Consider the following theorem and the part of its proof shown:
So let $R$ be a homogeneous Cohen-Macaulay $k$-algebra with canonical module $\omega_R$. Let $b$ be the smallest degree for which $\omega_R$ is non-zero. The authors want to show that there exists an element $x \in (\omega_R)_b$ with $\operatorname{ann}_R(x)=0$. The case where $R$ is a domain is clear so lets consider the case where $R$ is generically Gorenstein and a level ring. The authors embedded $\omega_R$ into a module $F$ and they show that $(\omega_R)_b$ is not contained in $\cup_{p \in \operatorname{Ass} R} p F$. So far so good. Then they say that if $x \in (\omega_R)_b - \cup_{p \in \operatorname{Ass} R} p F$, then $x$ has zero annihilator.
Question: Why is that true? Moreover, i don't understand what it is they are trying to achieve by embedding $\omega_R$ into $F$. Since $F= \operatorname{Hom}_R(G,\omega_R)$, and $G$ is free over $R$, then $F$ is just the direct sum of copies of $\omega_R$. So why not take an element $x \in (\omega_R)_b - \cup_{p \in \operatorname{Ass} R} p \omega_R$ in the first place? In summary, i don't understand what is the effect of $F$.
Suppose that $\operatorname{ann}_R(x)\ne0$ and let $a\in R$, $a\ne0$ such that $ax=0$. Since $x\in F$ we can write $x=(a_1,\dots,a_r)$ with $a_i\in R$, hence $aa_i=0$ for all $i$. This shows that $a_i\in\operatorname{ann}_R(a)$ for all $i$. Let $\mathfrak p$ be a minimal prime over $\operatorname{ann}_R(a)$. Then $\mathfrak p\in\operatorname{Ass}R$ and since all $a_i\in\mathfrak p$ we get $x\in\mathfrak pF$, a contradiction.