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 $n$th root of unity for some $n > m$.

Let $m$ be the degree of $K$ over $\mathbb{Q}$. If $\zeta \in K$ is a primitive $k$th 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 $k$th 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 $\mathbb{Q}$.