Struggling to understand example of Ideal which is not finitely generated

Question

I'm working through an algebraic number theory book, but I can't understand the example shown below:

I follow the example up till it assumes that $\frac{p_1}{q_1},...,\frac{p_n}{q_n}$ are the generators for the ideal $R$ of the abelian group $R$. "Then the only primes dividing the denominators of elements of $R$ will be those dividing $q_1,...,q_n$", I don't understand what it means by this or why it's a contradiction to $R$ being a finitely generated ideal.

Answer 1

$R= \Bbb Q$ has elements whose denominators are divisible by any given prime. Indeed, $1/p$ is such an element for any prime $p$. The ideal given in the text does not have this property, so it cannot be all of $R$.

Answer 2

IF $R$ were finitely generated, let $p_1/q_1,\ldots,p_n/q_n$ be that finite set of generators. Choose some prime $p$ that does not divide $q_1q_2\cdots q_n$. Then $\frac{1}{p}$ should be in $R$ but is not in $\langle p_1/q_1,\ldots,p_n/q_n\rangle$, which is supposedly $R$. Contradiction.

Note that $p_1/q_1+p_2/q_2$ has denominator that is some factor of $q_1q_2$, so inductively anything in $\langle p_1/q_1,\ldots,p_n/q_n\rangle$ has denominator that is some factor of the product $q_1q_2\cdots q_n$.