Tag Archives: field extension

Finite extensions of the rationals contain only finitely many roots of unity

Let K be a finite extension of \mathbb{Q}. Prove that K contains only finitely many roots of unity.

Suppose to the contrary that K contains infinitely many roots of unity. Now for each n, there are only finitely many primitive roots of unity (in fact \varphi(n) of them). So for each m, the number of primitive roots of unity of order at most m is finite. In particular, for any m, there exists a primitive nth root of unity for some n > m.

Let m be the degree of K over \mathbb{Q}. If \zeta \in K is a primitive kth root of unity, then [\mathbb{Q}(\zeta):\mathbb{Q}] = \varphi(k) by Corollary 42 on page 555, where \varphi denotes the Euler totient. By this previous exercise, since k is arbitrarily large, \varphi(k) is arbitrarily large. So there exists a primitive kth root \zeta such that \varphi(k) > m, a contradiction since \mathbb{Q}(\zeta) \subseteq K.

Argue that the regular 5-gon is constructible

Use the fact that \alpha = 2\mathsf{cos}(2\pi/5) satisfies the polynomial p(x) = x^2+x-1 (to be proved later) to argue that the regular 5-gon is constructible using a straightedge and compass.

Using the rational root test, we can see that p(x) is irreducible over \mathbb{Q}. Thus the roots of p(x) lie in a degree 2 extension of \mathbb{Q}, and we have seen that all such numbers are constructible by straightedge and compass. So 2\mathsf{cos}(2\pi/5), and hence \beta = \mathsf{cos}(2\pi/5), is constructible.

Recall that \mathsf{sin}^2\ \theta + \mathsf{cos}^2\ \theta = 1, and so (since 2\pi/5 is in the first quadrant) \mathsf{sin}(2\pi/5) = \sqrt{1 - \mathsf{cos}^2(2\pi/5)}. In particular, \mathsf{sin}(2\pi/5) is also constructible.

I’ll try to describe the rest in words, because my geometric diagrams tend to look like garbage unless I spend a couple of hours on them.

Suppose now that we have a line segment AB, which we want to be an edge of a regular 5-gon. Extend the line AB and construct the line perpendicular to AB at B. On AB, construct the point X such that ABX and BX has measure \mathsf{cos}(2\pi/5). On the perpendicular, construct the point Y such that BY has measure \mathsf{sin}(2\pi/5). Now construct the perpendicular to BX at X and to BY at Y, and let Z be the intersection of these two lines. Finally, construct the point C on BZ such that either BCZ or BZC and such that BC and AB have the same measure. By construction, \angle CBX has measure 2\pi/5, so that \angle ABC has measure 3\pi/5. Repeat this construction with BC (and so on), being careful with the orientation, to construct a regular 5-gon.

When the tensor product of finite field extensions is a field

Let K_1 and K_2 be finite extensions of a field F contained in a field K. Prove that the F-algebra K_1 \otimes_F K_2 is a field if and only if [K_1K_2 : F] = [K_1:F][K_2:F].

First, define \psi : K_1 \times K_2 \rightarrow K_1K_2 by (a,b) \mapsto ab. Clearly \psi is F-bilinear, and so induces an F-module homomorphism \Psi : K_1 \otimes_F K_2 \rightarrow K_1K_2. In fact \Psi is an F-algebra homomorphism. Using Proposition 21 in D&F, if A = \{\alpha_i\} and B = \{\beta_j\} are bases of K_1 and K_2 over F, then AB = \{\alpha_i\beta_j\} spans K_1K_2 over F. In particular, \Psi is surjective.

Suppose K_1 \otimes_F K_2 is a field. Now \mathsf{ker}\ \Psi is an ideal of K_1 \otimes_F K_2, and so must be trivial- so \Psi is an isomorphism of F-algebras, and thus an isomorphism of fields. Using Proposition 21 on page 421 of D&F, K_1 \otimes_F K_2 has dimension [K_1:F][K_2:F], and so [K_1K_2 : F] = [K_1:F][K_2:F] as desired.

Conversely, suppose [K_1K_2 : F] = [K_1:F][K_2:F]. That is, K_1 \otimes_F K_2 and K_1K_2 have the same dimension as F-algebras. By Corollary 9 on page 413 of D&F, \Psi is injective, and so K_1 \otimes_F K_2 and K_1K_2 are isomorphic as F-algebras, hence as rings, and so K_1 \otimes_F K_2 is a field.

A procedure for finding a polynomial satisfied by an element of a given algebraic field extension

Let F be a field, K an extension of F of finite degree, and let \alpha \in K. Show that if A is the matrix of the linear transformation \varphi_\alpha corresponding to ‘multiplication by \alpha‘ (described here) then \alpha is a root of the characteristic polynomial of A. Use this result to obtain monic polynomials of degree 3 satisfied by \alpha = \sqrt[3]{2} and \beta = 1 + \sqrt[3]{2} + \sqrt[3]{4}.

