Assume that $A$ and $B$ are two unital commutative Banach algebras. Assume that $\phi \in \mathcal{M} (A)$ is an element of the maximal Ideal space. Define $\alpha: A\hat{\otimes} B \to \mathbb{C}\otimes B\simeq B$ with $\alpha=\phi \otimes Id_{B}$. $\;$Put $I=\ker \alpha$. Assume that $J$ is an ideal in $A\hat{\otimes} B$ with $J+ I=A\hat{\otimes} B$.

1.Is there an ideal $K$ in $A$ with $K\hat{\otimes} B \subset J$ and $(K\hat{\otimes} B )+ I= A\hat{\otimes} B$?

2.Consider the same question for two (pure algebraic) complex unital algebras, so we remove $"\hat{}"$ from the above tensor products. and $\phi$ is an algebra morphism from $A$ to $\mathbb{C}$.

This question is motivated by the "tube lemma" in general topology. So the answer to this question is "yes" for commutative $C^{*}$ algebras $A$ and $B$

We can consider the same question for (noncommutative) $C^{*}$ algebras $A$ and $B$, an irreducible representation $\phi: A \to B(H)$ and $\alpha: A\otimes B \to B(H) \otimes B$ and $I=\ker \alpha$.

**I explain that why I think that this is a noncommutative version of the tube lemma:**

In general, assume that $X$ is a compact Hausdorff space and $F$ and $K$ are two closed sets in $X$, then they are two disjoint set iff $I_{F}+I_{K}=C(X)$ where $I_{F}$ is the ideal in $C(X)$ which consists all $g\in C(X)$ with $g(F)=0$ . Now assume that $F=\{x_{0}\}\times Y$ is a slice in $X\times Y$. Then $I_{F}$ has an algebraic description as the above $I=\ker \alpha$, in my question. If $U$ is an open set containing this slice, then $U^{c}$ and the slice are two disjoint closed set. So the inclusion $\{x_{0}\}\times Y \subset U$ implies that $J+I=C(X)$ where $J=I_{U^{c}}$.