Tag Archives: field

Over CC, matrices of finite multiplicative order are diagonalizable

Let A be an n \times n matrix over \mathbb{C}. Show that if A^k=I for some k, then A is diagonalizable.

Let F be a field of characteristic p. Show that A = \begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix} has finite order but cannot be diagonalized over F unless \alpha = 0.


Since A^k = I, the minimal polynomial m of A over \mathbb{C} divides x^k-1. In particular, the roots of m are distinct. Since \mathbb{C} contains all the roots of unity, by Corollary 25 on page 494 of D&F, A is diagonalizable over \mathbb{C}.

Note that \begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & \beta \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & \alpha+\beta \\ 0 & 1 \end{bmatrix}. By an easy inductive argument, then, \begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix}^t = \begin{bmatrix} 1 & t\alpha \\ 0 & 1 \end{bmatrix}, and in particular, A^p = I.

Suppose \alpha \neq 0. Now \frac{1}{\alpha}A is in Jordan canonical form, and is not diagonalizable. (See Corollary 24 on page 493 of D&F.) So A cannot be diagonalizable, for if it were, then so would \frac{1}{\alpha}A. (If P^{-1}AP = D is diagonal, then so is P^{-1}\frac{1}{\alpha}AP = \frac{1}{\alpha}D.)

Any field containing the nth roots of unity for odd n also contains the 2nth roots of unity

Let F be a field over which x^n-1 splits where n is odd. Show that x^{2n}-1 also splits over F.


Note that x^{2n}-1 = (x^n)^2-1 = (x^n+1)(x^n-1).

Since n is odd, if \zeta is a root of x^n-1, then (-\zeta)^n+1 = -\zeta^n+1 = 0. That is, the roots of x^n+1 are precisely the negatives of the roots of x^n-1 (note that these are all distinct, since the derivative of x^n-1 has no nonzero roots). So any field containing the roots of x^n-1 also contains the roots of x^{2n}-1.

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.

Exhibit a quadratic field as a field of matrices

Let K = \mathbb{Q}(\sqrt{D}), where D is a squarefree integer. Let \alpha = a+b\sqrt{D} be in K, and consider the basis B = \{1,\sqrt{D}\} of K over \mathbb{Q}. Compute the matrix of the \mathbb{Q}-linear transformation ‘multiplication by \alpha‘ (described previously) with respect to B. Give an explicit embedding of \mathbb{Q}(\sqrt{D}) in the ring \mathsf{Mat}_2(\mathbb{Q}).


We have \varphi_\alpha(1) = a+b\sqrt{D} and \varphi(\alpha)(\sqrt{D}) = bD + a\sqrt{D}. Making these the columns of a matrix M_\alpha, we have M_\alpha = \begin{bmatrix} a & bD \\ b & a \end{bmatrix}, and this is the matrix of \varphi_\alpha with respect to B. As we showed in the exercise linked above, \alpha \mapsto M_\alpha is an embedding of K in \mathsf{Mat}_2(\mathbb{Q}).

Compare to this previous exercise about \mathbb{Z}[\sqrt{D}].

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.

A class of biquadratic extensions

Let F be a field of characteristic not 2, and let a,b \in F with b not square. Prove that \sqrt{a + \sqrt{b}} = \sqrt{m} + \sqrt{n} for some m,n \in F if and only if a^2-b is square in F. Use this to give a sufficient condition for the extension \mathbb{Q}(\sqrt{a+\sqrt{b}}) to be biquadratic over \mathbb{Q}.


Suppose we have \sqrt{a + \sqrt{b}} = \sqrt{m} + \sqrt{n} = \zeta for some m,n \in F. Evidently, \zeta is a root of p(x) = x^4 - 2ax^2 + (a^2-b) and of q(x) = x^4 - 2(m+n)x^2 + (m-n)^2. Note that the degree of F(\sqrt{b}) over F is 2, and that the degree of F(\sqrt{a+\sqrt{b}}) over F(\sqrt{b}) is either 1 or 2. So the degree of F(\sqrt{a+\sqrt{b}}) over F is either 2 or 4.

Suppose the degree of \zeta over F is 2. Then there exist r,s \in F such that (r+s\sqrt{b})^2 = a+\sqrt{b}. Comparing coefficients, we have r^2+bs^2 = a and 2rs = 1.

