## 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 $i$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 $K$.

### 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 $p$.