Let \Psi : K \rightarrow \mathsf{Mat}_n(F) be the F-linear transformation described here. If c(x) is the characteristic polynomial of A, then we have c(A) = 0. On the other hand, 0 = c(A) = c(\Psi(\alpha)) = \Psi(c(\alpha)), and so c(\alpha) = 0 since \Psi is injective. So \alpha is a root of c(\alpha).

Consider the basis \{1,\sqrt[3]{2},\sqrt[3]{4}\} of \mathbb{Q}(\sqrt[3]{2}) over \mathbb{Q}. Evidently, with respect to this basis, the matrix of \varphi_\alpha is A = \begin{bmatrix} 0 & 0 & 2 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}. As we showed previously, the characteristic polynomial of A is x^3-2. So \sqrt[3]{2} satisfies x^3-2. (Surprise!)

Similarly, \varphi_\beta has the matrix B = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 1 \end{bmatrix}. Evidently, the characteristic polynomial of B is x^3-3x^2-3x-1. We can verify that \beta actually satisfies this polynomial (WolframAlpha agrees.)

Every degree n extension of a field F is embedded in the set of nxn matrices over F

Let F be a field, and let K be an extension of F of finite degree.

  1. Fix \alpha \in K. Prove that the mapping ‘multiplication by \alpha‘ is an F-linear transformation on K. (In fact an automorphism for \alpha \neq 0.)
  2. Deduce that K is isomorphically embedded in \mathsf{Mat}_n(F).

Let \varphi_\alpha(x) = \alpha x. Certainly then we have \varphi_\alpha(x+ry) = \alpha(x+ry) = \alpha x + r \alpha y = \varphi_\alpha(x) + r \varphi_\alpha(y) for all x,y \in K and r \in F; so \varphi_\alpha is an F-linear transformation. If \alpha \neq 0, then evidently \varphi_{\alpha^{-1}} \circ \varphi_\alpha = \varphi_\alpha \circ \varphi_{\alpha^{-1}} = 1.

Fix a basis for K over F; this yields a ring homomorphism \Psi : K \rightarrow \mathsf{Mat}_n(F) which takes \alpha and returns the matrix of \varphi_\alpha with respect to the chosen basis. Suppose \alpha \in \mathsf{ker}\ \Psi; then \varphi_\alpha(x) = 0 for all x \in K, and thus \alpha = 0. So \Psi is injective as desired.

On the degree of an extension of F(x)

Let F be a field, and consider the field F(x) of rational functions over F (that is, the field of fractions of the domain F[x]). Let t(x) = p(x)/q(x), with p,q \in F[x] and q \neq 0 such that the degree of q is strictly larger than the degree of p. In this exercise, we will compute the degree of F(x) over F(t(x)). (Note that if p has degree larger than or equal to the degree of q, then we can use the division algorithm to find p(x) = q(x)b(x) + r(x), and then F(t(x)) = F(r(x)/q(x)).)

  1. Show that p(y) - t(x)q(y) \in F(t(x))[y] is irreducible over F(t(x)) and has x as a root (in the extension F(x)).
  2. Show that the degree of p(y) - t(x)q(y) as a polynomial in y is the maximum of the degrees of p and q.
  3. Conclude that [F(x):F(t(x))] = \mathsf{max}(\mathsf{deg}\ p(x)), \mathsf{deg}\ q(x)).

(Fields of rational functions were introduced in Example 4 on page 264 of D&F.)

We claim that t(x) is indeterminate over F– that is, that t(x) does not satisfy any polynomial over F. Indeed, if \sum_{i=0}^n c_i t(x)^i = \sum c_i \dfrac{p(x)^i}{q(x)^i} = 0, then we have \sum c_i p(x)^iq(x)^{n-i} = 0. Let d_p and d_q denote the degrees of p and q, respectively. Since F has no zero divisors, the degree of the ith summand is d_pi + d_q(n-i). Suppose two summands have the same degree; then d_pj + d_q(n-i) = d_pj + d_q(n-j) for some i and j, which reduces to (d_p-d_q)i = (d_p-d_q)j. Since (as we assume) d_p \neq d_q, we have i = j. In this case, we can pick out the summand with the highest degree, and be guaranteed that no other summands contribute to its term of highest degree. This gives us that c_i = 0 for the highest degree summand; by induction we have c_i = 0 for all the coefficients c_i, a contradiction. (In short, no two summands have the same highest degree term. Starting from the highest of the highest degree terms and working down, we show that each c_i is 0.)

So t(x) is indeterminate over F, and in fact F[t(x)] is essentially a polynomial ring whose field of fractions is F(t(x)). By Gauss’ Lemma, the polynomial p(y) - t(x)q(y) is irreducible over F(t(x)) if and only if it is irreducible over F[t(x)]. Now F[t(x)][y] = F[y][t(x)], and our polynomial is irreducible in this ring since it is linear in t(x). So in fact p(y) - t(x)q(y) is irreducible over F(t(x)), and moreover has x as a root.

