Let $\alpha = \frac{1+\sqrt{-19}}{2}$. Let $A = \mathbb Z[\alpha]$. Let's assume that we know that its invertibles are $\{1,-1\}$. During an exercise we proved that:

Lemma: If $(D,g)$ is a Euclidean domain such that its invertibles are $\{1,-1\}$, and $x$ is an element of minimal degree among the elements that are not invertible, then $D/(x)$ is isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.

Now the exercise asks:

Prove that $A$ is not a Euclidean Domain.

Everything hints to an argument by contradiction: let $(A, d)$ be a ED and $x$ an element of minimal degree among the non invertibles we'd like to show that $A/(x)$ is not isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.

How do we do that? My problem is that, since I don't know what this degree function looks like, I don't know how to choose this $x$!

I know that the elements of $A/(x)$ are of the form $a+(x)$, with $a$ of degree less than $x$ or zero. By minimality of $x$ this means that $a\in \{0, 1, -1\}$. Now I'm lost: how do we derive a contradiction from this?

Bill Dubuque
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Jacopo Notarstefano
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5 Answers5


You might find enlightening the following sketched proof that $\, \mathbb Z[w],\ w = (1 + \sqrt{-19})/2\,$ is a non-Euclidean PID -- based on a sketch by the eminent number theorist Hendrik W. Lenstra.

Note that the proof in Dummit & Foote uses the Dedekind-Hasse criterion to prove it is a PID, and the universal side divisor criterion to prove it is not Euclidean is probably the simplest known. The so-called universal side divisor criterion is essentially a special case of research of Lenstra, Motzkin, Samuel, Williams et al. that applies in much wider generality to Euclidean domains. You can obtain a deeper understanding of Euclidean domains from the excellent surveys by Lenstra in Mathematical Intelligencer 1979/1980 (Euclidean Number Fields 1,2,3) and Lemmermeyer's superb survey The Euclidean algorithm in algebraic number fields. Below is said sketched proof of Lenstra, excerpted from George Bergman's web page.

Let $\,w\,$ denote the complex number $\,(1 + \sqrt{-19})/2,\,$ and $\,R\,$ the ring $\, \Bbb Z[w].$ We shall show that $\,R\,$ is a principal ideal domain, but not a Euclidean ring. This is Exercise III.3.8 of Hungerford's Algebra (2nd edition), but no hints are given there; the proof outlined here was sketched for me (Bergman) by H. W. Lenstra, Jr.

$(1)\ $ Verify that $\, w^2\! - w + 5 = 0,\,$ that $\,R = \{m + n\ a\ :\ m, n \in \mathbb Z\} = \{m + n\ \bar a\ :\ m, n \in \mathbb Z\},\,$ where the bar denotes complex conjugation, and that the map $\,x \to |x|^2 = x \bar x\,$ is nonnegative integer-valued and respects multiplication.

$(2)\ $ Deduce that $\,|x|^2 = 1\,$ for all units of $\,R,\,$ and using a lower bound on the absolute value of the imaginary part of any nonreal member of $\,R,\,$ conclude that the only units of $\,R\,$ are $\pm 1.$

$(3)\ $ Assuming $\,R\,$ has a Euclidean function $\,h,\,$ let $\,x\ne 0\,$ be a nonunit of $\,R\,$ minimizing $\, h(x).\,$ Show that $\,R/xR\,$ consists of the images in this ring of $\,0\,$ and the units of $\,R,\,$ hence has cardinality at most $\,3.\,$ What nonzero rings are there of such cardinalities? Show $\,w^2 - w + 5 = 0 \,$ has no solution in any of these rings, and deduce a contradiction, showing that $\,R\,$ is not Euclidean.

We shall now show that $\,R\,$ is a principal ideal domain. To do this, let $\,I\,$ be any nonzero ideal of $\,R,\,$ and $\,x\,$ a nonzero element of $\,I\,$ of least absolute value, i.e., minimizing the integer $\,x \bar x.\,$ We shall prove $\,I = xR.\,$ (Thus, we are using the function $\,x \to x \bar x\,$ as a substitute for a Euclidean function, even though it doesn't enjoy all such properties.)

For convenience, let us "normalize" our problem by taking $\,J = x^{-1}I.\,$ Thus, $\,J\,$ is an $\,R$-submodule of $\,\mathbb C,\,$ containing $\,R\,$ and having no nonzero element of absolute value $< 1.\,$ We shall show from these properties that $\, J - R = \emptyset,\,$ i.e. that $\,J = R.$

$(4)\ $ Show that any element of $\,J\,$ that has distance less than $\,1\,$ from some element of $\,R\,$ must belong to $\,R.\,$ Deduce that in any element of $\,J - R,\,$ the imaginary part must differ from any integral multiple of $\,\sqrt{19}/2\,$ by at least $\,\sqrt{3}/2.\,$ (Suggestion: draw a picture showing the set of complex numbers which the preceding observation excludes. However, unless you are told the contrary, this picture does not replace a proof; it is merely to help you find a proof.)

