QUADRATIC RINGS
PETE L. CLARK
1.
Quadratic fields and quadratic rings
Let
D
be a squarefree integer not equal to 0 or 1.
Then
√
D
is irrational, and
Q
[
√
D
], the subring of
C
obtained by adjoining
√
D
to
Q
, is a field.
From an abstract algebraic perspective, an explanation for this can be given as
follows: since
√
D
is irrational, the polynomial
t
2

D
is irreducible over
Q
. Since
the ring
Q
[
t
] is a PID, the irreducible element
t
2

D
generates a maximal ideal
(
t
2

D
), so that the quotient
Q
[
t
]
/
(
t
2

D
) is a field. Moreover, the map
Q
[
√
D
]
→
Q
[
t
]
/
(
t
2

D
) which is the identity on
Q
and sends
√
D
→
t
is an isomorphism of
rings, so
Q
[
√
D
] is also a field.
We may write
Q
[
√
D
] =
{
a
+
b
√
D

a, b
∈
Q
}
,
so that a basis for
Q
[
√
D
] as a
Q
vector space is 1
,
√
D
.
In particular
Q
[
√
D
] is
twodimensional as a
Q
vector space: we accordingly say it is a
quadratic field
.
It is also easy to check by hand that the ring
Q
[
√
D
] is a field.
For this and
for many other things to come, the key identity is
(
a
+
b
√
D
)(
a

b
√
D
) =
a
2

Db
2
.
For rational numbers
a
and
b
which are not both zero, the rational number
a
2

Db
2
is also nonzero:
equivalently there are no solutions to
D
=
a
2
b
2
, because
√
D
is
irrational. It follows that – again, for
a
,
b
not both 0 – we have
a
+
b
√
D
·
a
a
2

Db
2

b
a
2

Db
2
√
D
= 1
,
which gives a multiplicative inverse for
a
+
b
√
D
in
Q
[
√
D
].
We wish also to consider
quadratic rings
, certain integral domains whose frac
tion field is a quadratic field
Q
(
√
D
). Eventually we will want a more precise and
inclusive definition, but for now we consider
Z
[
√
D
] =
{
a
+
b
√
D

a, b
∈
Z
}
.
1
2.
Fermat’s Two Squares Theorem
The rings
Z
[
√
D
] occur naturally when we study Diophantine equations. E.g:
1
This equality is a fact which is not difficult to check; it is
not
the definition of
Z
[
√
D
]. By way
of comparison, we recommend that the reader check that the ring
Z
[
√
D
2
] is
not
of the form
Z
α
+
Z
β
for any two fixed elements
α, β
of
Z
[
√
D
2
]. In fact its additive group is not finitely generated as an
abelian group.
1
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2
PETE L. CLARK
Question 1.
Which prime numbers
p
can be expressed as a sum of two squares?
More precisely, for for which prime numbers
p
are there integers
x
and
y
such that
(1)
x
2
+
y
2
=
p
?
Remark: Evidently 2 is a sum of squares: 1
2
+1
2
= 2. Henceforth we assume
p >
2.
At the moment we have exactly one general technique
2
for studying Diophantine
equations: congruences.
So let’s try to apply it here: if we reduce the equation
x
2
+
y
2
=
p
modulo
p
, we get
x
2
+
y
2
≡
0 (mod
p
).
Suppose (
x, y
)
∈
Z
2
is a
solution to (1).
It cannot be that
p

x
, because then we would also have
p

y
,
and then
p
2

x
2
+
y
2
=
p
, a contradiction. So both
x
and
y
are nonzero modulo
p
(hence invertible in
Z
/p
Z
) and the congruence can be rewritten as
x
y
2
≡ 
1
(mod
p
)
.
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 Spring '11
 Staff
 Algebra, Number Theory, Prime number, Integral domain, square modulo, UFD, pete l. clark

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