Brian Bi
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## Section 15.3. The Degree of a Field Extension

Suppose $$F \subseteq K \subseteq L$$ is a tower of fields where $$K$$ is a finite extension of $$F$$, and suppose $$\alpha \in L$$ has degree $$m$$ over $$F$$, where $$m$$ is relatively prime to $$[K : F]$$. Then $$\alpha$$ has degree $$m$$ over $$K$$.

Clearly $$[K(\alpha) : F] = [K(\alpha) : K][K : F]$$. Also, by Corollary 15.3.8, $$[K(\alpha) : F] = m[K : F]$$. Therefore, $$[K(\alpha) : K] = m$$.

Exercise 15.3.1 Since $$\alpha$$ has degree 5 over $$F$$, $$\alpha^2 \notin F$$. Obviously, $$\alpha^2 \in F(\alpha)$$. By Corollary 15.3.7, $$F(\alpha^2) = F(\alpha)$$.

Exercise 15.3.2 By the Eisenstein criterion, the polynomial $$x^4 + 3x + 3$$ is irreducible over $$\Q$$. Let $$\alpha$$ be a root of $$x^4 + 3x + 3$$; then $$[\Q(\alpha) : \Q] = 4$$. We also have $$[\Q(\sqrt{2}) : \Q] = 3$$. By the Lemma, $$\alpha$$ also has degree 4 over $$\Q(\sqrt{2})$$, so $$x^4 + 3x + 3$$ is also irreducible over $$\Q(\sqrt{2})$$.

Exercise 15.3.3 By irreducibility of the cyclotomic polynomials, the field $$\Q(\zeta_7)$$ has degree 6 as an extension of $$\Q$$, while the number $$\zeta_5$$ has degree 4 over $$\Q$$. Since 4 does not divide 6, Corollary 15.3.6 implies that $$\zeta_5 \notin \Q(\zeta_7)$$.

Note from Brian: The irreducibility of the cyclotomic polynomials $$\Phi_n(x)$$ with $$n$$ a prime power was Exercise 12.4.18, so it should be assumed to be background knowledge at this point. However, for general $$n$$, all known proofs are fairly involved.

Exercise 15.3.4 Feel free to email me to contribute better solutions.

1. $$\zeta_4 = i$$ so clearly the minimal polynomial over $$\Q$$ is $$x^2 + 1$$.

Suppose for the sake of contradiction that $$i \in \Q(\zeta_3)$$. Then we have the chain of field extensions $$\Q \subseteq \Q(i) \subseteq \Q(\zeta_3)$$. Since $$[\Q(\zeta_3) : \Q] = 2$$, and the first inclusion is proper, we have $$\Q(\zeta_3) = \Q(i)$$. But $$\zeta_3$$ has irrational imaginary part, so $$\zeta_3 \notin \Q(i)$$, and a contradiction has been reached. So we conclude that $$i \notin \Q(\zeta_3)$$. The polynomial $$x^2 + 1$$ is therefore minimal for $$i$$ over $$\Q(\zeta_3)$$ as well.

2. $$\zeta_6$$ is a root of the polynomial $$x^6 - 1$$, but we may divide out the factors $$x^3 - 1$$ and $$x + 1$$ to obtain $$x^2 - x + 1$$. Since $$\zeta_6$$ is not a root of either of these factors, $$\zeta_6$$ is still a root of $$x^2 - x + 1$$. The only possible rational roots of this quadratic are $$\pm 1$$, which are not roots, so this polynomial is irreducible. So $$x^2 - x + 1$$ is the minimal polynomial for $$\zeta_6$$ over $$\Q$$.

Notice that $$\zeta_6 = -\zeta_3^2$$, so the minimal polynomial of $$\zeta_6$$ over $$\Q(\zeta_3)$$ is just $$x - \zeta_6$$.

3. We have $$\zeta_8^4 = -1$$, so $$\zeta_8$$ is a root of $$x^4 + 1$$. This is irreducible by Exercise 12.4.18, so it is the minimal polynomial for $$\zeta_8$$ over $$\Q$$.

