By Proposition 3.2.3 we get \((1)\) and by Corollary 3.2.6 we get \((2)\).

So we only need to prove \(\mathcal{O}_K\) is integrally closed.

Let \(\gamma =\alpha / \beta \in K\) satisfy a monic polynomial with coefficients in \(\mathcal{O}_K\), then I claim that it is in fact an algebraic integer and therefore it is in \(K \cap \overline{\mathbb {Z}}=\mathcal{O}_K\). (Note \(\mathrm{Frac}(\mathcal{O}_K)=K\).)

The claim is Exercise 2.1.18.

Assume for contradiction this is not the case. Let \(S\) denote the set of all ideals that do not contain a product of prime ideals. Then, since \(R\) is Noetherian, by Definition 3.2.1\((3)\), \(S\) must contain a maximal element \(\mathfrak {m}\) (not to be confused with maximal ideal, here maximal means with respect to the property of not containing a product of prime ideals). Now \(\mathfrak {m}\) cannot be prime (otherwise \(\mathfrak {m}\subseteq \mathfrak {m}\) gives a contradiction). Therefore we can find \(r,s \in R \backslash \mathfrak {m}\) such that \(rs \in \mathfrak {m}\). Now the ideals \((r)+\mathfrak {m}\) and \((s)+\mathfrak {m}\) are both larger then \(\mathfrak {m}\) so must contain a product of prime ideals, but then so does their product \(((r)+\mathfrak {m})((s)+\mathfrak {m})\). This product is contained in \(\mathfrak {m}\) (since \(rs \in \mathfrak {m}\)) so we have a contradiction.

Let \(a \in \mathfrak {a}\) be any non-zero element. By Lemma 3.3.3 the ideal \((a)\) contains a product of prime ideals. So lets take prime ideals \(\mathfrak {p}_i\) such that \(\prod _{i=1}^m \mathfrak {p}_i \subset (a)\) with \(m\) as small as possible. Now, since every proper ideal is contained in a maximal ideal \(\mathfrak {p}\) (see Proposition 1.1.22), which is also a prime ideal (as maximal ideals are always prime). Therefore \(\mathfrak {a}\subset \mathfrak {p}\), so \(\prod _i \mathfrak {p}_i \subset \mathfrak {p}\).

Now \(\mathfrak {p}\) must contain some \(\mathfrak {p}_i\) since if not, we can take \(a_i \in \mathfrak {p}_i \backslash \mathfrak {p}\) then \(\mathfrak {p}\) contains \(\prod _i a_i\) but none of the \(a_i\) which contradicts \(\mathfrak {p}\) being a prime ideal. So after possibly relabelling we have \(\mathfrak {p}_1 \subset \mathfrak {p}\). But by Definition 3.3.1, in a Dedekind domain all prime ideals are maximal, so \(\mathfrak {p}=\mathfrak {p}_1\).

Since \(m\) was taken as small as possible, \((a)\) can’t contain any smaller product of prime ideals, so we can find \(b \in (\prod _{i=2}^m \mathfrak {p}_i) \backslash (a)\).

We claim that taking \(x=\frac{b}{a}\) gives the result: First note \(x \in K \backslash R\) since otherwise \(\frac{b}{a}=c \in R\) therefore \(b=ac \in (a)\) which contradicts our choice of \(b\). Moreover, we claim that \(x \mathfrak {a}\subset R\). To see this note that since \(b \mathfrak {p}\subset \prod _i \mathfrak {p}_i \subset (a)\) we have \(x \mathfrak {p}\subset R\) and therefore since \(\mathfrak {a}\subset \mathfrak {p}\) we have \(x\mathfrak {a}\subset x\mathfrak {p}\subset R\).

Let \(\alpha \in \mathfrak {a}\backslash 0\) and let

This is again an ideal and it is non-zero since \(\alpha \neq 0\). By definition we have

so we need to show equality.

For this consider the set \(\mathfrak {c}=\frac{1}{\alpha }\mathfrak {a}\mathfrak {b}\). This is a subset of \(R\) by \((\dagger )\) and is in fact an ideal. Now, if \(\mathfrak {c}=R\) then \(\mathfrak {a}\mathfrak {b}=(\alpha )\) and we are done. So assume for contradiction that \(\mathfrak {c}\) is a proper ideal. Then by Lemma 3.3.4 we can find \(x \in K \backslash R\) with \(K\) the fraction field of \(R\) and \(x \mathfrak {c}\subset R\).

Next, we note that \(\mathfrak {c}\) contains \(\mathfrak {b}\) since \(\alpha \in \mathfrak {a}\), therefore \(x \mathfrak {b}\subset x \mathfrak {c}\subset R\). Moreover, since \(x\mathfrak {c}=\frac{x}{\alpha }\mathfrak {b}\mathfrak {a}\subset R\) we have \(x\mathfrak {b}\mathfrak {a}\subset (\alpha )\). Now, if we look back at the definition of \(\mathfrak {b}\) we see that since \(x \mathfrak {b}\subset R\) we have \(x\mathfrak {b}\subset \mathfrak {b}\).

