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?
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 $A \otimes^\min J$. $\Box$
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.
