Lemma 10.157.6. Let $R$ be a Noetherian normal domain with fraction field $K$. Then

1. for any nonzero $a \in R$ the quotient $R/aR$ has no embedded primes, and all its associated primes have height $1$

2. $R = \bigcap \nolimits _{\text{height}(\mathfrak p) = 1} R_{\mathfrak p}$
3. For any nonzero $x \in K$ the quotient $R/(R \cap xR)$ has no embedded primes, and all its associates primes have height $1$.

Proof. By Lemma 10.157.4 we see that $R$ has $(S_2)$. Hence for any nonzero element $a \in R$ we see that $R/aR$ has $(S_1)$ (use Lemma 10.72.6 for example) Hence $R/aR$ has no embedded primes (Lemma 10.157.2). We conclude the associated primes of $R/aR$ are exactly the minimal primes $\mathfrak p$ over $(a)$, which have height $1$ as $a$ is not zero (Lemma 10.60.11). This proves (1).

Thus, given $b \in R$ we have $b \in aR$ if and only if $b \in aR_{\mathfrak p}$ for every minimal prime $\mathfrak p$ over $(a)$ (see Lemma 10.63.19). These primes all have height $1$ as seen above so $b/a \in R$ if and only if $b/a \in R_{\mathfrak p}$ for all height 1 primes. Hence (2) holds.

For (3) write $x = a/b$. Let $\mathfrak p_1, \ldots , \mathfrak p_ r$ be the minimal primes over $(ab)$. These all have height 1 by the above. Then we see that $R \cap xR = \bigcap _{i = 1, \ldots , r} (R \cap xR_{\mathfrak p_ i})$ by part (2) of the lemma. Hence $R/(R \cap xR)$ is a submodule of $\bigoplus R/(R \cap xR_{\mathfrak p_ i})$. As $R_{\mathfrak p_ i}$ is a discrete valuation ring (by property $(R_1)$ for the Noetherian normal domain $R$, see Lemma 10.157.4) we have $xR_{\mathfrak p_ i} = \mathfrak p_ i^{e_ i}R_{\mathfrak p_ i}$ for some $e_ i \in \mathbf{Z}$. Hence the direct sum is equal to $\bigoplus _{e_ i > 0} R/\mathfrak p_ i^{(e_ i)}$, see Definition 10.64.1. By Lemma 10.64.2 the only associated prime of the module $R/\mathfrak p^{(n)}$ is $\mathfrak p$. Hence the set of associate primes of $R/(R \cap xR)$ is a subset of $\{ \mathfrak p_ i\}$ and there are no inclusion relations among them. This proves (3). $\square$

Comment #7456 by mi on

In the statement of (1), shall we also say a is not a unit?

Comment #7608 by on

We don't have to: the $0$ module has no embedded and no associated primes so (1) holds.

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