In commutative algebra, the extension and contraction of ideals are operations performed on sets of ideals.
Let A and B be two commutative rings with unity, and let f : A → B be a (unital) ring homomorphism. If
a
is an ideal in A, then
f
(
a
)
need not be an ideal in B (e.g. take f to be the inclusion of the ring of integers Z into the field of rationals Q). The extension
a
e
of
a
in B is defined to be the ideal in B generated by
f
(
a
)
. Explicitly,
a
e
=
{
∑
y
i
f
(
x
i
)
:
x
i
∈
a
,
y
i
∈
B
}
If
b
is an ideal of B, then
f
−
1
(
b
)
is always an ideal of A, called the contraction
b
c
of
b
to A.
Assuming f : A → B is a unital ring homomorphism,
a
is an ideal in A,
b
is an ideal in B, then:
b
is prime in B
⇒
b
c
is prime in A.
a
e
c
⊇
a
b
c
e
⊆
b
It is false, in general, that
a
being prime (or maximal) in A implies that
a
e
is prime (or maximal) in B. Many classic examples of this stem from algebraic number theory. For example, embedding
Z
→
Z
[
i
]
. In
B
=
Z
[
i
]
, the element 2 factors as
2
=
(
1
+
i
)
(
1
−
i
)
where (one can show) neither of
1
+
i
,
1
−
i
are units in B. So
(
2
)
e
is not prime in B (and therefore not maximal, as well). Indeed,
(
1
±
i
)
2
=
±
2
i
shows that
(
1
+
i
)
=
(
(
1
−
i
)
−
(
1
−
i
)
2
)
,
(
1
−
i
)
=
(
(
1
+
i
)
−
(
1
+
i
)
2
)
, and therefore
(
2
)
e
=
(
1
+
i
)
2
.
On the other hand, if f is surjective and
a
⊇
k
e
r
f
then:
a
e
c
=
a
and
b
c
e
=
b
.
a
is a prime ideal in A
⇔
a
e
is a prime ideal in B.
a
is a maximal ideal in A
⇔
a
e
is a maximal ideal in B.
Let K be a field extension of L, and let B and A be the rings of integers of K and L, respectively. Then B is an integral extension of A, and we let f be the inclusion map from A to B. The behaviour of a prime ideal
a
=
p
of A under extension is one of the central problems of algebraic number theory.
