In algebra, Zariski's lemma, introduced by Oscar Zariski (1947), states that if K is a finitely generated algebra over a field k and if K is a field, then K is a finite field extension of k.
An important application of the lemma is a proof of the weak form of Hilbert's nullstellensatz: if I is a proper ideal of
k
[
t
1
,
.
.
.
,
t
n
]
(k algebraically closed field), then I has a zero; i.e., there is a point x in
k
n
such that
f
(
x
)
=
0
for all f in I.
The lemma may also be understood from the following perspective. In general, a ring R is a Jacobson ring if and only if every finitely generated R-algebra that is a field is finite over R. Thus, the lemma follows from the fact that a field is a Jacobson ring.
Two direct proofs, one of which is due to Zariski, are given in Atiyah–MacDonald. For Zariski's original proof, see the original paper. Another direct proof in the context of Jacobson rings is given below. The lemma is also a consequence of the Noether normalization lemma. Indeed, by the normalization lemma, K is a finite module over the polynomial ring
k
[
x
1
,
…
,
x
d
]
where
x
1
,
…
,
x
d
are algebraically independent over k. But since K has Krull dimension zero, the polynomial ring must have dimension zero; i.e.,
d
=
0
.
The following characterization of a Jacobson ring contains Zariski's lemma as a special case. Recall that a ring is a Jacobson ring if every prime ideal is an intersection of maximal ideals. (When A is a field, A is a Jacobson ring and the theorem below is precisely Zariski's lemma.)
Proof: 2.
⇒
1.: Let
p
be a prime ideal of A and set
B
=
A
/
p
. We need to show the Jacobson radical of B is zero. For that end, let f be a nonzero element of B. Let
m
be a maximal ideal of the localization
B
[
f
−
1
]
. Then
B
[
f
−
1
]
/
m
is a field that is a finitely generated A-algebra and so is finite over A by assumption; thus it is finite over
B
=
A
/
p
and so is finite over the subring
B
/
q
where
q
=
m
∩
B
. By integrality,
q
is a maximal ideal not containing f.
1.
⇒
2.: Since a factor ring of a Jacobson ring is Jacobson, we can assume B contains A as a subring. Then the assertion is a consequence of the next algebraic fact:
(*) Let
B
⊃
A
be integral domains such that
B is finitely generated as
A-algebra. Then there exists a nonzero
a in
A such that every ring homomorphism
ϕ
:
A
→
K
,
K an algebrically closed field, with
ϕ
(
a
)
≠
0
extends to
ϕ
~
:
B
→
K
.
Indeed, choose a maximal ideal
m
of A not containing a. Writing K for some algebraic closure of
A
/
m
, the canonical map
ϕ
:
A
→
A
/
m
↪
K
extends to
ϕ
~
:
B
→
K
. Since B is a field,
ϕ
~
is injective and so B is algebraic (thus finite algebraic) over
A
/
m
. We now prove (*). If B contains an element that is transcendental over A, then it contains a polynomial ring over A to which φ extends (with a = 1) and so we can assume B is algebraic over A. Let
x
1
,
…
,
x
r
be the generators of B as A-algebra. Then each
x
i
satisfies the relation
a
i
0
x
i
n
+
a
i
1
x
i
n
−
1
+
⋯
+
a
i
n
=
0
where
a
i
j
's are all in A. Set
a
=
a
10
a
20
…
a
r
0
. Then B is integral over
A
[
a
−
1
]
. Now given
ϕ
:
A
→
K
, we first extend it to
ϕ
~
:
A
[
a
−
1
]
→
K
by setting
ϕ
~
(
a
−
1
)
=
ϕ
(
a
)
−
1
. Next, let
m
=
ker
ϕ
~
. By integrality,
m
=
n
∩
B
for some maximal ideal
n
of B. Then
ϕ
~
:
A
[
a
−
1
]
→
A
[
a
−
1
]
/
m
→
K
extends to
B
→
B
/
n
→
K
.
◻
