The ring $\Bbb Z\left [\frac{-1+\sqrt{-19}}{2}\right ]$ is not a Euclidean domain

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?

Solution 1:

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 $23$?

Solution 2:

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 $x$.

Solution 3:

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.)