$(5)\ $ Deduce that if $\, J - R\,$ is nonempty, it must contain an element $\,y\,$ with imaginary part in the range $\,[\sqrt{3}/2,\,\sqrt{19}/2 - \sqrt{3}/2],\,$ and real part in the range $\, (-1/2,\,1/2].$

$(6)\ $ Show that for such a $\, y,\,$ the element $\, 2y\,$ will have imaginary part too close to $\,\sqrt{19}/2\,$ to lie in $\, J - R.\,$ Deduce that $\,y = w/2\,$ or $\,-\bar w/2,\,$ and hence that $\,w\,\bar w/2\,\in J.$

$(7)\ $ Compute $\, w\,\bar w/2,\,$ and obtain a contradiction. Conclude that $\,R\,$ is a principal ideal domain.

$(8)\ $ What goes wrong with these arguments if we replace $19$ throughout by $17$? By $23$?

Bill Dubuque
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  • @BillDubuque "Thus, we are using the function $x\rightarrow x\bar x$ as a substitute for a Euclidean function, even though it doesn't enjoy all such properties." What other properties does Euclidean function enjoy? Also how to do step $4$? – Turbo Jan 06 '17 at 02:29
  • A simlar but slightly different approach can be found in the first section of [these notes](http://websites.math.leidenuniv.nl/algebra/ant.pdf), for the most part in the exercises. – This site has become a dump. Apr 06 '19 at 14:34
  • @Servaes Stevenhagen was a [student of Lenstra](https://www.genealogy.math.ndsu.nodak.edu/id.php?id=18964) - which likely explains the similarity. – Bill Dubuque Apr 06 '19 at 15:15
  • @BillDubuque I've read the D&F excerpt and can't understand how using the field norm $N(a+b(1+\sqrt{-19})/2) = a^2+ab+5b^2$ to show there are no universal side divisors, proves that $\mathbb{Z}[(1+\sqrt{-19})/2]$ is not a Euclidean Domain with respect to $\color{red}{any}$ norm? Also, I don't mean to be fussy, I just simply enjoy looking for these things, nothing on you, but there is missing a bracket in $(1+\sqrt{-19}\color{aqua}{)}$ in the third paragraph. – no lemon no melon Apr 25 '21 at 23:05
  • @nolemon What is not clear to you in the proof sketched in the first paragraph of $(3)$ above? (where the universal side divisor criterion is applied). – Bill Dubuque Apr 26 '21 at 00:46
  • @Bill to be completely honest, the whole thing(embarrassingly), I've read over the final example of section 8.1 of D&F multiple times now(3 as well), I get that there are no universal side divisors through the use of $N$(field), but how does one jump from there to $R=\mathbb{Z}[(1+\sqrt{-19})/2]$ is not a ED with respect to _any_ norm. My thing is how does one go about using one norm, deducing a contradiction, and then letting it suffice for all other possible norms? _What if there were some other norm, other than the field norm for which there might be universal side divisors?_ – no lemon no melon Apr 26 '21 at 00:56
  • @Bill better yet, some other norm where $R$ is a ED? – no lemon no melon Apr 26 '21 at 01:06
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    @nolemon The argument in $(3)$ chooses a nonunit $\,x\neq 0\,$ that is minimal wrt to the purported *Euclidean* norm $\,h,\,$ i.e. $x$ has the next smallest E-norm value besides units (value $=1$). So the only possible remainders mod $x$ are $0$ or units $= \pm1$ so $\,R/x\,$ has cardinalty $2$ or $3$, contra $\,\Bbb Z/2\,$ and $\Bbb Z/3\,$ are not images of $R$ since $\,x^2-x+5\,$ has no roots in either. – Bill Dubuque Apr 26 '21 at 01:10
  • @Bill So how is the example in D&F complete then? Roots are transferred through ring homomorphisms?? – no lemon no melon Apr 26 '21 at 01:26
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    D&F use the field norm to show that the only possible side divisors are $\,x = \pm2,\,\pm 3.\,$ Yes, hom images preserve roots, applying a hom $h$ to \,$w^2-w+5 = 0\,$ yields $\,(h(w))^2-h(w)+h(5) = 0\,$ so $h(w)$ remains a root in $h(R)\ \ $ – Bill Dubuque Apr 26 '21 at 01:39

What a coincidence, this was a recent homework problem for me as well. Here's an additional hint: Show that $X^2 + X + 5$ does not split over $\mathbb{F}_2$ or $\mathbb{F}_3$. Deduce a contradiction to the minimality of the degree of $x$.

