Journal ArticleDOI

# 101.15 An elementary proof that not all principal ideal domains are Euclidean domains

01 Jul 2017-The Mathematical Gazette (Cambridge University Press (CUP))-Vol. 101, Iss: 551, pp 289-293

### 1. Introduction

• A standard result in undergraduate algebra courses is that every Euclidean domain (ED) is a principal ideal domain (PID).
• It is routinely stated, but rarely proved, that the converse is false.
• A more elementary proof, accessible to advanced undergraduates, is given by Cámpoli [2] in 1988, though with a more restricted definition of Euclidean norm than Motzkin uses.
• Notice this does not include the property, usually included in the definition of Euclidean function, that if a|b then d(a) ≤ d(b).

### 4. Quasi-Euclidean domains

• The former contradicts the minimality of d(b), while the latter contradicts the assumption that a is not in bA.
• This suggests the following definition of a quasi-Euclidean domain (QED).
• (Note that this is not a standard definition, and there are other definitions of quasiEuclidean domains in the literature, not necessarily equivalent to this one.).

### In this situation, A is called a quasi-Euclidean domain (or Motzkin domain).

• The above proof then generalizes immediately to a proof that every QED is a PID.
• It is not obvious that the authors have gained anything, as it is not obvious that there exist quasi-Euclidean domains that are not Euclidean.

### 5. R is a principal ideal domain

• The large dots represent the elements of R in the Argand diagram.
• The small dots represent all the remaining values of a/b, that is a/b = q/2.
• Adding ordinary integers to a/b as necessary, the authors may assume that a/b = (±1 + √ 19)/4, and by symmetry these two cases are essentially the same.

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AN ELEMENTARY PROOF THAT NOT ALL PRINCIPAL IDEAL
DOMAINS ARE EUCLIDEAN DOMAINS
ROBERT A. WILSON
1. Introduction
A standard result in undergraduate algebra courses is that every Euclidean do-
main (ED) is a principal ideal domain (PID). It is routinely stated, but rarely
proved, that the converse is false. The ring R = Z[θ], where θ = (1 +
19)/2, so
that θ
2
= θ 5, is sometimes mentioned as a counterexample. The proof that R is
indeed a counterexample is due to Motzkin [1] in 1948, as a special case of much
uates, is given by ampoli [2] in 1988, though with a more restricted deﬁnition of
Euclidean norm than Motzkin uses.
The ﬁrst part of this proof, that R is not a Euclidean domain, ampoli attributes
to the referee of his paper. It is worth remarking, however, that that proof, with very
little modiﬁcation, actually proves Motzkin’s slightly more general result, namely
that R is not a Euclidean domain under the following deﬁnition.
Deﬁnition 1. A Euclidean function on an integral domain A is a function d :
A \ {0} N {0} with the property that, for all a, b A, with b 6= 0, there exist
q, r A such that a = bq + r and either d(r) < d(b) or r = 0. In this situation, A
is called a Euclidean domain.
Notice this does not include the property, usually included in the deﬁnition of
Euclidean function, that if a|b then d(a) d(b).
At the same time, it is possible to simplify the second part of ampoli’s proof,
that R is a PID, reducing his seven (or nine, depending on how you count them)
cases to three (or four). This simpliﬁcation comes from treating the problem as a
geometric problem in the Argand diagram, rather than an arithmetic problem in
the divisors of the coeﬃcients x, y of elements x + yθ of R.
2. Basic properties of R
As above, let θ = (1 +
19)/2, and R = Z[θ]. Since θ
2
= θ 5, we have that
R = {a + | a, b Z}. Since it is a subring of the complex numbers, it is an
integral domain, that is, a commutative ring in which αβ = 0 implies α = 0 or
β = 0. Since
¯
θ = 1 θ, the ring R is invariant under complex conjugation. The
ring R inherits from C the norm N(α) = α¯α = |α|
2
, which is multiplicative in the
sense that N(αβ) = N (α)N(β).
Next we show that the only units (that is, divisors of 1) in R are 1, 1. For if
a + (with a, b Z) is a divisor of 1 in R, then N(a + ) = (a + )(a + b
¯
θ) is a
Date: 11th November 2015; revised 19th January and 1st September 2016.
1