Substituting, we have a^2-b = s^4b^2 + (rs-1)b + r^4, which is quadratic in b. Letting p(x) = s^4x^2 + (rs-1)x + r^4, we see that the discriminant of p is (rs-1)^2 - 4s^4r^4 = 0. So p(x) is a perfect square trinomial, and in fact p(b) = (s^2b - \dfrac{1-rs}{2s^2})^2. Thus a^2-b is square in F.

Now suppose the degree of \zeta over F is 4. In this case, p(x) = q(x) is the (unique) monic minimal polynomial of \zeta, and we have a^2-b = (m-n)^2 square in F.

Conversely, suppose a^2-b = t^2 is square in F. Letting m = (a+t)/2 and n = (a-t)/2, evidently we have (\sqrt{m} + \sqrt{n})^2 = a+\sqrt{b}, and thus \sqrt{a+\sqrt{b}} = \sqrt{m} + \sqrt{n}.

Note that by a very similar argument, we can show that this theorem holds if we replace \sqrt{a+\sqrt{b}} = \sqrt{m} + \sqrt{n} by \sqrt{a - \sqrt{b}} = \sqrt{m} - \sqrt{n}.

Now consider the case F = \mathbb{Q}, which has characteristic 0, and let a,b \in \mathbb{Q}. Suppose p(x) = x^4 - 2ax^2 + (a^2-b) is irreducible over \mathbb{Q} and a^2-b is square in \mathbb{Q}. By the above result, \sqrt{a+\sqrt{b}} = \sqrt{m} + \sqrt{n} for some m,n \in \mathbb{Q}. So \mathbb{Q}(\sqrt{a+\sqrt{b}}) \subseteq \mathbb{Q}(\sqrt{m} + \sqrt{n}) \subseteq \mathbb{Q}(\sqrt{m},\sqrt{n}). Note that the first field in this chain has degree 4, and that the last has degree at most 4. So in fact \mathbb{Q}(\sqrt{a+\sqrt{b}}) = \mathbb{Q}(\sqrt{m},\sqrt{n}) is a biquadratic extension of \mathbb{Q}.

Biquadratic field extensions

Let F be a field of characteristic not 2, and let D_1 and D_2 be non-squares in F. Prove that F(\sqrt{D_1}, \sqrt{D_2}) has degree 4 over F if D_1D_1 is nonsquare in F, and degree 2 otherwise. (If the degree is 4, F(\sqrt{D_1},\sqrt{D_2}) is called a biquadratic extension of F.


Suppose D_1D_2 is nonsquare in F. We claim that p(x) = x^2 - D_2 is irreducible over F(\sqrt{D_1}). To see this, suppose to the contrary that (a+b\sqrt{D_1})^2 = D_2. Comparing coefficients, we have a^2 + b^2D_1 - D_2 = 0 and 2ab = 0. If b = 0, then D_2 = a^2, a contradiction. If a = 0, then D_1 = D_2/b^2. But then D_1D_2 = (D_2/b)^2 is square, a contradiction. So F(\sqrt{D_1},\sqrt{D_2}) has degree 2 over F(\sqrt{D_1}), and thus degree 4 over F.

Now suppose that D_1D_2 = \eta^2 is square in F. then (x - \eta/\sqrt{D_1})(x + \eta/\sqrt{D_1}) = x^2 - D_2 is reducible over F(\sqrt{D_1}), and we have F(\sqrt{D_1},\sqrt{D_2}) = F(\sqrt{D_1}) of degree 2 over F.

Every finite field has prime power order

Let F be a finite field of characteristic p. Prove that |F| = p^n for some n.


Suppose F has a nonzero element \alpha of additive order q. That is, q is minimal such that q\alpha = 0. Note that q \leq p, since p annihilates every element of F (being the characteristic of F). If q < p, then by the division algorithm in \mathbb{Z} we have p = qb + r for some b and r with 0 \leq r < q. If r \neq 0, then r \alpha = 0, violating the minimalness of q. If r = 0, then p = qb. Since p is prime, either q = 1 (so \alpha = 0) or b = 1 (so q = p), which both give a contradiction. Thus q = p.

So every element of F has additive order p, and thus by Cauchy’s theorem |F| is a power of p.