Algebraic Geometry: 0. Algorithms for Polynomials 5
Proof. (a) ⇒ (b): If a
1
⊂ a
2
⊂··· is an ascending chain, then ∪a
i
is again an
ideal, and hence has a finite set {a
1
,... ,a
n
} of generators. For some m,allthea
i
belong a
m
and then
a
m
= a
m+1
= ···= a.
(b) ⇒ (c): If (c) is false, then there exists a nonempty set S of ideals with no
maximal element. Let a
1
∈ S; because a
1
is not maximal in S, there exists an
ideal a
2
in S that properly contains a
1
. Similarly, there exists an ideal a
3
in S
properly containing a
2
, etc.. In this way, we can construct an ascending chain of
ideals a
1
⊂ a
2
⊂ a
3
⊂··· in S that never becomes stationary.
(c) ⇒ (a): Let a be an ideal, and let S be the set of ideals b ⊂ a that are finitely
generated. Let c =(a
1
,... ,a
r
) be a maximal element of S.Ifc = a, so that there
exists an element a ∈ a, a/∈ c,thenc
=(a
1
,... ,a
r
,a) ⊂ a and properly contains c,
which contradicts the definition of c.
AringA is Noetherian if it satisfies the conditions of the proposition. Note that,
in a Noetherian ring, every ideal is contained in a maximal ideal (apply (c) to the
set of all proper ideals of A containing the given ideal). In fact, this is true in any
ring, but the proof for non-Noetherian rings requires the axiom of choice (Atiyah and
MacDonald 1969, p3).
Algebras. Let A be a ring. An A-algebra is a ring B together with a homomorphism
i
B
: A → B.Ahomomorphism of A-algebras B → C is a homomorphism of rings
ϕ: B → C such that ϕ(i
B
(a)) = i
C
(a) for all a ∈ A.
An A-algebra B is said to be finitely generated (or of finite-type over A)ifthere
exist elements x
1
,... ,x
n
∈ B such that every element of B can be expressed as
a polynomial in the x
i
with coefficients in i(A), i.e., such that the homomorphism
A[X
1
,... ,X
n
] → B sending X
i
to x
i
is surjective.
A ring homomorphism A → B is finite,andB is a finite A-algebra, if B is finitely
generated as an A-module
4
.
Let k be a field, and let A be a k-algebra. If 1 =0inA, then the map k → A is
injective, and we can identify k with its image, i.e., we can regard k asasubringof
A. If 1 = 0 in a ring R,theR is the zero ring, i.e., R = {0}.
Polynomial rings. Let k be a field. A monomial in X
1
,... ,X
n
is an expression of
the form
X
a
1
1
···X
a
n
n
,a
j
∈ N.
The total degree of the monomial is
a
i
. We sometimes abbreviate it by X
α
,α=
(a
1
,... ,a
n
) ∈ N
n
.
The elements of the polynomial ring k[X
1
,... ,X
n
] are finite sums
c
a
1
···a
n
X
a
1
1
···X
a
n
n
,c
a
1
···a
n
∈ k, a
j
∈ N.
with the obvious notions of equality, addition, and multiplication. Thus the mono-
mials from a basis for k[X
1
,... ,X
n
]asak-vector space.
4
The term “module-finite” is used in this context only by the English-insensitive.