2 ROBERT A. WILSON
s s s s s s s s s
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-
x
6
y
0 1
θ
Figure 1. The proof that N is not a Euclidean function for R
divisor of 1 in Z. Since (a + )(a + b
¯
θ) = a
2
+ ab + 5b
2
= (a +
1
2
b)
2
+
19
4
b
2
, this
implies that b = 0, and therefore a = ±1.
A similar argument shows that the elements 2, 3 are irreducible in R, that is,
their only factors are unit multiples of themselves and 1. For, if a + is a proper
divisor of 2 or 3 in R, then (a +
1
2
b)
2
+
19
4
b
2
is a proper divisor of 4 or 9 in Z, so is
equal to 2 or 3. But then b = 0 so a
2
= 2 or 3, which is impossible.
3. R is not a Euclidean domain
It is easy to see that N is not a Euclidean function for R. Motzkin [1] attributes
this observation to Dedekind in 1894, but the reference is missing from [1]. We
give a proof here, as it will provide useful context for the proof below that R is
a principal ideal domain. The property that a = bq + r with either N(r) < N (b)
or r = 0 translates to saying that |a/b q| < 1, that is, every fraction a/b is at a
distance strictly less than 1 from some element q of the ring. In Figure 1 we have
drawn some circles of radius 1 centred on the elements of R to show that they do
not cover the whole plane. Algebraically, we see that, for example, if a =
19
and b = 4, then a/b has distance
19/4 > 1 from the nearest lattice point.
It is slightly more diﬃcult to show that there is no Euclidean function at all on
R. In fact Motzkin proves that Z[(1 +
1 4n)/2] is not a Euclidean domain for
any n 5. The proof we now give also generalises easily to this situation.
Theorem 1. The ring R = Z[θ], where θ = 1 +
19)/2, is not a Euclidean
domain.
Proof. Assume that d is any Euclidean function on R. Choose m A \ {0, ±1}
with d(m) as small as possible. Then there exist q, r A such that 2 = mq + r,
with either d(r) < d(m) or r = 0. The minimality of d(m) implies r = 0, 1, 1.
Hence mq = 2, 1, 3. Then irreducibility of 2, 3 implies that m = ±2, ±3.
Similarly, there exist q
0
, r
0
A such that θ = mq
0
+r
0
, with either d(r
0
) < d(m) or
r
0
= 0. Again, it follows that r
0
= 0, 1, 1, and therefore mq
0
= θ, θ1, θ+1. Now it
is easy to see that none of θ, θ ±1 is divisible by 2 or 3, so this is a contradiction.
4. Quasi-Euclidean domains
The proof given below that R is a PID is a slight generalisation of one of the
standard proofs that every Euclidean domain is a PID. Recall that an ideal I of a
commutative ring A is a nonempty subset with the property that for every a, b I

