A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
The algebra k [x1 , x2 , . . . , xn ]/I
Julius M Basilla
February 19, 2008
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
The elements of k [x1 , x2 , . . . , xn ]/I
The elements of k [x1 , x2 , . . . , xn ]/I are of the form
f + I,
where f is an element of k [x1 , x2 , . . . , xn ].
Two elements f + I and g + I are the same if and only if
f − g ∈ I.
If I = (f1 , f2 , . . . , fs ), then f + I = g + I means that f − g is
generated by {f1 , f2 , . . . , f2 }. That is,
f −g =
s
X
gi fi ,
i=1
where gi ∈ k [x1 , x2 , . . . , xn ].
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
The cosets of I and gröbner basis.
That is, f + (f1 , f2 , . . . , fs ) = g + (f1 , f2 , . . . , fs ) means that
f ,f ,...,fs
1 2
f −−
−−−→ g
If {f1 , f2 , . . . , fs } is a Gröbner basis for the ideal I, then for
each fink [x1 , x2 , . . . , xn ], there is a unique
r ∈ k [x1 , x2 , . . . , xn ] such that
f ,f ,...,fs
1 2
f −−
−−−→ r ,
and r is reduced with respect to {f1 , f2 , . . . , fs }.
This suggests that for every f + I ∈ k [x1 , x2 , . . . , xn ]/I there
is an element r ∈ k [x1 , x2 , . . . , xn ] which is reduces with
respect to a Gröbner basis for I such that
f + I = r + I.
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
A complete set of coset representatives
Hence, if S is a Gröbner basis for I, the set
{r + I : r ∈ k [x1 , x2 , . . . , xn ] and r is reduced with respect to S}
is a complete set of coset representatives for
k [x1 , x2 , . . . , xn ]/I.
That is the set
{r + I : r ∈ k [x1 , x2 , . . . , xn ] and r is reduced with respect to S}
exhausts all elements of k [x1 , x2 , . . . , xn ]/I and every
element of k [x1 , x2 , . . . , xn ]/I is found in
{r + I : r ∈ k [x1 , x2 , . . . , xn ] and r is reduced with respect to S}
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
Vector Spaces
Let k be a field. A vector space over k is an abelian group V
together with an operation
k ×V
(r , α)
→ V
7
→
rα
such that for all r , s ∈ k and α, β ∈ V , we have
r (α + β) = r α + r β
(r + s)α = r α + sα
r (sα) = (rs)(α).
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
Algebra
An algebra A over a field k is a vector space over k with an
additional operation called multiplication of vectors given by
V ×V
(α, β)
→ V
7
→
αβ
and satisfying
α(β + γ) = αβ + αγ,
(α + β)γ = αγ + βγ,
α(r β) = r (αβ) = (r α)β,
for all vectors α, β, γ in V and scalar r ∈ k .
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
k [x1 , x2 , . . . , xn ]/I is a vector space over k
The set k [x1 , x2 , . . . , xn ]/I is a vector space under the
addition of vectors given by
(f + I) + (g + I) = (fg) + I,
and scalar multiplication given by
r (f + I) = rf + I.
It can be shown that for all polynomials f , g and scalars r
and s, we have
r ((f + I) + (g + I)) = (rf + I) + (rg + I)
(r + s)(f + I) = (rf + I) + (sf + I)
r (s(f + I)) = (rs)(f + I).
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
k [x1 , x2 , . . . , xn ]/I is an algebra over k
Since k [x1 , x2 , . . . , xn ]/I is also a commutative ring with
unity with multiplication given by
(f + I)(g + I) = (fg + I),
we can make k [x1 , x2 , . . . , xn ]/I into an algebra over k
under the same multiplication operations.
The algebra k [x1 , x2 , . . . , xn ]/I is a commutative algebra
over k with unity 1 + I.
For brevity, we will denote the element
f + I ∈ k [x1 , x2 , . . . , xn ]/I by [f ].
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
Left to Right Proof
proof of Converse
Finiteness Theorem
Theorem
The algebra k [x1 , x2 , . . . , xn ]/I is finite dimensional if and only if
V (I) is finite.
