Mathematics 352 – Abstract Algebra II Solutions for Problem Set 5 February 18, 2008

Section 10.3

3. Following the example on page 349 of the text, consider the tower of field extensions

ℚ ⊂ ℚ(\sqrt{2}) ⊂ ℚ(\sqrt{2})(\sqrt{3}) ⊂ ℚ(\sqrt{2}, \sqrt{3})(\sqrt{5}).

We have [ℚ(\sqrt{2}) : ℚ] = 2 with basis \{1, \sqrt{2}\} since the minimal polynomial of \sqrt{2} over ℚ is x^{2} − 2. At the next step, we claim that [ℚ(\sqrt{2})(\sqrt{3}) : ℚ(\sqrt{2})] = 2 also and a basis is \{1, \sqrt{3}\}. Clearly x^{2} − 3 is a polynomial in ℚ(\sqrt{2})[x] with \sqrt{3} as a root. Moreover, it is irreducible because no a + b\sqrt{2} is a root of this polynomial. (If there were such a root, then (a + b\sqrt{2})^{2} = 3 would imply \sqrt{2} = \frac{3−a^{2}−2b^{2}}{2ab} ∈ ℚ, which we know is not the case.) Hence as in the proof of Theorem 10.4 (the multiplicativity of degree in towers), \{1, \sqrt{2}, \sqrt{3}, \sqrt{6}\} is a basis for ℚ(\sqrt{2})(\sqrt{3}) = ℚ(\sqrt{2}, \sqrt{3}) over ℚ. Now we argue the same way for the final extension above. \sqrt{5} is a root of x^{2} −5 ∈ ℚ(\sqrt{2}, \sqrt{3})[x], but no a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} is a root of this polynomial, so [ℚ(\sqrt{2}, \sqrt{3})(\sqrt{5}) : ℚ(\sqrt{2}, \sqrt{3})] = 2 and \{1, \sqrt{2}, \sqrt{3}, \sqrt{6}, \sqrt{5}, \sqrt{10}, \sqrt{15}, \sqrt{30}\} is a basis of ℚ(\sqrt{2}, \sqrt{3}, \sqrt{5}) over ℚ.

5. We follow the same procedure as in the previous problem. Consider

ℚ ⊂ ℚ(\sqrt{3}) ⊂ ℚ(\sqrt{3}, i).

We have [ℚ(\sqrt{3}) : ℚ] = 2 with basis \{1, \sqrt{3}\} since the minimal polynomial of \sqrt{3} over ℚ is x^{2} − 3. Next, i is a root of x^{2} + 1 ∈ ℚ(\sqrt{3})[x] and this is the minimal polynomial because no a + b\sqrt{3} can be a root of x^{2} + 1 (since a + b\sqrt{3} are all real numbers). Hence [Q(\sqrt{3}, i) : ℚ(\sqrt{3})] = 2 and \{1, i\} is a basis. As in the proof of Theorem 10.4, \{1, \sqrt{3}, i, i\sqrt{3}\} is a basis for ℚ(\sqrt{3}, i) over ℚ. Hence [ℚ(\sqrt{3}, i) : ℚ] = 4.

7. Assume [K : F] is finite and u is algebraic over K. By Theorem 10.7, [K(u) : K] is finite (this equals the degree of the minimal polynomial of u over K, p(x) ∈ K[x]. Now, we have F ⊆ K ⊆ K(u), so also F ⊆ F (u) ⊆ K(u). From the first tower [K(u) : F] = [K(u) : K][K : F] is finite. This implies [F (u) : F] is also finite from the second tower. By Theorem 10.7 again, [F (u) : F] equals the degree of the minimal polynomial q(x) ∈ F [x] of u over F. Since this polynomial has u as a root and also lies in K[x], it follows that p(x)|q(x) in K[x]. Therefore deg q(x) = [F (u) : F] ≥ [K(u) : K] = deg p(x).

8. Let [K : F] be finite and assume u is algebraic over K. By Theorem 10.7, every element v of K(u) is of the form

v = a_{0} + a_{1}u + ⋯ a_{n−1}u^{n−1},  (1)

with a ∈ K (where n is the degree of the minimal polynomial of u over K). Now let {b_{1}, …, b_{k}} be a basis of K over F. We can expand each coefficient

a_{i} = c_{i1}b_{1} + ⋯ + c_{ik}b_{k},

where c_{ij} ∈ F. Substituting these into (1), and regrouping terms we see that

v = (c_{01} + c_{11}u + ⋯ + c_{n−1,1}u^{n−1})b_{1} + ⋯ + (c_{0k} + c_{1k}u + ⋯ + c_{n−1,k}u^{n−1})b_{k}.

This shows that {b_{1}, …, b_{k}} also spans K(u) over F (u). Hence [K(u) : F (u)] ≤ [K : F] (with equality if and only if {b_{1}, …, b_{k}} is linearly independent over F (u), which is not automatic).

11. (a) Consider the towers of field extensions F ⊆ F (u) ⊆ F (u)(v) = F (u, v) and F ⊆ F (v) ⊆ F (v)(u) = F (u, v). By hypothesis and Theorem 10.7, [F (u) : F] = deg p(x) = m. So then by Theorem 10.4, m|[F (u, v) : F]. Similarly, [F (v) : F] = deg q(x) = n. So then by Theorem 10.4, n|[F (u, v) : F]. So this says [F (u, v) : F] is a common multiple of m, n. Since (m, n) = 1, then least common multiple is of m, n is the product mn. However, from problem 7 above with K = F (v), we also know [F (u, v) : F (v)] ≤ [F (u) : F] = m. Therefore [F (u, v) : F] = [F (u, v) : F (v)][F (v) : F] ≤ mn. Therefore [F (u, v) : F] = mn.

