# Criterion on whether a given ideal of a quadratic order is regular or not

Let $K$ be an algebraic number field of degree $n$. Let $\mathcal{O}_K$ be the ring of algebraic integers. Let $R$ be an order of $K$, i.e. a subring of $K$ which is a free $\mathbb{Z}$-module of rank $n$. Let $\alpha_1, \cdots, \alpha_n$ be a basis of $R$ as a $\mathbb{Z}$-module. Let $D =$ det$(Tr_{K/\mathbb{Q}}(\alpha_i\alpha_j))$. It is easy to see that $D$ is independent of a choice of a basis of $R$. We call $D$ the discriminant of $R$. Let $I$ be an ideal of $R$. Let $\mathfrak{f} = \{x \in R | x\mathcal{O}_K \subset R\}$. If $I + \mathfrak{f} = R$, we call $I$ regular. Since regular ideals have nice properties as shown here, it is desirable to have a handy criterion of whether a given ideal of $R$ is regular or not. I came up with the following proposition.

Proposition Let $K$ be a quadratic number field, $d$ its discriminant. Let $R$ be an order of $K$, $D$ its discriminant. It is easy to see that there exists an integer $f \gt 0$ such that $D = f^2d$. Let $I$ be a non-zero ideal of $R$. By the result of this question, there exist unique integers $a, b, c$ such that $I = \mathbb{Z}a + \mathbb{Z}(b + c\frac{(D+ \sqrt D)}{2}), a \gt 0, c \gt 0, 0 \le b \lt a, a \equiv 0$ (mod $c$), $b \equiv 0$ (mod $c$). Then $I$ is regular if and only if gcd$(a, f) = 1$.

Outline of my proof I used the result of this question and this question. A full proof was posted as an answer below.

My question How do you prove the proposition? I would like to know other proofs based on different ideas from mine. I welcome you to provide as many different proofs as possible. I wish the proofs would be detailed enough for people who have basic knowledge of introductory algebraic number theory to be able to understand.

• Why don't you just set up a blog? Anyone interested will surely look at it!
– Pedro
Nov 9, 2013 at 22:13
• @PedroTamaroff I'm asking the alternative proofs. Nov 9, 2013 at 22:15
• While the SE policy encourages a user to ask and answer his/her own question, why bother setting up a blog? Nov 9, 2013 at 23:49
• The point is that you don't have a question, you already know the answer and all, so you can just set up a blog, share this to anyone who is interested and make it is detailed or lengthy as you want to.
– Pedro
Nov 9, 2013 at 23:54
• @PedroTamaroff I'm asking the alternative proofs. Even if I didn't have a question as you say(this is not correct), what's wrong with it? As I wrote in the remark, one of my intentions is to use the result to answer other questions(not necessarily mine) in this site. Nov 10, 2013 at 0:15

Let $K, R, f$ be as in the proposition. By this question, $f$ is the order of the group $\mathcal{O}_K/R$. Hence, by this question, $\mathfrak{f} = f\mathcal{O}_K$

Lemma 1(Lying-over theorem) Let $B$ be a commutative ring. Let $A$ be a subring of $B$. Suppose $B$ is integral over $A$. Let $P$ be a prime ideal of $A$. Then there exists a prime ideal $Q$ of $B$ lying over $P$, i.e. $P = Q \cap A$.

Proof: See any introductory textbook on commutative algebra or Lemma 7.33.15 of this online book.

Lemma 2 Let $R$ be an order of an algebraic number field. Let $I$ be a non-zero ideal of $R$. Then $R/I$ is a finite ring.

Proof: Let $\alpha$ be a non-zero element of $I$. Since $\alpha R \subset I$, $I$ is a free $\mathbb{Z}$-module of the same rank as $R$. Hence $R/I$ is a finite ring.

Lemma 3 Let $R, f$ be as in the proposition. Let $P \ne 0$ be a prime ideal of $R$. Suppose $f \in P$. Then $f\mathcal{O}_K \subset P$.

Proof: By Lemma 1, there exists a prime ideal $Q$ of $\mathcal{O}_K$ lying over $P$. Then $f\mathcal{O}_K \subset Q$. Since $f\mathcal{O}_K$ is an ideal of $R$, $f\mathcal{O}_K \subset Q \cap R = P$.

