this is the zero ideal. If $\widetilde{I} \neq 0$, then, again by Proposition 4.5 of [ASS], $\widetilde{I}$ contains a non-zero elementary tensor, say, $a \otimes (b +J)$. Let $K$ be the closed ideal in $B$ generated by $b$. Since $A$ is topologically simple, it equals the closed ideal generated by $a$ and we obtain $$A \otimes^\min K \subseteq I,$$ a contradiction to the maximality of $J$ as $A \otimes^\min K$ is not contained in $A \otimes^\min J$. $\Box$

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  • $\begingroup$ Does your proof go via Zacharias's result on intermediate subalgebras? $\endgroup$
    – Yemon Choi
    Jul 27 '20 at 19:51
  • $\begingroup$ No, it just uses the hypothesis, the fact that a closed ideal in a tensor product contains elementary tensors and exactness of certain sequence w.r.t. max tensor product. $\endgroup$
    – Ved
    Jul 29 '20 at 11:09
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When you think something might be true for all C${}^*$-algebras, the first thing to do is to check it on abelian C${}^*$-algebras. Then check it on the algebra of $2\times 2$ matrices, and if it also works there then you've got a good chance.

Your conjecture is absurdly false in the abelian case. Let $A = C(X)$ and $B = C(Y)$ --- these are both nuclear --- and then $A\otimes B \cong C(X\times Y)$. The ideals of $C(X\times Y)$ correspond to closed subsets of $X\times Y$, which obviously need not have the special form you describe.

I realize that you are obviously missing some hypotheses, but next time try checking the commutative case yourself first.

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  • $\begingroup$ Thank you! Probably we need to assume that $A$ is topologically simple? $\endgroup$
    – Math Lover
    Jul 25 '20 at 9:06

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