NOT EVERY PID IS A EUCLIDEAN DOMAIN 3
and r A, both a b and ar lie in I. The set bA of all multiples of b is easily
seen to be an ideal, and is called the principal ideal generated by b. If every ideal
is principal, then A is called a principal ideal domain.
Given any non-zero ideal I in a Euclidean domain A, we must ﬁnd an element
b I such that I = bA. So, pick b I such that b 6= 0 and d(b) is as small as
possible, and assume, for a contradiction, that a I \ bA. Then by the Euclidean
property, there exist q, r A with r = a bq I and either d(r) < d(b) or
r = 0. The former contradicts the minimality of d(b), while the latter contradicts
the assumption that a is not in bA.
Notice that the argument almost works more generally if we allow r = ap bq
and not insisting that p = 1. However, if p 6= 1 then the case when r = 0 does not
immediately lead to a contradiction, and an alternative argument is required. Our
method is to exclude the case r = 0 completely, except in the case p = 1, when we
cannot avoid it.
This suggests the following deﬁnition of a quasi-Euclidean domain (QED). (Note
that this is not a standard deﬁnition, and there are other deﬁnitions of quasi-
Euclidean domains in the literature, not necessarily equivalent to this one.)
Deﬁnition 2. A quasi-Euclidean function (or Motzkin function) on an integral
domain A is a function d : A \ {0} N {0} with the property that, for all
a, b A, with b 6= 0, there exist p, q, r A, with p 6= 0, such that ap = bq + r and
either
d(r) < d(b); or
p = 1 and r = 0.
In this situation, A is called a quasi-Euclidean domain (or Motzkin domain).
The above proof then generalizes immediately to a proof that every QED is a
PID. However, it is not obvious that we have gained anything, as it is not obvious
that there exist quasi-Euclidean domains that are not Euclidean. We shall resolve
this by showing that R is indeed a QED.
5. R is a principal ideal domain
In the light of the previous section, it suﬃces to show that R is a QED in the
sense deﬁned above. Working in the ﬁeld of fractions Q(
19), we can divide by
b, and use the fact that N(a)/N (b) = |a/b|
2
, so that the property d(r) < d(b)
translates to
0 <
a
b
p q
< 1.
After some initial reductions, the proof divides into three cases, in which we
shall take p = 1, p = 2, and p = θ or
¯
θ respectively. Figure 2 illustrates these
three cases. The large dots represent the elements of R in the Argand diagram.
The interiors (excluding the centres) of the large circles represent points a/b with
0 < |a/b q| < 1, for some q R, and the small circles similarly represent points
with 0 < |2a/b q| < 1, or equivalently 0 < |a/b q/2| < 1/2. The small dots
represent all the remaining values of a/b, that is a/b = q/2. The horizontal line
represents points with imaginary part
3/2, which we may take as the boundary
betweeen the cases p = 1 and p = 2.
The picture tells us what we have to do, and it only remains to ﬁll in the algebraic
details. The required result is the following.

4 ROBERT A. WILSON
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s s s s s s s s s
s s s s s s s s s
q q q q q q q q q q q q q q q q q
q q q q q q q q q q q q q q q q q
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

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


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
-
x
6
y
0 1
θ
Figure 2. A graphical illustration of the proof that R is a PID
Lemma 1. For every a, b R, with b 6= 0, and b not an exact divisor of a, there
exist p, q R such that
0 <
a
b
p q
< 1.
Proof. Since we may replace a by a
0
= a + bt, for any t R, we may add any
desired element of R to a/b, and in particular assume that the imaginary part y of
a/b = x + iy lies (weakly) between ±
19/4. By symmetry it suﬃces to consider
the case 0 y
19/4.
Now if 0 y <
3/2, then p = 1 will suﬃce, as then a/b is at distance less
than 1 from an ordinary integer. Otherwise,
3/2 y
19/4, and we try p = 2.
Writing y
0
for the imaginary part of 2a/b θ we calculate y
0
= 2y
19/2 and
then
3
19/2 y
0
0. In order to show that the distance from 2a/b to the
nearest element of R is strictly less than 1, it suﬃces to show that y
0
>
3/2.
But
19 <
27 = 3
3, so
3/2 <
3
19/2 < 0, as required. The result now
follows unless 2a/b R, which can only happen when y =
19/4.
In this special case, adding ordinary integers to a/b as necessary, we may assume
that a/b = (±1 +
19)/4, and by symmetry these two cases are essentially the
same. We now choose p = (1 +
19)/2, so that ap/b = 5/2 and if q = 2 then
|ap/b q| = 1/4.
Acknowledgements
I thank er´emy Blanc for his interest in this proof, and especially for pointing
out that in an earlier version, among other deﬁciencies, I had failed to consider the
ﬁnal case 2a/b R in the proof of Lemma 1.
References
[1] T. Motzkin, The Euclidean algorithm, Bull. Amer. Math. Soc. 55 (1949), 1142–1146.
[2] Oscar A. ampoli, A principal ideal domain that is not a Euclidean domain, Amer. Math.
Monthly 95 (1988), 868–871.
School of Mathematical Sciences, Queen Mary University of London, Mile End Road,
London E1 4NS, U.K.
##### References
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