## If m and n are relatively prime, the product of a primitive mth root of unity and a primitive nth root of unity is a primitive mnth root of unity

Suppose $m$ and $n$ are relatively prime, and let $\zeta_m$ and $\zeta_n$ be primitive $m$th and $n$th roots of unity, respectively. Show that $\zeta_m\zeta_n$ is a primitive $mn$th root of unity.

Note first that $(\zeta_m\zeta_n)^{mn} = (\zeta_m^m)^n (\zeta_n^n)^m = 1$, so that $\zeta_m\zeta_n$ is an $mn$th root of unity.

Now let $t$ be the order of $\zeta_m\zeta_n$; we have $(\zeta_m\zeta_n)^t = 1$, so that $\zeta_m^t = \zeta_n^{-t}$. In particular, $\zeta_m^t$ and $\zeta_n^{-t}$ have the same order, which must be a divisor of both $m$ and $n$. Since $m$ and $n$ are relatively prime, the order of $\zeta_m^t$ is 1, so $\zeta_m^t = 1$. Likewise $\zeta_n^t = 1$. So $m|t$ and $n|t$, and again since $m$ and $n$ are relatively prime, $mn|t$. So $|\zeta_m\zeta_n| = mn$, and $\zeta_m\zeta_n$ is a primitive $mn$th root of unity.

## A fact about polynomials over a perfect field

Let $K$ be an extension of $F$, with $F$ a perfect field. Suppose $p(x) \in F[x]$ has no repeated irreducible factors; prove that $p(x)$ has no repeated irreducible factors in $K[x]$.

Suppose $p$ has no repeated irreducible factors. Since $F$ is perfect, the irreducible factors of $p$ are separable (Prop. 37 on 549 in D&F), and so $p$ is separable. (If not, then it would have a repeated irreducible factor.) That is, $p$ has no repeated roots.

Now if $p$ has a repeated irreducible factor over $K$, then it has repeated roots- a contradiction.

## A fact about multivariate polynomials over ZZ

Let $f(x_1,\ldots,x_n) \in \mathbb{Z}[x_1,\ldots,x_n]$, and let $p$ be a prime. Prove that $f(x_1,\ldots,x_n)^p = f(x_1^p,\ldots,x_n^p)$ mod $p$.

Remember that $\alpha^p \equiv \alpha$ mod $p$ for all $\alpha$. Say $f = \sum_i c_i \prod_j x_j^{e_{i,j}}$.

Then $f(x_1,\ldots,x_n)^p = (\sum_i c_i \prod_j x_j^{e_{i,j}})^p$ $= \sum_i c_i^p \prod_j x_j^{e_{i,j}p}$ $= \sum_i c_i \prod_j (x_j^p)^{e_{i,j}}$ $= f(x_1^p,\ldots,x_n^p)$ as desired.

## pn choose pk and n choose k are congruent mod p

Show that $\binom{pn}{pk} \equiv \binom{n}{k}$ mod $p$.

Note that $(1+x)^{pn} = \sum_{t=0}^{pn} \binom{pn}{t}x^t$ by the binomial theorem. The coefficient of $x^{pk}$ is $\binom{pn}{pk}$. Over $\mathbb{F}_p$, we have $(1+x)^{pn} = ((1+x)^p)^n$ $= (1+x^p)^n$ $= \sum_{t=0}^n \binom{n}{t}(x^p)^t$, and now the coefficient of $x^{pk}$ is $\binom{n}{k}$. In particular, we have $\binom{pn}{pk} \equiv \binom{n}{k}$ mod $p$.

## f(x)ᵖ = f(xᵖ) over ZZ/(p)

Prove that $f(x)^p = f(x^p)$ for all $f(x) \in \mathbb{F}_p[x]$.

Let $f(x) = \sum c_ix^i$. Remember that the elements of $\mathbb{F}_p$ are precisely the roots of $x^p-x$; in particular, $\alpha^p = \alpha$ for all $\alpha \in \mathbb{F}_p$.

Then $f(x)^p = (\sum c_ix^i)^p$ $= \sum (c_ix^i)^p$ $= \sum c_i^p(x^i)^p$ $= \sum c_i (x^p)^i$ $= f(x^p)$ as desired.

## Over an imperfect field of characteristic p, there exist irreducible inseparable polynomials

Let $K$ be an imperfect field of characteristic $p$. Prove that there exist irreducible inseparable polynomials over $K$. Conclude that there exist inseparable finite extensions of $K$.

Let $\alpha \in K$ be an element which is not a $p$th power in $K$, and let $\zeta$ be a root of $x^p-\alpha$. In particular, $\zeta \notin K$. Now consider $q(x) = (x-\zeta)^p = x^p-\alpha$. Suppose $t(x)$ is an irreducible factor of $q$ over $K$. Now $t$ cannot be linear, since $\zeta \notin K$. But any factor of $q$ of degree at least 2 has a multiple root (since all the roots are $\zeta$). So $t$ is an irreducible inseparable polynomial over $K$.

Then $K(\zeta)$ is an inseparable extension of $K$.

## Wilson’s Theorem

Prove that $x^{p^n-1}-1 = \prod_{\alpha \in \mathbb{F}_{p^n}^\times} (x-\alpha)$. Conclude that $\prod_{\alpha \in \mathbb{F}_{p^n}^\times} \alpha = (-1)^{p^n}$. Deduce Wilson’s Theorem: if $p$ is an odd prime, then $(p-1)! \equiv -1$ mod $p$.

