MATH 392 – Seminar in Computational Commutative Algebra Solutions for Midterm Exam March 22, 2019

I. A) (20) Prove that every ideal I in the polynomial ring k[x] (one variable) is principal

(that is, I = ⟨g(x)⟩ for some single polynomial).

Solution: Let I be an ideal in k[x]. If I = {0}, then we can take g to be the zero polynomial. If I contains nonzero polynomials, let g(x) be any nonzero element of I of minimal degree. We must show that every f(x) ∈ I is a multiple of g(x). So using the division algorithm in k[x], write f(x) = q(x)g(x) + r(x), where either r(x) is identically zero, or else deg(r(x)) < deg(g(x)). We have r(x) = f(x) − q(x)g(x). f(x) ∈ I by assumption. g(x) is also in I by construction. I is an ideal, hence closed under products by arbitrary polynomials, so q(x)g(x) ∈ I. Similarly I is closed under sums, so r(x) = f(x) − q(x)g(x) ∈ I. But g(x) was a nonzero element of I of minimal degree and r(x) is either identically zero, or else deg(r(x)) < deg(g(x)). Since the second alternative is not possible, r(x) = 0, which shows f(x) = q(x)g(x). This shows I ⊆ ⟨g(x)⟩. The opposite inclusion is also true because g(x) ∈ I, so ⟨g(x)⟩ ⊆ I since I is closed under multiplication by arbitrary polynomials.

B) (10) Find g(x) as in part A for the ideal I = ⟨x^{2} + 7x + 10, x^{3} + 2x^{2} + 4⟩ in Q[x].

Solution 1: The polynomial g(x) must be the gcd of the two generators for I. We see by factoring that x^{2} + 7x + 10 = (x + 2)(x + 5) and x^{3} + x^{2} + 4 = (x + 2)(x^{2} − x + 2) and x^{2} − x + 2 does not factor farther in Q[x] because this polynomial has no rational (or even real) roots. Hence the gcd is x + 2.

Solution 2: We can also find the gcd using the Euclidean Algorithm in Q[x]. Dividing:

\begin{aligned}x^{3} + 2x^{2} + 4 & = (x − 6)(x^{2} + 7x + 10) + 32x + 64 \\ x^{2} + 7x + 10 & = (\frac{x}{32} + \frac{5}{32}) (32x + 64) + 0 \end{aligned}

The last nonzero remainder is the gcd, up to a constant multiple. Note 32x + 64 = 32(x + 2) so the monic gcd is x + 2 as in the first solution.

II. A) (15) Define: G is a Gröbner basis for an ideal I ⊂ k[x_{1}, …, x_{n}] with respect to a

monomial order >.

Solution: A Gröbner basis for I with respect to > is a finite collection of polynomials G = {g_{1}, …, g_{t}} ⊂ I such that the monomial ideal ⟨LT_{>}(I)⟩ is generated by LT_{>}(g_{1}), …, LT_{>}(g_{t}). (Equivalently you could also say it is a finite collection of polynomials G as above such that for every nonzero f ∈ I, LT_{>}(f) is divisible by LT_{>}(g_{i}) for some i.)

B) (10) Assuming the statement of Dickson’s Lemma, prove that Gröbner bases exist for every ideal I in k[x_{1}, …, x_{n}] and with respect to every monomial order.

Solution: We may assume that I ≠ {0}, since in that case we can take G = ∅. Dickson’s Lemma is the statement that every monomial ideal in k[x_{1}, …, x_{n}] is generated by a finite set of monomials. We apply that result to the monomial ideal ⟨LT_{>}(I)⟩ = ⟨LT_{>}(f) | f ∈ I⟩. This says ⟨LT_{>}(I)⟩ is generated by some finite collection of monomials x^{α(1)}, …, x^{α(t)}. Moreover, by definition each of those is equal to the leading term of some element in I:

x^{α(i)} = LT_{>}(g_{i}),

for some g_{i} ∈ I. By the definition (part A), this says {g_{1}, …, g_{t}} is a Gröbner basis for I.