Zhen Lin
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  • $X^2-X+5$ doesn't split in $\mathbb F_{2}[X]$ or $\mathbb F_{3}[X]$ because it has no roots in those fields. But it does split in $A[X]$, thus in $A/(x)[X]$. Contradiction! I hope :) – Jacopo Notarstefano Feb 26 '11 at 15:56
  • @Jacopo: Yes, something like that. Essentially what happens is that you are forced to conclude that $A/(x)$ is the trivial ring, so $x$ must be a unit, but we already excluded all the units. – Zhen Lin Feb 26 '11 at 16:19

You don't need to know what the degree function looks like to choose $x$: you already chose it when you said "$x$ an element of minimal degree." (By the well-ordering principle, the set of degrees of nonzero, non-unital elements has a least element.)

Qiaochu Yuan
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  • I guess I should have phrased my question better. I do understand that this $x$, given a degree function, exists. I'll try to edit my question to better reflect what I was trying to say. – Jacopo Notarstefano Feb 26 '11 at 14:22
  • @Jacopo: oh. That's a completely different question. Then in fact Zhen Lin gave you the correct hint. – Qiaochu Yuan Feb 26 '11 at 15:46

I have a solution that seems different from the ones as above, I'm no too sure about whether is correct or not.

We shall show that for any $a \neq \pm 1 \in A$, we have $|A/(a)| > 3$.

Lemma: Let $R = \mathbb{Z}[\beta]$, where $\beta \in \mathbb{C}-\mathbb{R}$ and $|\beta| > 1$. Then fix $a \in R$, and consider the parallelipid $P$ spanned by $0, a, \beta a, a + \beta \alpha $, and let $b = |\partial P \cap R|$ (the boundary of $P$ interesected with $R$) and $i = |P^0 \cap R|$ (the interior of $P$ intersected with $R$. Then $|R/(a)| = \frac{b}{2} - 1 + i$

Proof: Let $r_1$ be the closed line segment joining $0$ and $a$, and $r_2$ be the half-open line segment joining $0$ and $\beta a$. Then a simple double counting shows $\frac{b}{2} - 1 = |(r_1 \cup r_2) \cap R|$. It's not hard to see for any $c \in R$, we can find one and only one $b \in r_1 \cup r_2 \cup P^0$ such that $b-c \in (a)$ (the proof is geometrically obvious but annoying to write), and thus $|R/(a)| = \frac{b}{2} - 1 + i$. $\square$

Now, returning to the main problem, let $a = m + n \alpha$. Then $\alpha a = -10n + \alpha(m+2)$. Consider the $\mathbb{R}$-linear (not $\mathbb{C}$-linear) transformation $T$ from $\mathbb{C}$ to $\mathbb{R}^2$ sending $\alpha$ to $(0,1)$ and $1$ to $(1,0)$. Then $T(a) = (m,n)$ and $T(\alpha a) = (-10n, m+2)$. Also, note that $T(A) = \{ T(r) | r \in A \} = \mathbb{Z}^2$.

By lemma (note that $|\partial (TP) \cap TR| = |\partial P \cap R|$ and $|(TP)^0 \cap TR| = |P^0 \cap R|$), let $b$ be the lattice points on the boundary of the parallelipid $P'$ spanned by $(0,0), (m,n), (-10n, m+2), (-10n+m, m+n+2)$, and $i$ be the number of interior lattice points of $P'$.

Then $|A/(a)| = \frac{b}{2}-1 + i = Area(P') = \det \begin{pmatrix} m & n \\ -10n & m+2 \end{pmatrix} = m^2 + 2m + 10n^2$ (where we have used Pick's theorerm for the second equality). Now, $|A/(a)| > 3 \Rightarrow (m+1)^2 + 10n^2 > 4$, which is true whenever $n = 0, m \not \in \{0, 1, -1, -2 \}$ or $n \neq 0$.

Thus the only case that's remaining to be checked is $(m,n) = (2,0)$, or $a = 2$. But then $A/(2)$ contains $5$ elements - viz $0, \alpha, 2 \alpha, 1, 1 + \alpha$, which is greater than $3$, so done !

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Here is a solution that may turn out to be helpful: http://www.maths.qmul.ac.uk/~raw/MTH5100/PIDnotED.pdf

Ivan Mathman
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