I’m learning about algebraic number theory and I stumbled upon the following statement:

Let RR be a number ring (i.e. a subring of a finite extension of Q\mathbb{Q}) with normalization ˜R\tilde{R}. Then RR is singular over a prime number pp if and only if p|[˜R:R]p|[\tilde{R}:R].

Using the fact that a prime ideal p\mathfrak{p} of RR is regular if and only if its mutliplier ring r(p)={x∈Quot(R):xp⊂p}r(\mathfrak{p})=\{x\in \mathrm{Quot}(R):x\mathfrak{p}\subset\mathfrak{p}\} equals RR, I did one implication:

If RR has a singular prime p\mathfrak{p} over pp, then there is some x∈r(p)∖Rx\in r(\mathfrak{p})\setminus R. Now p\mathfrak{p} is finitely generated as RR is Noetherian, so xp⊂px\mathfrak{p}\subset\mathfrak{p} implies that x∈˜Rx\in\tilde{R}. But then px∈p⊂Rpx\in\mathfrak{p}\subset R shows that ¯x∈˜R/R\overline{x}\in\tilde{R}/R has order pp, so that p|[˜R:R]p|[\tilde{R}:R].

Conversely, if pp divides the index then the finite abelian group ˜R/R\tilde{R}/R has an element of order pp by Cauchy, i.e. there is some x∈˜R∖Rx\in\tilde{R}\setminus R such that px∈Rpx\in R. As pxpx is not a unit in RR it is contained in a prime p\mathfrak{p} of RR. If now i can show that p∈pp\in\mathfrak{p}, this implies that p\mathfrak{p} is singular since then x∈r(p)∖Rx\in r(\mathfrak{p})\setminus R.

Is it true that p∈pp\in\mathfrak{p}? I have the feeling I’m missing something silly here. I know that p\mathfrak{p} must contain a unique prime number, but I don’t see why it has to be pp.

If someone is aware of some other proof for this then I would also love to hear it.

Thanks in advance!

=================

=================

=================