We now turn to the question of whether $$x^4 + 1$$ is irreducible over $$\Q(\zeta_3)$$ as well. The four roots of this polynomial are $$\zeta_8, \zeta_8^3, \zeta_8^5, \zeta_8^7$$ and their squares are $$\pm i$$. We already argued in part (a) that $$i \notin \Q(\zeta_3)$$, so it must be that none of the roots of $$x^4 + 1$$ are in $$\Q(\zeta_3)$$ either, or, in other words, $$x^4 + 1$$ doesn't have any linear factor in $$\Q(\zeta_3)[x]$$. If $$x^4 + 1$$ is a product of two irreducible quadratics in $$\Q(\zeta_3)[x]$$, the factor of which $$\zeta_8$$ is a root is one of $$(x-\zeta_8)(x-\zeta_8^3) = x^2 - i\sqrt{2}x - 1$$, $$(x-\zeta_8)(x-\zeta_8^5) = x^2 - i$$, or $$(x-\zeta_8)(x-\zeta_8^7) = x^2 - \sqrt{2}x + 1$$. Since $$\Q(\zeta_3) \subseteq \Q(\sqrt{3}, i)$$, it is clear that $$\sqrt{2}$$ and $$i\sqrt{2}$$ don't lie in $$\Q(\zeta_3)$$, so in fact $$x^4 + 1$$ cannot split over $$\Q(\zeta_3)$$ at all, and is the minimal polynomial for $$\zeta_8$$ over $$\Q(\zeta_3)$$.

4. $$\zeta_9$$ is a root of $$x^9 - 1$$, but is not a root of $$x^3 - 1$$. So it is a root of the quotient $$\frac{x^9 - 1}{x^3 - 1} = x^6 + x^3 + 1$$. According to Exercise 12.4.18, this polynomial is irreducible over $$\Q$$, so it is the minimal polynomial for $$\zeta_9$$.

Now $$\Q(\zeta_3) \subseteq \Q(\zeta_9)$$ so we can write $$[\Q(\zeta_9) : \Q] = [\Q(\zeta_9) : \Q(\zeta_3)][\Q(\zeta_3) : \Q]$$ which establishes that $$[\Q(\zeta_9) : \Q(\zeta_3)] = 6/2 = 3$$; the minimal polynomial for $$\zeta_9$$ over $$\Q(\zeta_3)$$ is cubic. Obviously that polynomial must be $$x^3 - \zeta_3$$.

5. $$\zeta_{10}$$ is a root of $$x^{10} - 1$$, but we may divide out factors of $$x^5 - 1$$ and $$x + 1$$, leaving $$x^4 - x^3 + x^2 - x + 1$$. There is an automorphism of polynomial rings that maps $$x$$ to $$-x$$, so this polynomial is irreducible iff $$x^4 + x^3 + x^2 + x + 1$$ is. This latter polynomial is indeed irreducible by Exercise 12.4.18. So the minimal polynomial for $$\zeta_{10}$$ over $$\Q$$ is $$x^4 - x^3 + x^2 - x + 1$$.

It's possible to show that $$x^4 + x^3 + x^2 + x + 1$$ is also irreducible over $$\Q(\zeta_3)$$, although I am not sure how Artin intended for this to be done at this point in the book. A fairly simple method uses the techniques of Chapter 16. On Galois-theoretic grounds, the field $$\Q(\zeta_5)$$ contains a single quadratic subfield, namely $$\Q(\sqrt{5})$$; it does not contain $$\Q(\sqrt{-3})$$, therefore it does not contain $$\zeta_3$$. Therefore $$\zeta_3$$ has degree 2 over $$\Q(\zeta_5)$$, which in turn implies that $$\zeta_5$$ has degree 4 over $$\Q(\zeta_3)$$, so its minimal polynomial over $$\Q(\zeta_3)$$ is indeed $$x^4 + x^3 + x^2 + x + 1$$. Thus, this polynomial is irreducible, and so is $$x^4 - x^3 + x^2 - x + 1$$, so the latter is also the minimal polynomial for $$\zeta_{10}$$ over $$\Q(\zeta_3)$$. See Exercise 16.10.1 for more details on this approach.

