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
more general results. A more elementary proof, accessible to advanced undergrad
uates, is given by C´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, C´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 ab then d(a) ≤ d(b).
At the same time, it is possible to simplify the second part of C´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 + bθ  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 + bθ (with a, b ∈ Z) is a divisor of 1 in R, then N(a + bθ) = (a + bθ)(a + b
¯
θ) is a
Date: 11th November 2015; revised 19th January and 1st September 2016.
1
2 ROBERT A. WILSON
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Figure 1. The proof that N is not a Euclidean function for R
divisor of 1 in Z. Since (a + bθ)(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 + bθ 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, θ−1, 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. QuasiEuclidean 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 nonzero 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 quasiEuclidean 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 quasiEuclidean 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 quasiEuclidean 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 quasiEuclidean 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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q q q q q q q q q q q q q q q q q
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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 J´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. C´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.
Email address: R.A.Wilson@qmul.ac.uk