Invertibility of ideals in Dedekind domains

Based on arguments in Marcus’s Number Fields and K. Conrad’s expository paper on ideal factorization. I assume familiarity with the concepts related to ideals and fractional ideals in a commutative ring.

Recall that an integral domain R with the field of fractions K is called a Dedekind domain if the following conditions hold:

  • R is Noetherian, so every nonempty set of ideals has a maximal element, or equivalently, every ideal is finitely generated,
  • every prime ideal in R is a maximal ideal, and
  • R is integrally closed in K, so that every root of a monic polynomial from R[x] lying in K lies in R.

Our goal is to prove the following proposition:

Proposition: Assume R is a Dedekind domain. For any nonzero (i.e. containing a nonzero element) fractional ideal I in K there exists a fractional ideal J such that IJ=R.

Proof: We can construct the inverse explicitly: define \widetilde I=\{\gamma\in K:\gamma I\subseteq R\}. It is straightforward to check that \widetilde I is a fractional ideal. From its definition we see immediately I\widetilde I\subseteq R, so it is an ideal (since it’s a fractional ideal contained in R). If the equality holds, then we are home, since we can just take J=\widetilde I. Otherwise, it is a proper ideal. Hence we can apply the following lemma:

Lemma 1: If A is a proper ideal in R, then \widetilde A\not\subseteq R.

Before we prove it, let’s see why it is enough. We apply the lemma to A=\widetilde I I. Taking \gamma\in\widetilde A\setminus R we get \gamma\widetile I I\substeq R, so that \gamma\widetilde I\subseteq\widetilde I (recall the definition). We will reach a contradiction if we show \gamma is integral over R, i.e. satisfies a monic polynomial equation over R, since we assumed that R is integrally closed. This proceeds by a standard linear algebra argument, which we include here for completeness.

Recall that every ideal of R is finitely generated. It follows that every fractional ideal is also finitely generated (since it’s essentially just a scaled ideal). Say \widetilde I=(a_1,\dots,a_n). Since \gamma\widetilde I\subseteq\widetilde I, we can express \gamma a_i as a linear combination of a_i with coefficients in R. We can arrange these into a square matrix M, and setting v to be the vector consisting of a_1,\dots,a_n, we obtain Mv=\gamma v, so matrix \gamma I-M (I is the identity matrix here, not an ideal!) is singular. Expanding its determinant we find a monic polynomial equation satisfied by \gamma.

We only need to prove lemma 1 now. We will require two further lemmas.

Lemma 2: Let \frak p_1,\dots,\frak p_k and \frak p be prime ideals. If \frak p contains the product \frak{p}_1\dots\frak p_k, then \frak{p}=\frak p_i for some i.

Proof of lemma 2: Suppose \frak{p}\neq\frak p_i for all i. Since prime ideals are maximal, we can’t have \frak p_i\subsetneq\frak p, so there is an element a_i\in\frak p_i\setminus\frak p. Then clearly a_1\dots a_k\in\frak{p}_1\dots\frak p_k, yet, since \frak{p} is prime and a_i\not\in\frak pa_1\dots a_k\not\in\frak{p}, contradicting our assumption. \square

Lemma 3: Every nonzero fractional ideal contains a product of prime ideals.

Proof of lemma 3: Note that we can restrict our attention to ideals of R, since every fractional ideal contains an ideal as a subset.
Suppose there is an ideal which doesn’t contain such a product. Because R is Noetherian, there must be a maximal such ideal M. M clearly can’t be prime, so there are a,b\in R\setminus M such that ab\in M. Ideals M+(a),M+(b) must then contain M properly, so by choice of M they contain products of prime ideals, thus so does their product. But their product is (M+(a))(M+(b))=M+aM+bM+(ab)\subseteq M, contradicting the fact that M contains no product of prime ideals. \square

Proof of lemma 1: Choose any \alpha\in A. Then, by lemma 3, (\alpha) contains a product of prime ideals, say \frak{p}_1\dots\frak p_k. We may assume k is minimal for which it is possible, i.e. (\alpha) doesn’t contain a product of any k-1 prime ideals. Since R is Noetherian, A is contained in a maximal ideal, which is also prime. Call it \frak p. Then \frak p\supseteq A\supseteq (\alpha)\supseteq\frak{p}_1\dots\frak p_k, so, by lemma 1, \frak p=\frak p_i for some i, say i=1. By our minimality assumption, \frak{p}_2\dots\frak p_k\not\subseteq(\alpha), say \beta\in\frak{p}_2\dots\frak p_k\not\setminus(\alpha). Then \frac{\beta}{\alpha}\not\in R. However, \beta A\subseteq \beta\frak p\subseteq\frak{p}_2\dots\frak p_k\frak p=\frak{p}_1\dots\frak p_k\subseteq(\alpha), so that \frac{\beta}{\alpha}A\subseteq R thus \frac{\beta}{\alpha}\in\widetilde A\setminus R. \square

This completes the proof of I\widetilde I=R. \square

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2 thoughts on “Invertibility of ideals in Dedekind domains

  1. Pingback: Discriminant and different – Abstraction

  2. Pingback: Discriminant and different – Abstraction

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