If we let \(\beta _1,\dots ,\beta _n\) be a generating set for \(\mathfrak {b}\), we can use \(x\mathfrak {b}\subset \mathfrak {b}\) to think of multiplication by \(x\) on \(\mathfrak {b}\) as a matrix \(A_x\) defined by

\(A_x\) is a \(n \times n\) matrix over \(R\) which has \(x\) as an eigenvalue. This means, \(x\) satisfies a monic polynomial with coefficients in \(R\). But \(R\) is integrally closed, so \(x \in R\) which gives a contradiction as we took \(x \in K\backslash R\). This completes the proof.

As we had before, if \(\mathfrak {a},\mathfrak {b}\) are fractional ideals, then \(\mathfrak {a}\mathfrak {b}\) is defined as the fractional ideal generated by products of elements in \(\mathfrak {a}\) and \(\mathfrak {b}\). This is clearly associative, so we only need to check the existence of inverses.

If \(\mathfrak {a}\) is a proper ideal of \(R\), then by Theorem 3.3.5 we can find an ideal \(\mathfrak {b}\) such that \(\mathfrak {a}\mathfrak {b}=(\alpha )\) for some \(\alpha \). Then setting \(\mathfrak {a}^{-1}=\frac{1}{\alpha }\mathfrak {b}\) gives an inverse to \(\mathfrak {a}\). If \(\mathfrak {a}\) is a fractional ideal, then we can find some \(x \in R\) such that \(x \mathfrak {a}\subset R\) is an ideal (not just a fractional ideal), call it \(\mathfrak {c}\), then \(\mathfrak {a}^{-1}=x\mathfrak {c}^{-1}\) is the inverse.

Multiplying on the left by \(\mathfrak {a}^{-1}\) gives the result.

If \(\mathfrak {a}\mid \mathfrak {b}\) then by definition we have \(\mathfrak {b}=\mathfrak {c}\mathfrak {a}\), which means \(\mathfrak {b}\subset \mathfrak {a}\). Conversely, assume that \(\mathfrak {b}\subset \mathfrak {a}\). Then by Theorem 3.3.5 we can find an ideal \(\mathfrak {m}\) such that \(\mathfrak {a}\mathfrak {m}=(\alpha )\). Now, \(\mathfrak {m}\mathfrak {b}\subset \mathfrak {a}\mathfrak {m}=(\alpha )\) therefore \(\mathfrak {c}=\frac{1}{\alpha }\mathfrak {m}\mathfrak {b}\) is again in \(R\) and moreover it is an ideal (since, for example it is the product of two fractional ideals). The result follows from then noting that \(\mathfrak {a}\mathfrak {c}=\mathfrak {b}\).

Lets begin by showing that every ideal can be written as a product of prime ideal. Assume for contradiction that this is not the case. Then the set of proper ideals which are not a product of prime ideals, must have a maximal element by Definition 3.2.1 \((3)\). Let \(\mathfrak {b}\) be this maximal element. Then \(\mathfrak {b}\) must be contained in some maximal ideal \(\mathfrak {p}\) of \(R\) (which we recall is also prime). Then by Corollary 3.3.14 we have \(\mathfrak {b}=\mathfrak {p}\mathfrak {c}\).

This implies \(\mathfrak {b}\subset \mathfrak {c}\) and this must be a strict containment as otherwise if \(\mathfrak {c}=\mathfrak {b}\) then \(\mathfrak {b}=\mathfrak {b}\mathfrak {p}\) which by Corollary 3.3.13 would mean \(\mathfrak {p}=R\), which cannot happen as \(\mathfrak {p}\) is a proper ideal.

Now, since \(\mathfrak {c}\) is larger than \(\mathfrak {b}\) we must have \(\mathfrak {c}\) being a product of prime ideals. So then \(\mathfrak {p}\mathfrak {c}=\mathfrak {b}\) is also a product of prime ideals contradicting our assumption.

Lets now prove that the representation as a product of prime ideals is unique. Suppose we have

with \(\mathfrak {p}_i,\mathfrak {q}_i\) not necessarily distinct prime ideals. Then \(\mathfrak {q}_1\dots \mathfrak {q}_m \subset \mathfrak {p}_1\), which means \(\mathfrak {p}_1\) contains some \(\mathfrak {q}_i\) (see the proof of Lemma 3.3.4 to see why this is true). By relabelling the \(\mathfrak {q}_i\) we can assume \(\mathfrak {q}_1 \subset \mathfrak {p}_1\). But since all prime ideals in a Dedekind ring are maximal we must have \(\mathfrak {q}_1=\mathfrak {p}_1\). So we can cancel this from each side of the equality to get

Continuing like this we get \(n=m\) and after relabelling \(\mathfrak {p}_i=\mathfrak {q}_i\), which completes the proof.

We know \(\mathcal{O}_K\) is a Dedekind domain, so the result follows.