Lemma 4 Let $R, f$ be as in the proposition. Let $P \ne 0$ be a prime ideal of $R$. Then $P$ is regular if and only if $P$ does not contain $f$.

Proof: By Lemma 2, $R/P$ is a field. Hence $P$ is a maximal ideal of $R$.

Suppose $P$ is not regular. Then $P + f\mathcal{O}_K \ne R$. Hence there exists a maximal ideal $M$ of $R$ such that $P + f\mathcal{O}_K \subset M$. Since $P$ is maximal, $P = M$. Hence $f \in P$.

Conversely suppose $f \in P$. By Lemma 3, $P$ is not regular. QED

Lemma 5 Let $R, f$ be be as in the proposition. Let $I$ be a non-zero ideal of $R$. Then $I$ is regular if and only if every prime ideal $P$ containing $I$ is regular.

Proof: Suppose $I$ is regular. Let $P$ be a prime ideal of $R$ containing $I$. Suppose $P$ is not regular. By Lemma 4, $f \in P$. Then $f\mathcal{O}_K \subset P$ by Lemma 3. Hence $I + f\mathcal{O}_K \subset P$. Hence $I$ is not regular. This is a contradiction.

Conversely suppose $P$ is a non-regular prime ideal containing $I$. Since $I + f\mathcal{O}_K \subset P + f\mathcal{O}_K \ne R$, $I$ is not regular. QED

Proof of the proposition Let $a = ca', b = cb'$. Then $I = cJ$, where $J = \mathbb{Z}a' + \mathbb{Z}(b' + \omega)$.

Suppose $I$ is regular. We claim that gcd$(a, f) = 1$. Suppose this is not the case. Let $p$ be a prime divisor of gcd$(a, f)$. We first consider the case that $c$ is divisible by $p$. By Lemma 1, there exists a prime ideal $P$ lying over $p$. Since $c \in P$, $I \subset P$. Since $P$ is not regular by Lemma 4, $I$ is not regular by Lemma 5. This is a contradiction. Nexe we consider the case that $a'$ is divisible by $p$. Since $I \subset J$, $J$ is regular. Hence, by this question, there exists a prime ideal $P$ dividing $J$ and lying over $p$. Since $P$ is not regular by Lemma 4, $I$ is not regular by Lemma 5. This is a contradiction. Hence gcd$(a, f) = 1$.

Conversely suppose $I$ is not regular. By Lemma 5, there exists a non-regular prime ideal $P$ such that $I \subset P$. Let $p$ be the unique prime number such that $p\mathbb{Z} = P \cap \mathbb{Z}$. Since $P$ is non-regular, $f$ is divisible by $p$ by Lemma 4. Since $a \in I \subset P$, $a$ is divisle by $p$. Hence gcd$(a, f)$ is divisible by $p$. QED

• This is an example of a style of writing (multiple formal lemmas and proofs, no other prose or explanation) that is generally not in line with the norms of this site. This site is not intended to be a place to record long sequences of lemmas and theorems. I also note that you deleted another answer of yours to this question, which was voted -3, when you could have edited that instead of deleting it and posting a new answer. Nov 12, 2013 at 23:48
• @CarlMummert As you know we had discussion on this matter on a meta thread. [I also note that you deleted another answer of yours to this question, which was voted -3, when you could have edited that instead of deleting it and posting a new answer.] It was deleted by the system because it was too short. I could edit it but could not undelete it. Nov 14, 2013 at 1:41
• I am certain I have pointed out before the problem with this type of answer, although not in that meta thread. Would you prefer in the future if I refrain from commenting if I downvote for this reason? Nov 14, 2013 at 1:45
• I have explained it before in comments to previous answers that had the same issue. Perhaps you might look at a broad collection of answers not written by you and emulate the overall style you see there - which will be more explanatory, and convey the "big ideas" better, while not dwelling on trivialities. The style in which this answer is written might be suitable for a formal paper or book, but it's not ideal for an online question-and-answer site. So, if this is the sort of answer you are interested in writing, you should explore other publication options, such as a blog or the arXiv. Nov 14, 2013 at 12:44
• I would say that downvoted based on the entire context visible on this page, knowing that I have raised this issue (using the site as a notebook) in comments before. See my comment dated Dec 21 '12 at 1:58 in meta.math.stackexchange.com/questions/6927/…, and the following comments there, and see math.stackexchange.com/questions/261114/… . I have nothing more to add. Nov 14, 2013 at 13:14