MATH 401, EXAM 2

DANIEL R. GRAYSON

Problem 1 (35 points). True or false? (No support needed for your answer.)

(a) 𝔽_{9}^{×} is a cyclic group. [4 points] Solution: True. This was done in exercise 13.2 #2.

(b) Any prime ideal is maximal. [4 points] Solution: False. An example is the zero ideal of ℤ.

(c) Any field is noetherian. [4 points] Solution: True. Since there are only two ideals, any family of ideals has a maximal member.

(d) \sqrt[3]{100} is contained in a field of degree 100 over ℚ. [4 points] Solution: False. We know x^{3} − 100 has no rational roots, so is irreducible over ℚ, and thus ℚ(\sqrt[3]{100}) has degree 3 over ℚ, but 3 doesn’t divide 100.

(e) ℤ[(1 + \sqrt{−3})/2] is a unique factorization domain. [4 points] Solution: True, since in exercise 8.1 #8a it was shown to be a Euclidean domain.

(f) Any unique factorization domain is noetherian. [5 points] Solution: False. Consider ℚ[x_{1}, x_{2}, x_{3}, …].

(g) ℤ[\sqrt{−3}] is a unique factorization domain. [5 points] Solution: False, since 2^{2} = (1 + \sqrt{−3})(1 − \sqrt{−3}), and (1 + \sqrt{−3}) is irreducible. (1 + \sqrt{−3} is not divisible by 2, and other factors would have smaller modulus, so we only have to check divisibility by \sqrt{−3}.

(h) [ℚ(cos \frac{π}{15}) : ℚ] = 15 [5 points] Solution: False. Let ζ = exp(πi/15). Since cos \frac{π}{15} = (ζ + ζ^{−1})/2, we see that ℚ(cos \frac{π}{15}) ⊆ ℚ(ζ), and since ζ^{15} = −1 we see that [ℚ(ζ) : ℚ] ≤ 15, but cos \frac{π}{15} is real and ζ is not, so [ℚ(cos \frac{π}{15}) : ℚ] < [ℚ(ζ) : ℚ].

Problem 1 (15 points). Prove that in a commutative ring where every ideal is finitely generated, there is no infinite strictly ascending chain of ideals I_{1} ⊊ I_{2} ⊊ I_{3} ⊊ . . .. Solution: Let I be the union of the ideals. It’s an ideal — for example, it’s closed under addition since any two elements are contained in some I_{n}. It’s finitely generated, so the generators are also contained in I_{n} for some n, hence I ⊆ I_{n} ⊆ I, forcing I_{n} = I_{n+1} = ⋯ = I.

Date: November 22, 1999.

Problem 3 (25 points). Prove that 𝔽_{243} is a splitting field over 𝔽_{3} of a polynomial of degree 5.

Solution: We know that 𝔽_{243} is the splitting field over 𝔽_{3} of x^{243} − x. Consider the prime factorization of x^{243} − x in 𝔽_{3}[x]; not all of its prime factors are of degree 1, so let g(x) be one not of degree 1, let α ∈ 𝔽_{243} be one of its roots, and let K = 𝔽_{3}(α). (An alternate approach would be to start by choosing α ∈ 𝔽_{243} − 𝔽_{3} and letting g(x) be its irreducible polynomial over 𝔽_{3}.) We know K ≠ 𝔽_{3}, and we know [K : 𝔽_{3}] divides [𝔽_{243} : 𝔽_{3}] = 5, so since 5 is a prime number, [K : 𝔽_{3}] = 5, and hence K = 𝔽_{243}, showing that g(x) has degree 5. We know g(x) is a divisor of x^{243} − x because α^{243} − α = 0, and g(x) is the irreducible polynomial of α over 𝔽_{3}. Since x^{243} − x splits completely into linear factors over 𝔽_{243}, we know that g(x) also splits completely into linear factors over 𝔽_{243}, so that 𝔽_{243} is its splitting field.

Problem 4 (25 points). Suppose R is a commutative ring, and T is a set of (Zariski) closed subsets of Spec(R) whose intersection ⋂_{W ∈T} W is empty. Show there is a finite subset T′ ⊆ T whose intersection ⋂_{W ∈T′} W is empty.

Solution: For each W ∈ T let I_{W} ⊆ R be an ideal such that W = Z(I_{W}). Then ϕ = ⋂_{W ∈T} W = ⋂_{W ∈T} Z(I_{W}) = Z(⋃_{W ∈T} I_{W}) = Z(∑_{W ∈T} I_{W}). Since any proper ideal is contained in some prime ideal, it follows that ∑_{W ∈T} I_{W} = R, or equivalently, that 1 ∈ ∑_{W ∈T} I_{W}, so we can find W_{1}, …, W_{n} ∈ T and write 1 = f_{1} + ⋯ + f_{n} with each f_{i} ∈ I_{W_{i}}. Reversing the reasoning above, we see that ⋂_{i=1}^{n} W_{i} = ϕ.

University of Illinois at Urbana-Champaign

   E-mail address: dan@math.uiuc.edu
   URL: http://www.math.uiuc.edu/~dan