Tag Archives: degree

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.

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.

Field extensions of prime degree have no intermediate subfields

Let K/F be a field extension with prime degree p. Show that any subfield of K containing F is either K or F.

Let F \subseteq E \subseteq K. Then [K:E][E:F] = [K:F] = p by Theorem 14 in D&F. Since the degree of a (finite) field extension is an integer, either [K:E] = 1 (so E = K) or [E:F] = 1 (so E = F).

Compute the degree of a given extension of the rationals

Compute the degrees of \mathbb{Q}(\sqrt{3+4i} + \sqrt{3-4i}) and \mathbb{Q}(\sqrt{1+\sqrt{-3}} + \sqrt{1-\sqrt{-3}}) over \mathbb{Q}.

Let \zeta = \sqrt{3+4i} + \sqrt{3-4i}. Evidently, \zeta^2 = 16. (WolframAlpha agrees.) That is, \zeta is a root of p(x) = x^2 - 16 = (x+4)(x-4). Thus the minimal polynomial of \zeta over \mathbb{Q} has degree 1, and the extension \mathbb{Q}(\zeta) has degree 1 over \mathbb{Q}.

Similarly, let \eta = \sqrt{1+\sqrt{-3}} + \sqrt{1-\sqrt{-3}}. Evidently, \eta^2 = 6 (WolframAlpha agrees), so that \eta is a root of q(x) = x^2-6. q is irreducible by Eisenstein’s criterion, and so is the minimal polynomial of \eta over \mathbb{Q}. The degree of \mathbb{Q}(\eta) over \mathbb{Q} is thus 2.

Compute the degree of a given extension of the rationals

Find the degree of \mathbb{Q}(\sqrt{3+2\sqrt{2}}) over \mathbb{Q}.

We have \sqrt{3+2\sqrt{2}} = \sqrt{3 + \sqrt{8}}, and 3^2 - 8 = 1^2 is square over \mathbb{Q}. By this previous exercise, \sqrt{3+2\sqrt{2}} = 1 + \sqrt{2}. So \mathbb{Q}(\sqrt{3 + 2\sqrt{2}}) = \mathbb{Q}(1+\sqrt{2}) = \mathbb{Q}(\sqrt{2}) has degree 2 over \mathbb{Q}.

Compute the degree of a given extension of the rationals

Prove that \mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3}). Conclude that \mathbb{Q}(\sqrt{2}+\sqrt{3}) has degree 4 over \mathbb{Q}. Find an irreducible polynomial satisfied by \sqrt{2} + \sqrt{3}.

The (\supseteq) inclusion is clear. To see the (\subseteq) inclusion, let \zeta = \sqrt{2} + \sqrt{3}. Then evidently \frac{1}{2}(\zeta^3 - 9\zeta) = \sqrt{2} and -\frac{1}{2}(\zeta^3 - 11\zeta) = \sqrt{3}. (WolframAlpha agrees; see here and here.) Thus the extensions are equal.

Recall that \mathbb{Q}(\sqrt{2}) has degree 2 over \mathbb{Q}. Now we claim that \mathbb{Q}(\sqrt{2},\sqrt{3}) has degree 2 over \mathbb{Q}(\sqrt{2}). To see this, suppose to the contrary that \sqrt{3} \in \mathbb{Q}(\sqrt{2}). Say (a+b\sqrt{2})^2 = 3 for some rationals a and b. Coparing coefficients, we have a^2 + 2b^2 = 3 and 2ab = 0; if a = 0, then b^2 = \frac{3}{2}, and if b = 0, then a^2 = 3. Either case yields a contradiction.

So \mathbb{Q}(\sqrt{2} + \sqrt{3}) has degree 4 over \mathbb{Q}.

The minimal polynomial of \zeta has degree 4 over \mathbb{Q}. We can solve the system \zeta^4 + a\zeta^3 + b\zeta^2 + c\zeta + d = 0 over \mathbb{Q}, and in so doing we see that p(x) = x^4-10x^2+1 is the minimal polynomial of \sqrt{2} + \sqrt{3}.

Compute the degree of a given extension of QQ

Compute the degree of \mathbb{Q}(\alpha) over \mathbb{Q}, where \alpha is 2+\sqrt{3} or 1 + \sqrt[3]{2} + \sqrt[3]{4}.

Since 2+\sqrt{3} \in \mathbb{Q}(\sqrt{3}), the degree of \mathbb{Q}(2+\sqrt{3}) over \mathbb{Q} is at most 2. We can solve the linear system \alpha^2 + a\alpha + b = 0 in \mathbb{Q}(\sqrt{3}) (as a vector space over \mathbb{Q}) to find a polynomial satisfied by \alpha; evidently 2+\sqrt{3} is a root of p(x) = x^2 - 4x + 1. (WolframAlpha agrees.) Evidently, p(x+1) = x^2-2x-2 (WolframAlpha agrees), which is irreducible by Eisenstein’s criterion; so p(x) is irreducible. Thus p(x) is the minimal polynomial of 2+\sqrt{3} over \mathbb{Q}, and so the degree of \mathbb{Q}(2+\sqrt{3}) over \mathbb{Q} is 2.

In a similar fashion, \beta = 1+\sqrt[3]{2} + \sqrt[3]{4}, as an element of \mathbb{Q}(\sqrt[3]{2}), has degree at most 3 over \mathbb{Q}. Evidently, \beta is a root of q(x) = x^3 - 3x^2 - 3x - 1 (WolframAlpha agrees). Evidently, q(x+1) = x^3-6x-6, which is irreducible by Eisenstein. So q(x) is the minimal polynomial of \beta, and thus \mathbb{Q}(\beta) has degree 3 over \mathbb{Q}.