6. By similar techniques as in previous parts we find the polynomial $$x^4 - x^2 + 1$$ which is satisfied by $$\zeta_{12}$$. If any of the primitive twelfth roots of unity had degree 1 then so would $$\zeta_6$$, which we already know not to be the case by part (b). If any of them had degree 2, then we would have $$\Q(\zeta_{12}) = \Q(\zeta_6)$$, but we already know that $$i \notin \Q(\zeta_3) = \Q(\zeta_6)$$ by parts (a) and (b), yet $$i \in \Q(\zeta_{12})$$, so this is impossible. Therefore $$x^4 - x^2 + 1$$ is irreducible over $$\Q$$ and is the minimal polynomial for $$\zeta_{12}$$.

Now $$\Q(\zeta_3) \subseteq \Q(\zeta_{12})$$ so $$[\Q(\zeta_{12}) : \Q] = [\Q(\zeta_{12}) : \Q(\zeta_3)][\Q(\zeta_3) : \Q]$$ whereupon we find $$[\Q(\zeta_{12}) : \Q(\zeta_3)] = 2$$. The minimal polynomial for $$\zeta_{12}$$ over $$\Q(\zeta_3)$$ is therefore $$x^2 + \zeta_3$$.

Exercise 15.3.5 By Exercise 12.4.18, the degree of $$\zeta_p$$ over $$\Q$$ is $$p - 1$$ whenever $$p$$ is prime. In general $$\Q(\zeta_m) \subseteq \Q(\zeta_n)$$ whenever $$m \divides n$$, so the degree of $$\zeta_n$$ is at least $$p - 1$$ for each prime $$p$$ that divides $$n$$. The only possibilities if the degree is to be at most 3 are therefore $$n = 2^a 3^b$$. As argued in Exercise 15.3.4, $$[\Q(\zeta_8) : \Q] = 4$$ and $$[\Q(\zeta_9) : \Q] = 6$$, so in fact we must have $$0 \le a \le 2$$ and $$0 \le b \le 1$$, that is, $$n \in \{1, 2, 3, 4, 6, 12\}$$. By the results in Exercise 15.3.4, the solutions are $$n \in \{1, 2, 3, 4, 6\}$$ while $$\zeta_{12}$$ has degree 4.

Exercise 15.3.6 We are given that $$\sqrt{a} \notin \Q$$, so $$\Q(\sqrt{a})$$ is a quadratic extension of $$\Q$$. Clearly $$\Q(\sqrt{a})$$ is an extension of $$\Q(\sqrt{a})$$. Is it possible that $$\sqrt{a} \in \Q(\sqrt{a})$$? No. Suppose $$\sqrt{a} = p + q\sqrt{a}$$ with $$p, q \in \Q$$. Then $$(p + q\sqrt{a})^2 = \sqrt{a}$$, and expanding out the LHS gives $$p^2 + aq^2 + 2pq\sqrt{a} = \sqrt{a}$$. Since $$\{1, \sqrt{a}\}$$ is a $$\Q$$-basis of $$\Q(\sqrt{a})$$, this implies $$p^2 + aq^2 = 0$$. But $$a, p^2, q^2 \ge 0$$, so $$p = q = 0$$, but obviously this is not a solution. Therefore, $$\sqrt{a} \notin \Q(\sqrt{a})$$, so that $$\Q(\sqrt{a})$$ is a quadratic extension of $$Q(\sqrt{a})$$. Therefore $$[\Q(\sqrt{a}) : \Q] = [\Q(\sqrt{a}) : \Q(\sqrt{a})] [\Q(\sqrt{a}) : \Q] = 4$$.

