Tag Archives: rationals

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.

Compute the splitting field of x⁶-4 over QQ

Compute the splitting field of p(x) = x^6-4 over \mathbb{Q} and its degree.


Note that x^6-1 factors as x^6-1 = (x+1)(x-1)(x^2+x+1)(x^2-x+1). (Using only the difference of squares and sum of cubes formulas familiar to middle schoolers.) Using the quadratic formula (again familiar to middle schoolers) we see that the roots of p(x) are \pm 1 and \pm \dfrac{1}{2} \pm \dfrac{i\sqrt{3}}{2}.

Now if \zeta is a 6th root of 1, then \sqrt[6]{4}\zeta is a root of p(x), where \sqrt[6]{4} = \sqrt[3]{2} denotes the positive real 6th root of 4 (aka the positive cube root of 2). (Middle schoolers could verify that.) Evidently, then, the splitting field of p(x) is \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2}(1+i\sqrt{3}, \sqrt[3]{2}(1-i\sqrt{3}) = \mathbb{Q}(\sqrt[3]{2}, i\sqrt{3}).

Now \mathbb{Q}(\sqrt[3]{2}) has degree 3 over \mathbb{Q}, and \mathbb{Q}(i\sqrt{3}) has degree 2 over \mathbb{Q}. (Use Eisenstein for both.) By Corollary 22 on page 529 of D&F, then, \mathbb{Q}(\sqrt[3]{2}, i\sqrt{3}) has degree (as a middle schooler could compute) 3 \cdot 2 = 6 over \mathbb{Q}.

So we have proved not only that the splitting field of p(x) has degree 6 over \mathbb{Q}, but, in a metamathematical twist, also that a middle schooler could prove this as well.

Compute the splitting field of x⁴+x²+1 over QQ

Compute the splitting field of p(x) = x^4 + x^2 + 1 over \mathbb{Q}, and its degree.


Note that p = g \circ h, where g(y) = y^2+y+1 and h(x) = x^2. Evidently, the roots of g(y) are \eta = \dfrac{-1 \pm i\sqrt{3}}{2}, and \zeta is a root of p if h(\zeta) = \eta. This yields the (distinct) roots \zeta = \pm \sqrt{\dfrac{-1 \pm i\sqrt{3}}{2}}. There are four of these, and so we have completely factored p.

The splitting field of p(x) over \mathbb{Q} is thus \mathbb{Q}(\pm \sqrt{\dfrac{-1 \pm i\sqrt{3}}{2}}) = \mathbb{Q}(\sqrt{-1/2 \pm \sqrt{-3/4}}).

Using this previous exercise, we have \sqrt{-1/2 + \sqrt{-3/4}} = 1/2 + i\sqrt{3}/2 and \sqrt{-1/2 - \sqrt{-3/4}} = 1/2 - i\sqrt{3}/2.

Evidently, then, the splitting field of p(x) is \mathbb{Q}(i\sqrt{3}). Since i\sqrt{3} is a root of the irreducible x^2+3 (Eisenstein), this extension of \mathbb{Q} has degree 2.

Compute the splitting field of x⁴+2 over QQ

Compute the splitting field of p(x) = x^4+2 over \mathbb{Q} and its degree.


The roots of p(x) exist in \mathbb{C} (if we don’t know this yet, just assume some roots are in \mathbb{C}). Let \zeta = a+bi be such a root; then \zeta^4 = -2.

Evidently, \zeta^4 = [a^4-6a^2b^2 + b^4] + [4ab(a^2-b^2]i. Comparing coefficients, we see that either a = 0, b = 0, or a^2 = b^2. If a = 0, then b^4 = -2, a contradiction in \mathbb{R}. Likewise, if b = 0 we get a contradiction. Thus a^2 = b^2. Substituting, we have b^4 - 6b^4 + b^4 = -2, so -4b^4 = -2, and so b^4 = 1/2. There is a unique positive 4th root of 1/2, which we denote by 2^{-1/4}; so b = \pm 2^{-1/4} and a = \pm 2^{-1/4}, and hence \zeta = \pm 2^{-1/4} \pm 2^{-1/4}i. There are 4 such roots, and so we have completely factored p(x). (WolframAlpha agrees.)

So the splitting field of p(x) is \mathbb{Q}(\pm 2^{-1/4} \pm 2^{-1/4}i). Note that if \zeta_1 = 2^{-1/4} + 2^{-1/4}i and \zeta_2 = \pm 2^{-1/4} - 2^{-1/4}i, then (\zeta_1 + \zeta_2)/2 = 2^{-1/4}, and (\zeta_1 - \zeta_2)/22^{-1/4} = i. Thus \mathbb{Q}(\pm 2^{-1/4} \pm 2^{-1/4}i) = \mathbb{Q}(i,2^{-1/4}).

Note that 2^{-1/4} is a root of q(x) = 2x^4 - 1, and that the reverse of q(x) is t(x) = -x^4+2. Now t(x) is irreducible over the UFD \mathbb{Z}[i], since it is Eisenstein at the irreducible element 1+i. So t(x) is irreducible over \mathbb{Q}(i), the field of fractions of \mathbb{Z}[i] as a consequence of Gauss’ Lemma. As we showed previously, the reverse of t, namely q, is also irreducible over \mathbb{Q}(i). So \mathbb{Q}(i,2^{-1/4}) has degree 4 over \mathbb{Q}(i), and thus has degree 8 over \mathbb{Q}.

Compute the splitting field of x⁴-2 over QQ

Compute the splitting field of p(x) = x^4-2 over \mathbb{Q}, as well as its degree.


Let \sqrt[4]{2} denote the positive real fourth root of 2. Evidently, p(x) = (x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2}). So the splitting field of p(x) over \mathbb{Q} is \mathbb{Q}(\sqrt[4]{2}, i\sqrt[4]{2}) = \mathbb{Q}(i,\sqrt[4]{2}).

Note that \mathbb{Q}(i) has degree 2 over \mathbb{Q}. Over the UFD \mathbb{Z}[i], p(x) is Eisenstein at 1+i, hence irreducible, and so by Gauss’ Lemma, is irreducible over \mathbb{Q}(i) (the field of fractions of \mathbb{Z}[i]). So \mathbb{Q}(i,\sqrt[4]{2}) has degree 4 over \mathbb{Q}(i). We can visualize this scenario in the following diagram.

A field diagram

So \mathbb{Q}(i,\sqrt[4]{2}) has degree 8 over \mathbb{Q}.

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.

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}].

Show that a given family of extensions of the rationals does not contain a cube root of 2

Suppose F = \mathbb{Q}(\alpha_1,\ldots,\alpha_k), where \alpha_i^2 \in \mathbb{Q} for each i. Prove that \sqrt[3]{2} \notin F.


We can see, by an inductive argument, that F has degree 2^k over \mathbb{Q} for some k. (In the chain of fields \mathbb{Q} \subseteq \mathbb{Q}(\alpha_1) \subseteq \ldots F, each entry has degree 1 or 2 over its predecessor, and Theorem 14 in D&F applies.)

Suppose now that \sqrt[3]{2} \in F. Then \mathbb{Q}(\sqrt[3]{2}) \subseteq F, and we have the following diagram of field extensions.

A field diagram

Again using Theorem 14, we have that 3 divides 2^k, a contradiction. So in fact \sqrt[3]{2} \notin 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}.