(b) For a simple example, consider F = ℚ, u = \sqrt{2} and v = 1 + \sqrt{2}. We have [ℚ(u) : ℚ] = [ℚ(v) : ℚ] = 2 (look at the minimal polynomials). So the degrees are not relatively prime. Moreover v ∈ ℚ(u), so in fact ℚ(u, v) = ℚ(u) and [ℚ(u, v) : ℚ] = 2, not 2 · 2 = 4.

(c) [ℚ(\sqrt{2}, \sqrt[3]{2}) : ℚ] = 6 by part (a) since [ℚ(\sqrt{2}) : ℚ] = 2 and [ℚ(\sqrt[3]{2}) : ℚ] = 3 are relatively prime.

12. Method 1: (A “brute force” method not following the Hint.) Since we are given that D is a ring, the only thing “missing” is that we must show that every u ∈ D has an inverse in D. So let u ∈ D. Since K is algebraic over F, every element of K is algebraic over F, and it follows that u is a root of a nonzero polynomial c_{n}x^{n} + ⋯ + c_{1}x + c_{0} ∈ F [x], or

c_{n}u^{n} + ⋯ + c_{1}u + c_{0} = 0.

We can assume that zero is not also a root of this polynomial (why?). Hence we may assume c_{0} ≠ 0. But then rearranging the last equation,

(−c_{0}^{−1}(c_{n}u^{n−1} + ⋯ + c_{1}))u = 1.

This shows u^{−1} = −c_{0}^{−1}(c_{n}u^{n−1} +⋯+c_{1}), which is in D since D is closed under sums and products and u ∈ D. Hence D is a field.

Method 2: (Following the Hint.) Let u ∈ D. Since u is in K, which is a field, we have the tower F ⊆ F (u) ⊂ K. Since K is algebraic over F, by Theorem 10.7 we know F (u) has a basis {1, u, …, u^{n−1}} if the minimal polynomial of u over F has degree n. Therefore all the elements of F (u) look like v = a_{0} +a_{1}u+⋯+a_{n−1}u^{n−1} where a_{i} ∈ F. But u ∈ D and D is a ring (hence closed under sums and products). It follows that F (u) ⊆ D. Therefore D contains u^{−1}, the multiplicative inverse of u in K. Hence D is a field. Comment: I happen to prefer proofs like the first one, because they give a method for finding u^{−1} in addition to establishing the existence of u^{−1} in D.

Section 10.4

2. For x^{2} − 3, we have the roots x = ±\sqrt{3}. Hence the splitting field is ℚ(\sqrt{3}, −\sqrt{3}) = ℚ(\sqrt{3}). For x^{2} − 2x − 2, the quadratic formula gives

x = \frac{2 ± \sqrt{4 + 8}}{2} = \frac{2 ± 2\sqrt{3}}{2} = 1 ± \sqrt{3}.

Therefore the splitting field of this polynomial is ℚ(1 + \sqrt{3}, 1 − \sqrt{3}) = ℚ(1 + \sqrt{3}). Since 1 + \sqrt{3} is an element of ℚ(\sqrt{3}) and conversely every element of ℚ(\sqrt{3}) is contained in ℚ(1 + \sqrt{3}), it follows that the splitting fields of these two polynomials are equal.

6. We have u ∈ K, so F (u) ⊆ K. Since K is a splitting field of a polynomial in F [x] it is a finite (algebraic) extension of F. By Theorem 10.4, [K : F] = [K : F (u)][F (u) : F]. Since [K : F] = p is prime, but we are given u ≠ F, it follows that [F (u) : F] = p and [K : F (u)] = 1. Therefore K = F (u).

7. Since F (u) is normal over F, by definition, if p(x) is any irreducible polynomial in F [x] and p(x) has one root in F (u), then p(x) splits completely over F (u) (has all its roots in F (u)). Apply this to p(x) = the minimal polynomial of u over F. It follows that all the roots u = u_{1}, u_{2}, …, u_{n} of p(x) are contained in F (u). Therefore F (u, u_{2}, …, u_{n}) ⊆ F (u). The other inclusion is automatic, so F (u, u_{2}, …, u_{n}) = F (u). This shows that F (u) is the splitting field of p(x), which is what we wanted to show.

8. We use Theorem 10.15 for all the parts here.

(a) ℚ(\sqrt{3}) is normal over ℚ since this is the splitting field of the polynomial x^{2} − 3 over ℚ.

(b) ℚ(\sqrt[3]{3}) is not a normal extension of ℚ since this number is a root of the irreducible polynomial x^{3} − 3 ∈ ℚ[x], but the other two roots of this polynomial are not contained in ℚ(\sqrt[3]{3}) (they are the nonreal numbers ρ\sqrt[3]{3}, ρ^{2} \sqrt[3]{3} where ρ = \frac{−1}{2} + \frac{i\sqrt{3}}{2}).

(c) ℚ(\sqrt{5}, i) is a normal extension of ℚ, since this is the splitting field of p(x) = (x^{2} −5)(x^{2} +1). (Note that the polynomial need not be irreducible in F [x] in the theorem.)

18. Let u_{1}, …, u_{t} be the roots of f(x) ∈ F [x] adjoined to obtain E, and let u_{1}, …, u_{n}, n ≥ t be all of the roots of f(x). By the definition, K is a splitting field of f(x) over f if and only if K = F (u_{1}, …, u_{n}). But this is true if and only if

K = F (u_{1}, …, u_{n})
= F (u_{1}, …, u_{t})(u_{t+1}, …, u_{n})
= E(u_{t+1}, …, u_{n})
= E(u_{1}, …, u_{n}) since u_{1}, …, u_{t} are already in E .

Finally, K = E(u_{1}, …, u_{n}) if and only if K is a splitting field of f(x) over E, again by the definition.