# Euclidean Domains - Dummit and Foote - Chapter 8 - Section 8.1 - Example on Quadratic

• Oct 12th 2012, 05:22 AM
Bernhard
Euclidean Domains - Dummit and Foote - Chapter 8 - Section 8.1 - Example on Quadratic
I am reading Dummit and Foote Chapter 8, Section 8.1 - Euclidean DOmains

I am working through Example 2 on page 273 (see attachment)

Example 2 demonstrates that the quadratic integer ring $\mathbb{Z} [ \surd -5 ]$ is not a Euclidean domain.

I can follow the argument down to the point where D&F state (see attachment)

"Multiplying both sdes by $2 - \surd -5$ would then imply that $2 - \surd -5$ is a multiple of 3 in R, a contradiction"

================================================== ====================================

Peter
• Oct 12th 2012, 07:55 AM
johnsomeone
Re: Euclidean Domains - Dummit and Foote - Chapter 8 - Section 8.1 - Example on Quadr
$1 \in I = (3, 2 + \sqrt{-5})$ implies $\exists \gamma, \delta \in \mathbb{Z}[\sqrt{-5}]$ such that $3 \gamma + (2 + \sqrt{-5}) \delta = 1$.

Multiply both sides by $2 - \sqrt{-5}$, getting:

$3(2 - \sqrt{-5})\gamma + (2 + \sqrt{-5}) (2 - \sqrt{-5}) \delta = 2 - \sqrt{-5}$, so

$3(2 - \sqrt{-5})\gamma + 9 \delta = 2 - \sqrt{-5}$, so

$3 \{(2 - \sqrt{-5})\gamma + 3 \delta \} = 2 - \sqrt{-5}$.

Thus $3 \alpha = 2 - \sqrt{-5}$, where $\alpha = (2 - \sqrt{-5})\gamma + 3 \delta \in \mathbb{Z}[\sqrt{-5}]$.

Thus $3$ divides $(2 - \sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$.

But that's impossible because $3x = 2$ has no solution in $\mathbb{Z}$.

(In detail, if $\alpha = x + y\sqrt{-5}, x, y\in \mathbb{Z}$, then $3\alpha = 3x + 3y\sqrt{-5}$, so

$3 \alpha = 2 - \sqrt{-5}$ implies $3x + 3y\sqrt{-5} = 2 - \sqrt{-5}$ implies $3x = 2, 3y =-1$.)

Therefore $1 \notin I$.



There's a tiny mistake in the proof of Proposition1. It should read "by the Well Ordering of $\mathbb{N}$".
• Oct 12th 2012, 11:19 AM
Deveno
Re: Euclidean Domains - Dummit and Foote - Chapter 8 - Section 8.1 - Example on Quadr
Euclidean domains are a very restrictive class of rings. They are "almost" fields. In particular, they are: unique factorization domains, greatest common divisor domains, and principal ideal domains.

So if a given ring lacks one of these properties, we can conclude it is NOT a Euclidean domain. In this case, D&F choose to show that $\mathbb{Z}[\sqrt{-5}]$ is not a PID.

One can also show R is not a UFD:

9 = 3*3
9 = (2+√(-5))(2-√(-5))

are two distinct factorizations of 9 (that is 3 is not a factor of either 2+√(-5) or 2-√(-5)), which is equivalent to showing that 3 is not prime in R (3 divides a product ab, but divides neither a nor b). However, 3 IS irreducible in R, and in a Euclidean domain "irreducibles = primes" (the same norm N can be used to show that 3 is irreducible:

if 3 = (a+b√(-5))(c+d√(-5)), then N(3) = 9, so we have either:

N(a+b√(-5)) = 1,3 or 9. if N(a+b√(-5)) = 1, then a = ±1, b = 0, in which case a+b√(-5) = ±1 is a unit. A similar proof show c+d√(-5) = ±1 if N(a+√(-5)) = 9. so if both a+b√(-5) and c+d√(-5) are to be non-units, we must have N(a+b√(-5)) = 3. this means a2+5b2 = 3, for INTEGERS a,b, so |b| < 1, and is thus 0, and a2 = 3 has no integer solution).