Corollary 3.3.14 tells us that division turns into containment for ideals, so the greatest common divisor of \(\mathfrak {a},\mathfrak {b}\) is the smallest ideal containing both \(\mathfrak {a},\mathfrak {b}\) which by definition is \(\mathfrak {a}+\mathfrak {b}\).

Similarly, the least common multiple is the largest ideal contained in both of them, which by definition is \(\mathfrak {a}\cap \mathfrak {b}\).

Consider the ring homomorphism

given by \(\phi (x)=(x \pmod\mathfrak {a}, x \pmod\mathfrak {b})\). Then the kernel is given by \(\mathfrak {a}\cap \mathfrak {b}\).

Now, since \(\mathfrak {a}+\mathfrak {b}=(1)\) then

Moreover, its easy to see \(\mathfrak {a}\mathfrak {b}\subset \mathfrak {a}\cap \mathfrak {b}\), therefore \(\mathfrak {a}\mathfrak {b}=\mathfrak {a}\cap \mathfrak {b}\). So we have an injective ring homomorphism

It remains to check it is surjective. Write \(1=a+b\) for \(a \in \mathfrak {a},b \in \mathfrak {b}\), then for any \((r,s) \in R^2\) the element \(x=as+br \in R\) is such that \(x \equiv r \pmod\mathfrak {a}\) and \(x \equiv s \pmod\mathfrak {b}\) which gives us subjectivity.

We will find a \(\beta \) such that \(\mathfrak {a}=\gcd ((\alpha ),(\beta ))=(\alpha )+(\beta )=(\alpha ,\beta )\).

Let

with \(\mathfrak {p}_i\) distinct prime ideals. Then since \((\alpha ) \subset \mathfrak {a}\) we have \((\alpha ) \subset \mathfrak {p}_i^{n_i}\) for all \(i\), in other words \(\alpha \) is divisible by the \(\mathfrak {p}_i^{n_i}\). Now, let \(\mathfrak {q}_1,\dots ,\mathfrak {q}_s\) be any other prime ideals dividing \((\alpha )\) (if any exist). So we have

We construct \(\beta \) as follows. Take \(\beta _i \in \mathfrak {p}_i^{n_i} \backslash \mathfrak {p}_i^{n_i+1}\) and then use the Chinese remainder theorem 3.3.21 to find \(\beta \in R\) such that \(\beta \equiv \beta _i \pmod{\mathfrak {p}_i^{n_i+1}}\) for \(i \in \{ 1,\dots ,r\} \) and \(\beta \equiv 1 \pmod{\mathfrak {q}_j}\) for \(j \in \{ 1,\dots ,s\} \). Note that we can do this, and the \(\mathfrak {p}_i,\mathfrak {q}_j\) are all pairwise distinct so \(\mathfrak {q}_j+\mathfrak {p}_i^{n_i}=(1)\) (as \(\mathfrak {q}_j\) is maximal) for all \(i,j\) and moreover, \(\mathfrak {p}_i^{n_i}+\mathfrak {p}_j^{n_j}=\gcd (\mathfrak {p}_i^{n_i},\mathfrak {p}_j^{n_j})\) which by Definition 3.3.18 is just \((1)\).

So now, Definition 3.3.18 gives us \(\gcd ((\alpha ),(\beta ))=\prod _i \mathfrak {p}_i^{n_i}=\mathfrak {a}\) and therefore \(\beta \in \mathfrak {a}\) which finishes the proof.

If \(R\) is a Dedekind domain that is a PID, then by Proposition 1.2.6 it is a UFD.

Now, let \(R\) be a UFD and assume for contradiction it is not a PID. Then by Theorem 3.3.15 there must exist at least one non-principal prime ideal, call it \(\mathfrak {p}\). Now, let \(S\) be the set of ideals \(\mathfrak {a}\) such that \(\mathfrak {a}\mathfrak {p}\) is principal. By Theorem 3.3.5 \(S\) is non-empty, so we can do the usual trick and find a maximal element, \(\mathfrak {m}\). Let \(\mathfrak {m}\mathfrak {p}=(\alpha )\), we claim \(\alpha \) must be irreducible. Assuming this for the moment, if we take \(a \in \mathfrak {p}\backslash (\alpha )\) and \(b \in \mathfrak {m}\backslash (\alpha )\) (which we can do in the first case as \(\mathfrak {p}\) is not principal and in the second by Exercise 3.3.6) then \(ab \in (\alpha )\) but \(\alpha \mid ab\) but \(\alpha \nmid a\) and \(\alpha \nmid b\) which cannot happen in a UFD.

So it remains to prove the claim that \(\alpha \) is irreducible. For this we note that if \(\alpha =\beta \gamma \) then one of \((\beta )\) or \((\gamma )\) would be of the form \(\mathfrak {p}\mathfrak {b}\) for some \(\mathfrak {b}\) dividing \(\mathfrak {m}\), but \(\mathfrak {m}\) is maximal so \(\mathfrak {b}=\mathfrak {m}\) and therefore one of \(\beta ,\gamma \) is a unit.