I am trying to understand some lines in the proof of following theorem, which I am not getting.
Theorem: Let $\mathfrak{p}_i$ ($i=1,2,\ldots,d$) be prime ideals in a commutative ring $A$ and $\mathfrak{a}$ be any ideal. If for some $y\in A$, $y+\mathfrak{a}\subseteq \cup_{i=1}^d \mathfrak{p}_i$, then $\mathfrak{a}\subseteq \mathfrak{p}_i$ for some $i$.
Here are few lines of the proof.
Since $y=y+0$, so it is in some prime ideal $\mathfrak{p}_i$.
Case 1. Assume that $y$ is in some, but not all prime ideals, say $$ y\in \mathfrak{p}_1\cap \cdots \cap \mathfrak{p}_j, \,\,\,\, y\notin \mathfrak{p}_{j+1}\cup \cdots \cup \mathfrak{p}_d. $$ Assume that no prime ideals are contained in other prime ideals here.
Choose $x\in \mathfrak{a}\cap \mathfrak{p}_{j+1}\cap \cdots \cap \mathfrak{p}_d$ with $x\notin \mathfrak{p}_1\cup \cdots \cup \mathfrak{p}_j$.
Then $y+x\in\cup_i \mathfrak{p}_i$, contradiction.
Question: I do not get why we can guarantee the existence of $x$ with requirement in last but second step?
Ref: Lecture notes on Local Rings, by Iversen, Page 3,
For each $l\in\{1,...,d\}$ choose $\alpha_l \in \mathfrak{a}\setminus\mathfrak{p}_l$. This is possible because otherwise $\mathfrak{a}\subseteq\mathfrak{p}_l$ for some $l$ and the proof of the theorem is complete.
For each $k,l\in\{1,...,d\}$ such that $k\neq l$ choose $\beta_{kl}\in\mathfrak{p}_k\setminus\mathfrak{p}_l$. This is possible because of the non-containment assumption on the prime ideals.
For each $l\in\{1,...,d\}$ let $$\gamma_l=\prod_{k\neq l}\beta_{kl}$$
Then $\gamma_l\not\in\mathfrak{p}_l$ but $\gamma_l\in\mathfrak{p}_k$ for $k\neq l$.
For each $l\in\{1,...,j\}$ let $$x_l=\alpha_l\prod_{k=j+1}^d \beta_{kl}$$ Then $x_l\in\mathfrak{a}\cap \mathfrak{p}_{j+1}\cap \cdots \cap \mathfrak{p}_d$ but $x_l\not\in\mathfrak{p}_l$.
Now let $$x=\sum_{l=1}^j \gamma_l x_l$$ Then $x\in\mathfrak{a}\cap \mathfrak{p}_{j+1}\cap \cdots \cap \mathfrak{p}_d$ and $x\notin\mathfrak{p}_1\cup \cdots \cup \mathfrak{p}_j$.