Euclidean domain

Euclidean domain

In abstract algebra, a Euclidean domain (also called a Euclidean ring) is a type of ring in which the Euclidean algorithm applies.

A Euclidean domain is a specific type of integral domain, and can be characterized by the following (not necessarily exhaustive) chain of class inclusions:

* integral domains ⊃ unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields


Formally, a Euclidean domain is an integral domain "D" on which one can define a function "v" mapping nonzero elements of "D" to non-negative integers that satisfies the following division-with-remainder property:

*If "a" and "b" are in "D" and "b" is nonzero, then there are "q" and "r" in "D" such that "a" = "bq" + "r" and either "r" = 0 or "v"("r") < "v"("b").

The function "v" is called a "valuation" or "norm" or "gauge" and the key point here is that the remainder "r" has "v"-size smaller than the "v"-size of the divisor "b". The operation mapping ("a", "b") to ("q", "r") is called the Euclidean division, whereas q is called the Euclidean quotient.

Nearly all algebra textbooks which discuss Euclidean domains include the following extra property in the definition:

*for all nonzero "a" and "b" in "D", "v"("a") &le; "v"("ab").

This property does not have to be assumed since it is not needed to prove the most basic facts about Euclidean domains (see below). However, this inequality can always be arranged to occur by changing the choice of "v", as follows: if ("D","v") is a Euclidean domain as given above then the function "w" defined on nonzero elements of "D" by "w"("a") = least value of "v"("ax") as "x" runs over nonzero elements of "D" also makes "D" a Euclidean domain according to the above definition and it satisfies "w"("a") &le; "w"("ab") for all nonzero "a" and "b" in "D".

To check that "w" is a norm, suppose that "b" does not divide "a" and, amongst all expressions of the form "a" = "bq" + "r", choose one for which "v"("r") is minimal. If "w"("r") ≥ "w"("b"), then"v"("r") ≥"v"("bc") for some "c". We can write "a" = "bcQ" + "R" with "v"("R") < "v"("bc") ≤ "v"("r"), which contradicts the minimality of "v"("r").


Examples of Euclidean domains include:
*Z, the ring of integers. Define "v"("n") = |"n"|, the absolute value of "n".
*Z ["i"] , the ring of Gaussian integers. Define "v"("a"+"bi") = "a"2+"b"2, the norm of the Gaussian integer "a"+"bi".
* Z [&omega;] (where &omega; is a cube root of 1), the ring of Eisenstein integers. Define "v"("a"+"b"&omega;) = "a"2-"ab"+"b"2, the norm of the Eisenstein integer "a"+"b"&omega;.
*"K" ["X"] , the ring of polynomials over a field "K". For each nonzero polynomial "f", define "v"("f") to be the degree of "f".
*"K""X", the ring of formal power series over the field "K". For each nonzero power series "f", define "v"("f") as the degree of the smallest power of "X" occurring in "f".
*Any discrete valuation ring. Define "v"("x") to be the highest power of the maximal ideal "M" containing "x" (equivalently, to the power of the generator of the maximal ideal that "x" is associated to). The case "K""X" is a special case of the above.
*Any field. Define "v"("x") = 1 for all nonzero "x".

The examples of polynomial and power series rings in one variable are the reason that the function "v" in the definition of a Euclidean domain is not assumed to be defined at 0.


The following properties of Euclidean domains do not require the inequality "v"("a") &le; "v"("ab"):

*The extended Euclidean algorithm is applicable (which is the source of the name Euclidean domain).

*Every Euclidean domain is a principal ideal domain. In fact, if "I" is a nonzero ideal of a Euclidean domain "D" and "a" is chosen to minimize "v"("a") over all nonzero elements of "I", then "I" = "aD".

*The principal ideals of elements with minimal Euclidean valuation are the entire ring, i.e. they are units. (If the inequality "v"("a") &le; "v"("ab") is assumed, all the units have this minimal valuation.)

*Every nonzero nonunit is a product of irreducibles. This follows from the corresponding result for any principal ideal domain (or Noetherian domain), though assuming the inequality "v"("a") &le; "v"("ab") would enable a direct inductive argument.

Conversely, not every PID is Euclidean, though exceptions are not easy to find.For example, for "d" = -19, -43, -67, -163, the ring of integers of Q(sqrt{d}) is a PID which is not Euclidean, but the cases "d" = -1, -2, -3, -7, -11 are Euclidean. [Citation
last = Motzkin | first = Theodore | author-link = Theodore Motzkin
title = The Euclidean algorithm
journal = Bulletin of the American Mathematical Society
volume = 55 | issue = 12 | pages = 1142-1146 | year = 1949
url =

However, many finite extensions of Q with trivial class group do have Euclidean integral rings.Assuming the extended Riemann hypothesis, if "K" is a finite extension of Q and the ring of integers of "K" is a PID with an infinite number of units, then the ring of integers is Euclidean. [Citation
last = Weinberger | first = Peter J. | authorlink = Peter J. Weinberger
title = On Euclidean rings of algebraic integers
journal = Proceedings of Symposia in Pure Mathematics
publisher = AMS
volume = 24 | pages = 321-332 | year = 1973
] In particular this applies to the case of totally real quadratic number fields with trivial class group.In addition (and without assuming ERH), if the field "K" has trivial class group and unit rank strictly greater than three, then the ring of integers is Euclidean. [Citation
last = Harper | first = Malcolm
last2 = Murty | first2 = M. Ram | author2-link = M. Ram Murty
title = Euclidean rings of algebraic integers
journal = Canadian Journal of Mathematics
volume = 56 | issue = 1 | pages = 71-76 | year = 2004
url =
] An immediate corollary of this is that if the class group is trivial and the extension has degree greater than 8 then the ring of integers is necessarily Euclidean.

ee also

*Ordinal number - these allow a kind of Euclidean division: for all "α" and "β", if "β" > 0, then there are unique "γ" and "δ" such that "α" = "β" · "γ" + "δ" and "δ" < "β"; however, the ordinals are not a Euclidean domain, since they are not even a ring (addition of ordinals, for example, is not commutative).


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