## 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 $n$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 $\mathbb{Q}(\zeta) \subseteq K$.

### 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 $i$th 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.