In general, it is more convenient to characterize rings by the properties of ideals, rather than elements (by analogy to groups, where we characterize groups by the behavior of normal subgroups: that is, which factor groups we can form from them). In fact, the word "ideal" comes from the term "ideal numbers" which were first studied in quadratic extension rings of the integers (perhaps motivated by a desire to solve Fermat's Last Theorem) as "generalizations" of "prime numbers" in ordinary arithmetic (integers). The general construction is this:

One starts with Q, the rational numbers, and adjoins a root of a quadratic polynomial with integer coefficients, so one gets Q(a). then one considers the sub-ring Z[a]. The ring-theoretic properties depend on a, for some choices we get a Euclidean domain, for some we do not. The general idea is to extend "number theory" to such rings as much as possible. The "euclidean" definition of primes: p is a prime iff p|ab implies p|a or p|b generalizes to a prime ideal: ab in P implies a in P or b in P. If R is a Euclidean domain (the nicest situation), then R is a PID, and the prime ideals P are generated by prime elements p: P = (p).

In the case at hand, the polynomial is x2 + 5 in Q[x]. Z[√(-5)] are the "integers" of the field Q(a), where a is a root of that polynomial. because this ring is non-euclidean, "factoring" isn't as helpful as it could be (we cannot say that just because something is irreducible, it is prime, so divisibility arguments can go astray).
• Oct 12th 2012, 02:57 PM
Bernhard
Re: Euclidean Domains - Dummit and Foote - Chapter 8 - Section 8.1 - Example on Quadr
Thank you for these posts

Most helpful for those like me engaged in self-study of mathematics

Peter
• Oct 12th 2012, 03:01 PM
Bernhard
Re: Euclidean Domains - Dummit and Foote - Chapter 8 - Section 8.1 - Example on Quadr
Deveno ,

thanks for the considerable help

Working through the detail of your post now - your post is much appreciated

Peter
• Oct 12th 2012, 03:46 PM
Bernhard
Re: Euclidean Domains - Dummit and Foote - Chapter 8 - Section 8.1 - Example on Quadr
Deveno,

I followed your post except for the following point - you write:

"N(a+b√(-5)) = 1,3 or 9. if N(a+b√(-5)) = 1, then a = ±1, b = 0, in which case a+b√(-5) = ±1 is a unit. A similar proof show c+d√(-5) = ±1 if N(a+√(-5)) = 9. so if both a+b√(-5) and c+d√(-5) are to be non-units, we must have N(a+b√(-5)) = 3. this means a2+5b2 = 3, for INTEGERS a,b, so |b| < 1, and is thus 0, and a2 = 3 has no integer solution)."

My question is "Why do both a+b√(-5) and c+d√(-5) have to be non-units?"

[Apologies ... I suspect my question is rather basic ... but I have only just now skimmed the material on UFDs and have not covered them properly]

Peter
• Oct 13th 2012, 12:39 PM
Deveno
Re: Euclidean Domains - Dummit and Foote - Chapter 8 - Section 8.1 - Example on Quadr
in a unique factorization domain, the factorizations are only unique up to units.

for example, in Z, we have: 6 = 2*3 = 1*1*2*3, but we don't really consider these "different" because 1 is a unit. this is the logic behind the rule: "1 is not a prime number".

again, in say, Q[x], when we factor 4x2 - 4, we have (2x + 2)(2x - 2), AND (4)(x + 1)(x - 1), but these aren't considered "different" because 4 is a unit in Q.

in a ring with unity (which you have to have to even define units (invertible multiplicative elements)), any element r can ALWAYS be written r = u(u-1r), for any unit u.

the definition of an irreducible in a ring R is something that cannot be written as the product of 2 non-units. that is:

u in R is irreducible if u = ab with a,b in R, implies a or b is a unit.

prime elements are irreducible, but it is not always true that irreducible elements are prime. for example: 3 in Z[√(-5)].