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I'm working through Peter Stevenhagen's notes on Algebraic Number Theory, and the third section starts:

In order to factor an ideal $I$ in a number ring $R$ [into its primary composition $I = \prod_{\mathfrak p \supset I} I_{(\mathfrak p)}$], we have to determine for all prime ideals $\mathfrak p \supset I$ the $\mathfrak p$-primary part $I_{(\mathfrak p)}$ of $I$. As $I$ is of finite index in $R$, a prime $\mathfrak p$ of $R$ divides the integer $[R : I]$, hence a prime number $p$.

I don't see how the second sentence follows. Why does $\mathfrak p$ contain $[R : I]$?

Mark
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In fact $I$ contains the norm $[R : I]$.

If we take $x \in R$, then the additive order of $x + I$ in $R/I$ is a divisor of $\#(R/I) = [R : I]$. Hence $[R : I](x + I) = [R : I]x + I = I$ in $R/I$, which means that $[R : I]x \in I$. Now take $x = 1$.

($I = 0 + I$ is the additive identity in $R/I$.)

Ricardo Buring
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