If every irreducible elements in Noetherian ring is prime, it is UFD

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I have problem with the proof of theorem 4 in this link

It says if $R$ is an Noetherian ring, we construct $\mathbb U$, the set of ideals generated by each element of $R$ that cannot be written as a product of irreducible elements of $R$. then $\mathbb U$ have an maximal element $(r)$.

But I'm not sure what is mean by the "set of ideals generated by each element of $R$"

I think it means $\mathbb U = \{(a) : a \text{ is not a product of irreducibles} \}$ But in that case, how can I make an ascending chain that gives maximal element of $\mathbb U$?