Computer Algebra
Michael Sagraloff
Summer 2011
To be handed in on July, 5th.
Discussion on July, 6th.
Exercise 11
11.1
Operations on ideals
Recall the definition of an ideal from the lecture:
Definition (Ideal). Let R be a ring (commutative, with 1). A subset I ⊂ R of R is called an ideal
if, for all a, b ∈ I and r ∈ R, we have a + b ∈ I and a · r ∈ I.
Let I, J ⊂ R be ideals. We define the ideal operations
I ∩ J = {c ∈ R : c ∈ I and c ∈ J}
set union: I ∪ J = {c ∈ R : c ∈ I or c ∈ J}
addition: I + J = {a + b : a ∈ I and b ∈ J}
set intersection:
product:
quotient:
I · J = {a1 b1 + a2 b2 + · · · + an bn : n ∈ N, ai ∈ I and bi ∈ J for all i = 1, . . . , n}
I : J = {r ∈ R : r · J ⊂ I}, where r · J := {r · b : b ∈ J}
1. Which of the above operations yield ideals in R again?
2. Show that the product of ideals is generated by the pairwise products of the elements of the
ideals, that is I · J = h{ab : a ∈ I and b ∈ J}i.
3. Prove that I · J ⊂ I ∩ J.
4. Show: If R = F [x1 , . . . , xn ] is the polynomial ring over some field F and I, J are ideals in R,
then V(I) ∪ V(J) = V(I · J) = V(I ∩ J).
11.2
Properties of vanishing ideals and vanishing sets
Let F be a field and n ∈ N. For a subset M ⊂ F n , we define the vanishing ideal I(M ) of M as
I(M ) := {f ∈ F [x1 , . . . , xn ] : f (m) = 0 for all m ∈ M } ⊂ F [x1 , . . . , xn ].
Let M, N ⊂ F n and I ⊂ F [x1 , . . . , xn ] be an ideal. Prove:
1. I(M ) is in fact an ideal in F [x1 , . . . , xn ].
2. M ⊂ V(I(M )) and I ⊂ I(V(I)).
Give examples where the inclusions are proper (that is, equality does not hold).
3. If N ⊂ M, then I(M ) ⊂ I(N ).
4. I(M ) is radical, that is f r ∈ I(M ) implies f ∈ I(M ) for all r ∈ N and f ∈ F [x1 , . . . , xn ].
1 of 2
11.3
Hilbert’s Nullstellensatz
Hilbert’s Nullstellensatz is a fundamental theorem in the study of vanishing sets and vanishing
ideals. It establishes the link between algebraic sets (that is the vanishing loci of sets of polynomials)
and the corresponding ideals.
Theorem (Hilbert’s Nullstellensatz over the complex numbers). Let n ∈ N and I ⊂ R :=
C[x1 , . . . , xn ] be an ideal.
• Weak version: The vanishing locus of I in Cn is empty if and only if I is the ring R itself:
V(I) = ∅
⇔
1 ∈ I.
• Strong version: The radical of I is the vanishing ideal of the vanishing set of I:
rad I := {f ∈ R : f r ∈ I for some r ∈ N} = I(V(I)).
In this exercise, we aim to show the weak formulation.
1. State a reason why the weak Nullstellensatz holds for n = 1.
2. Next, turn to the case n = 2. Let I = hf1 , f2 , f3 i ⊂ C[x, y] and write f = f1 , g = uf2 + vf3 .
Consider the resultant
X
Res(f, g; y) = r(u, v, x) =
ri,j (x)ui v j ∈ C[u, v, x].
i,j
Show that ri,j ∈ I ∩ C[x].
Hint: Remember that the resultant can be expressed in terms of f and g and their cofactors.
3. Use the fact that r(u, v, c) ≡ 0 for all c ∈ V(I ∩ C[x]) to show that there exists an h ∈ C[y]
such that h | fk (c, y) for k = 1, 2, 3.
4. Prove that V(f1 , f2 , f3 ) = ∅ implies I = h1i.
5. Generalize your proof for arbitrary dimension n and number of generators of I.
6. (Bonus) Show the strong version of the Nullstellensatz.
Hint: For the inclusion rad I ⊃ I(V(I), assume f ∈ I(V(I)) and apply the weak result on
I˜ := hf1 , . . . , fs , yf − 1i ⊂ C[x1 , . . . , xn , y].
Have fun with the solution!
2 of 2