This means that if I = (f1 , f2 , · · · , fs ), as is always the
case(Hilbert Basis Theorem), k [x1 , x2 , . . . , xn ]/I is a finite
dimensional algebra if and only if the system of polynomial
equations
f1 (x1 , x2 , . . . , xn ) = 0
f2 (x1 , x2 , . . . , xn ) = 0
..
.
.
= ..
fs (x1 , x2 , . . . , xn ) = 0
has finitely many solutions.
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
Left to Right Proof
proof of Converse
Proof ⇒
For each i = 1, 2, . . . , n the set {[1], [xi ], [xi2 ], . . .} is linearly
dependent.
Hence, there exist an mi > 0 and constants
ci0 , ci1 , ci2 , . . . , cimi which are not all zero (cimi 6= 0, in part.)
such that
mi
X
cij [xij ] = [0].
j=0
Take pi (xi ) ∈ k [x1 , x2 , . . . , xn ] to be the polynomial given by
pi (xi ) = ci0 + ci1 xi + ci2 xi2 + · · · cimi ximi .
It follows that [pi (xi )] = [0]. That is, pi (xi ) ∈ I.
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
Left to Right Proof
proof of Converse
Proof( ⇒ ) Continued
Since pi (xi ) ∈ I, Then for all (y1 , y2 , . . . , yn ) ∈ V (I), we have
pi (yi ) = pi (y1 , y2 , . . . , yn ) = 0.
But there can only be finitely ( to be more precise, at most
mi ) many elements yi such that pi (yi ) = 0.
Hence,
there are only finitely (to be more precise, at most
Qs
m
i=1 i ) elements in V (I).
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
Left to Right Proof
proof of Converse
Proof ⇐
Consider the polynomials qi (xi ) ∈ k [x1 , x2 , . . . , xn ] such
that qi (yi ) = 0 for all (y1 , y2 , . . . , yn ) ∈ V (I).
We are sure that qi (xi ) exists be cause we can
take
Y
qi (xi ) =
(xi − yj )
∀(y1 ,y2 ,...,yn )∈V (I)
By this choice, we have qi (y1 , y2 , . . . , yn ) = qi (yi ) = 0 for all
(y1 , y2 , . . . , yn ) ∈ V (I).
Hence, Hilbert Nullstellensatz tells us that there exists an
integer Ni such that (qi (xi )Ni ∈ I.
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
Left to Right Proof
proof of Converse
Proof ⇐, continued
Let {g1 , g2 , . . . , gt } be a Gröbner basis for I, say with
respect to the lexicographical order. Then we have the
following properties,
I = (g1 , g2 , . . . , gt ).
Every nonzero polynomial f ∈ I has a leading term lt(f )
which is divisible by lt(gj ) for some j = 1, 2, . . . , t.
The set
B := {[x α ] = [x1α1 x2α2 · · · xnαn ] : lt(gj ) 6| x α , ∀j = 1, 2, . . . , t} is
a basis for k [x1 , x2 , . . . , xn ]/I.
It suffices therefore to show that B is finite.
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I
A factor ring involving polynomials
An algebra over a field
Finiteness Theorem
Left to Right Proof
proof of Converse
B is finite
Since, qi (xi )Ni ∈ I and lt(qi (xi )Ni = x Mi for some Mi ∈ N
then there exist a j ∈ {1, 2, . . . , t} such that lt(gj )|x Mi . That
M0
is lt(gj ) = xi i , Mi0 < Mi .
Hence, for each i = 1, 2, . . . , n there exists a ji such that
M0
lt(gji ) = xi i .
Therefore, if [x α ] = [x1α1 x2α2 · · · xnαn ] ∈ B, we must have
αi < Mi0 . Otherwise, lt(gji ) divides x α .
Q
Hence, B is finite. More precisely ]B ≤ si=1 (Mi0 + 1)
Julius M Basilla
The algebra k [x1 , x2 , . . . , xn ]/I