C) (5) What else do you need to know in order for the result from part B to give a proof of the Hilbert Basis Theorem? (You don’t need to give the proof, just say what else must be proved.)

Solution: The other fact that must be proved to get a proof of the Hilbert Basis Theorem (the statement that every ideal in k[x_{1}, …, x_{n}] has a finite basis) is that a Gröbner basis for I is also an ideal basis for I, or equivalently that if G is a Gröbner basis for I and f ∈ I, then the remainder on division of f by G is zero: \overline{f}^{G} = 0. This follows from the definition of a Gröbner basis and the properties of the Division Algorithm.

III. A) (20) State and prove the Elimination Theorem.

Solution: Let I be an ideal in k[x_{1}, …, x_{n}] and let I_{ℓ} = I ∩ k[x_{ℓ+1}, …, x_{n}] be the elimination ideals for ℓ = 1, …, n − 1. The Elimination Theorem states that if G is a Gröbner basis for I with respect to the lexicographic order with x_{1} > x_{2} > … > x_{n}, then G_{ℓ} = G ∩ k[x_{ℓ+1}, …, x_{n}] is a Gröbner basis of I_{ℓ}, for all ℓ = 1, …, n − 1. Proof: We must show that if f is any element of I_{ℓ}, then LT_{lex}(f) is divisible by one of the leading terms of the elements of G_{ℓ}. But if f ∈ I_{ℓ}, then f depends only on the variables x_{ℓ+1}, …, x_{n}, and the same is true for LT_{lex}(f). Since G is a Gröbner basis for I, LT_{lex}(f) is divisible by LT_{lex}(g) for some g ∈ G. But this means that LT_{lex}(g) can only depend on the variables x_{ℓ+1}, …, x_{n}. By the properties of the lex order with

x_{1} > x_{2} > ⋯ > x_{ℓ} > x_{ℓ+1} > ⋯ x_{n},

any monomial containing any of the variables x_{1}, …, x_{ℓ} is greater than all monomials containing only the variables x_{ℓ+1}, …, x_{n}. This means that no term in g can contain any of the variables x_{1}, …, x_{ℓ}. Hence by definition g ∈ G_{ℓ}. We have shown that for every nonzero f ∈ I_{ℓ}, LT_{lex}(f) is divisible by LT_{lex}(g) for some g ∈ G_{ℓ}. This shows that G_{ℓ} is a Gröbner basis for I_{ℓ} by the definition.

B) (10) A certain ideal J ⊂ Q[x, y, z] has a Gröbner basis

B = {x^{3} − 3x^{2} + 2x, x^{2}y − xy, y^{2} − y, z − xy}

with respect to the lexicographic order with z > y > x. What are bases for the elimination ideals

J_{1} = J ∩ Q[y, x] and J_{2} = J ∩ Q[x]?

Solution: J_{1} = ⟨x^{3} − 3x^{2} + 2x, x^{2}y − xy, y^{2} − y⟩ and J_{2} = ⟨x^{3} − 3x^{2} + 2x⟩.

C) (10) Use the information in part B to determine all of the points in V (J).

Solution: We begin by setting the generator of J_{2} equal to zero and by factoring we find x(x − 1)(x − 2) = 0, so x = 0, x = 1, or x = 2. If we substitute x = 0 in the rest of the Gröbner basis, we find:

B|_{x=0} = {0, 0, y^{2} − y, z}

From this we see y(y − 1) = 0 so y = 0 or y = 1. And then with either of those y-values, z = 0. So we have two points with x = 0, namely (0, 0, 0) and (0, 1, 0). If we substitute x = 1 in the rest of the Gröbner basis, we find:

B|_{x=1} = {0, 0, y^{2} − y, z − y}

Hence we find two more points (1, 0, 0) and (1, 1, 1). Finally, if we substitute x = 2 into the rest of the Gröbner basis, we find

B|_{x=2} = {0, 2y, y^{2} − y, z − 2y}

The only solution is (2, 0, 0). This means that V (J) consists of five points in all:

V (J) = {(0, 0, 0), (0, 1, 0), (1, 0, 0), (1, 1, 1), (2, 0, 0)}.