Spectrum: For Banach algebra $A$ spectrum is denoted by $\sigma(A)$ and defined as the set of all non-zero bounded linear multiplicative function from $A$ to $\Bbb C$.(Function $\psi:A\to\Bbb C$ is multiplicative if for any $a,b\in A$ we have $\psi(ab)=\psi(a)\psi(b)$).
Let $A$ be a Banach algebra, $X$ be a left Banach $A$-module and $\phi: X\to A$ be $A$-module homomorphism i.e., for any $a\in A, x\in X$ the equation $\phi(a.x)=a\phi(x)$ holds. Define $$\left\{\begin{array}{l}\circ: X\times X\to X\cr x\circ y=:\phi(x).y \end{array}\right.$$ By this product $X$ is Banach algebra.
What we can tell about relation between spectrum of $X$ and $A$ or spectrum of $A$? Any idea can help.
Let $B$ be any Banach algebra with non-empty spectrum, e.g. $B=\mathbb{C}$. Let $A=X=B\oplus_1 B$ with outer multiplication $(a_1\oplus a_2)(x_1\oplus x_2)=a_1x_1\oplus 0$ and consider $\varphi:X\to A:x_1\oplus x_2\mapsto 0\oplus x_2$. One can easily check that $\varphi$ is a morphism of $A$-modules and $$ x\circ y=\varphi(x)y=(0\oplus x_2)(y_1\oplus y_2)=0 $$ Thus $(X,\circ)$ is a Banach algebra with zero multiplication. Its spectrum is empty, while $\operatorname{Spec}(A)=\operatorname{Spec}(B)\neq \varnothing$. Thus, in general there is no relations between $\operatorname{Spec}(X)$ and $\operatorname{Spec}(A)$.