The degree of p(y) - t(x)q(y) in y is the maximum of the degrees of p and q because the coefficient of each term (in y) is a linear polynomial in t(x), which is nonzero precisely when one of the corresponding terms in p or q is nonzero. That is, we cannot have nonzero terms in p(y) and -t(x)q(y) adding to give a zero term.

To summarize, p(y) - t(x)q(y) is an irreducible polynomial over t(x) with x as a root, and so must be (essentially) the minimal polynomial of F(x) over F(t(x)). By the preceding paragraph, the degree of this extension is the larger of the degrees of p(x) and q(x).

On the degrees of the divisors of a composite polynomial

Let f(x) be an irreducible polynomial of degree n over a field F, and let g(x) be any polynomial in F[x]. Prove that every irreducible factor of the composite (f \circ g)(x) has degree divisible by n.

Let \beta be a root of f \circ g, and let m be the degree of \beta over F. Now g(\beta) \in F(\beta), and moreover g(\beta) is a root of the irreducible polynomial f(x). So F(g(\beta)) has degree n over F. Graphically, we have the following diagram of fields.

A field diagram

By Theorem 14 in D&F, n|m.

Subrings of an algebraic field extension which contain the base field are subfields

Let K/F be an algebraic field extension and let R \subseteq K be a subring containing F. Show that R is a subfield of K.

It suffices to show that R is closed under inversion. To this end, let \alpha \in R. Since K is algebraic over F, every element of K is algebraic over F. Let p(x) be the minimal polynomial of \alpha over F; if p(x) = \sum c_ix^i, then we have \sum c_i\alpha^i = 0. Note that c_0 \neq 0 since p is irreducible. Rearranging, we see that \alpha (\frac{-1}{c_0} \sum_{i \neq 0} c_i \alpha^{i-1}) = 1, and \frac{-1}{c_0} \sum_{i \neq 0} c_i \alpha^{i-1} \in R. So \alpha^{-1} \in R, and R is closed under inversion. Thus R is a subfield of K.

Odd degree extensions of formally real fields are formally real

A field F is called formally real if -1 cannot be expressed as a sum of squares in F. Let F be a formally real field and let p(x) be irreducible over F of odd degree with a root \alpha. Show that F(\alpha) is formally real.

Suppose to the contrary that there exist formally real fields F and irreducible, odd degree polynomials p over F with roots \alpha such that F(\alpha) is not formally real. Choose F and p from among these fields such that the degree of p is minimal, let \alpha be a root of p, and consider (1) F(\alpha) \cong F[x]/(p(x)).

Since (as we assume) F(\alpha) is not formally real, we have -1 = \sum \theta_i for some \theta_i \in F(\alpha). Let p_i be the residue in F[x] corresponding to \theta_i under the isomorphism (1); then we have -1 \equiv \sum p_i(x)^2 in F[x]/(p(x)), so that -1 + p(x)q(x) = \sum p_i(x)^2 in F[x], for some q(x).

Note that, since each p_i is a residue mod p (using the division algorithm), each p_i has degree strictly less than the degree of p. Since F is a domain, \sum p_i(x)^2 has even degree less than 2 \mathsf{deg}\ p. So that pq, has even degree, and since p has odd degree, q has odd degree, and in fact the degree of q is strictly less than the degree of p. Moreover, since the degree of q is the sum of the degrees of its irreducible factors, some irreducible factor of q, say t, must also have odd degree (less than that of p).

Let \beta be a root of t(x). In F(\beta) = F[x]/(t(x)), we have that -1 \equiv \sum p_i(x)^2. So F is formally real, t irreducible over F, \beta a root of t, and F(\beta) is not formally real – but F(\beta) has degree (over F) strictly less than the degree of F(\alpha), a contradiction.

So no such fields and polynomials exist, and in fact every odd extension of a formally real field is formally real.

If [F(α):F] is odd, then F(α²) = F(α)

Let F be a field, and let \alpha be algebraic over F. Prove that if [F(\alpha):F] is odd, then F(\alpha^2) = F(\alpha).

The inclusion F(\alpha^2) \subseteq F(\alpha) is immediate.

Let p(x) = \sum c_ix^i be the minimal polynomial of \alpha over F. Now write p(x) = \sum c_{2i}x^{2i} + \sum c_{2i+1}x^{2i+1}, separating the even and odd terms. Now p(x) = \sum c_{2i}(x^2)^i + \sum c_{2i+1}x(x^2)^i = r(x^2) + xs(x^2), and we have r(\alpha^2) + \alpha s(\alpha^2) = 0. Since p(x) has odd degree and the powers of \alpha are linearly independent over F, s(\alpha^2) \neq 0. Thus \alpha = -r(\alpha^2)/s(\alpha^2) \in F(\alpha^2), and we have F(\alpha) \subseteq F(\alpha^2).

Note that our proof only needs some nonzero odd-degree term in the minimal polynomial of \alpha, so in some cases the proof holds if \alpha has even degree.