Exercise 15.3.7

1. Suppose for the sake of contradiction that $$i \in \Q(\sqrt{-2})$$. Then we have the inclusions $$\Q \subseteq \Q(\sqrt{-2}) \subseteq \Q(\sqrt{-2})$$ and $$\Q \subseteq \Q(i) \subseteq \Q(\sqrt{-2})$$. Now, $$\sqrt{-2} = i\sqrt{2} \notin \Q(i)$$ so $$\Q(i, \sqrt{-2})$$ is a subfield of $$\Q(\sqrt{-2})$$ that is strictly larger than $$\Q(i)$$. But $$[\Q(\sqrt{-2}) : \Q(i)] = [\Q(\sqrt{-2}) : \Q]/[\Q(i) : \Q] = 2$$, so this subfield must be $$\Q(\sqrt{-2})$$ itself. This implies that $$\sqrt{-2} \in \Q(i, \sqrt{-2})$$. We can therefore write $$(x + y\sqrt{-2})^2 = \sqrt{-2}$$ where $$x, y \in \Q(i)$$. Expanding gives $$x^2 - 2y^2 + 2xy\sqrt{-2} = \sqrt{-2}$$. As $$\{1, \sqrt{-2}\}$$ is a basis for $$Q(i, \sqrt{-2})$$ over $$\Q(i)$$, this implies $$x^2 - 2y^2 = 0$$, which can be factorized to yield $$(x + \sqrt{2}y)(x - \sqrt{2}y) = 0$$. This is clearly impossible to satisfy when $$x, y \in \Q(i)$$, unless $$x = y = 0$$, which doesn't give a solution to the original equation. So we have reached a contradiction, and we conclude that $$i \notin \Q(\sqrt{-2})$$.
2. Some approaches to showing that $$\sqrt{5} \notin \Q(\sqrt{2})$$ are given here. It is not clear what the intended solution is, given that Galois theory hasn't been covered yet at this point in the text.

Exercise 15.3.8 By Corollary 15.3.6(d), the element $$(\alpha + \beta)^2 - 4(\alpha\beta)$$ is algebraic. But this is just $$(\alpha - \beta)^2$$. Since $$(\alpha - \beta)^2$$ is algebraic, so is $$\alpha - \beta$$. By adding and subtracting $$\alpha + \beta$$ and applying Corollary 15.3.6(d), we find that $$\alpha$$ and $$\beta$$ are both algebraic.

Exercise 15.3.9 Observe that \begin{align*} [\Q(\alpha, \beta) : \Q] &= [\Q(\alpha, \beta) : \Q(\alpha)] [\Q(\alpha) : \Q] \\ [\Q(\alpha, \beta) : \Q] &= [\Q(\alpha, \beta) : \Q(\beta)] [\Q(\beta) : \Q] \end{align*} from which it follows that \begin{equation*} \frac{[\Q(\alpha, \beta) : \Q(\alpha)]}{[\Q(\beta) : \Q]} = \frac{[\Q(\alpha, \beta) : \Q(\beta)]}{[\Q(\alpha) : \Q]} \end{equation*} $$f$$ is irreducible over $$L[x]$$ if and only if the RHS equals 1, which will be true if and only if the LHS equals 1, which will be true if and only if $$g$$ is irreducible over $$K[x]$$.

Exercise 15.3.10 Let $$\lambda \in L$$. We are given that $$L/K$$ is algebraic, so we can write $$f(\lambda) = 0$$ where $$f(x) = k_0 + k_1 x + \ldots + k_n x^n$$ with $$k_0, \ldots, k_n \in K$$ and $$k_n \ne 0$$. Then \begin{equation*} [F(\lambda) : F] \le [F(\lambda, k_0, \ldots, k_n) : F] = [F(\lambda, k_0, \ldots, k_n) : F(k_0, \ldots, k_n)] [F(k_0, \ldots, k_n) : F] \end{equation*} Now $$[F(\lambda, k_0, \ldots, k_n) : F(k_0, \ldots, k_n)] \le n$$ and $$[F(k_0, \ldots, k_n) : F]$$ is finite by Corollary 15.3.6(c) since $$K/F$$ is algebraic. Therefore $$[F(\lambda) : F]$$ is finite, that is, $$\lambda$$ is algebraic over $$F$$.