Recall that $x^{p^n}-x = \prod_{\alpha \in \mathbb{F}_{p^n}^\times}(x-\alpha)$ by definition, and that $0 \in \mathbb{F}_{p^n}$ is merely the root having minimal polynomial $x$. So $x^{p^n-1}-1 = \prod_{\alpha \in \mathbb{F}_{p^n}^\times} (x-\alpha)$. Comparing constant coefficients, we have $\prod_{\alpha \in \mathbb{F}_{p^n}^\times} (-\alpha) = -1$, so that \$latex $(-1)^{p^n-1} \prod_{\alpha \in \mathbb{F}_{p^n}^\times} \alpha = -1$, and hence $\prod_{\alpha \in \mathbb{F}_{p^n}^\times} \alpha = (-1)^{p^n}$.

Restrict now to the field $\mathbb{F}_p \cong \mathbb{Z}/(p)$ with $p$ odd. Then $\prod_{\alpha \in \mathbb{F}_p} \alpha = (p-1)!$, and $(-1)^p = -1$. Thus $(p-1)! \equiv -1$ mod $p$.

(I’d like to point out that this is a really roundabout way to prove Wilson’s Theorem. The easy(ier) way is to note that in $\mathbb{Z}/(p)^\times$, every element but -1 is distinct from its inverse.)

## xᵖ-x+a is irreducible and separable over ZZ/(p)

Prove that for any prime $p$ and any nonzero $a \in \mathbb{F}_p$, $q(x) = x^p-x+a$ is irreducible and separable over $\mathbb{F}_p$.

Note that $D_x(q) = px^{p-1}-1 = -1$, so that $q$ and $D(q)$ are relatively prime. So $q$ is separable.

Now let $\alpha$ be a root of $q$. Using the Frobenius endomorphism, $(\alpha+1)^p-(\alpha+1)+a = \alpha^p - \alpha + a = 0$, so that $\alpha+1$ is also a root. By induction, $\alpha_k$ is a root for all $k \in \mathbb{F}_p$, and since $q$ has degree $p$, these are all of the roots.

Now $\mathbb{F}_p(\alpha+k) = \mathbb{F}_p(\alpha)$, and in particular the minimal polynomials of $\alpha$ and $\alpha+k$ have the same degree over $\mathbb{F}_p$ – say $d$. Since $q$ is the product of the minimal polynomials of its roots, we have $p = dt$ for some $t$. Since $p$ is prime, we have either $d=1$ (so that $\alpha \in \mathbb{F}_p$, a contradiction) or $d=p$, so that $q$ itself is the minimal polynommial of $\alpha$, hence is irreducible.

## If a is an integer greater than 1, then aᵈ-1 divides aⁿ-1 if and only if d divides n

Fix an integer $a > 1$. Prove that for all $d,n \in \mathbb{N}$, $d$ divides $n$ if and only if $a^d-1$ divides $a^n-1$. Conclude that $\mathbb{F}_{p^d} \subseteq \mathbb{F}_{p^n}$ if and only if $d|n$.

If $d|n$, then by this previous exercise, $x^d-1$ divides $x^n-1$ in $\mathbb{Z}[x]$, and so $a^d-1$ divides $a^n-1$ in $\mathbb{Z}$.

Conversely, suppose $a^d-1$ divides $a^n-1$. Using the Division Algorithm, say $n = qd+r$. Now $a^n-1 = (a^d)^qa^r - a^r + a^r - 1$ $= a^r((a^d)^q-1) + (a^r-1)$. If $q = 1$, then we have $a^r-1 = 0$, so that $r = 0$ and $d|n$. If $q > 1$, then $a^n-1 = a^r(a^d-1)(\sum_{i=0}^{q-1} (a^d)^i) + (a^r-1)$, and again we have $r = 0$, so that $d|n$ as desired.

Recall that $\mathbb{F}_{p^k}$ is precisely the roots (in some splitting field) of $x^{p^k}-x$. Now $d|n$ if and only if $p^d-1$ divides $p^n-1$ by the above argument, if and only if $x^{p^d-1}-1$ divides $x^{p^n-1}-1$ by a previous exercise, if and only if $x^{p^d}-x$ divides $x^{p^n}-x$, if and only if $\mathbb{F}_{p^d} \subseteq \mathbb{F}_{p^n}$.

## xᵈ-1 divides xⁿ-1 if and only if d divides n

Prove that $x^d-1$ divides $x^n-1$ if and only if $d$ divdes $n$.

Suppose $d|n$; say $n = dt$.

If $t = 1$, there is nothing to show. Suppose $t > 1$; then $x^n-1 = (x^d)^t - 1$ $= (x^d-1)(\sum_{i=0}^{t-1} (x^d)^i)$, so that $x^d-1$ divides $x^n-1$.

Now suppose $x^d-1$ divides $x^n-1$. By the Division Algorithm in $\mathbb{N}$, we have $q$ and $r$ such that $n = qd+r$. Now $x^n-1 = x^{qd}x^r - 1$ $= x^{qd}x^r -x^r + x^r - 1$ $= x^r(x^{qd}-1) + (x^r-1)$. If $q=1$, then (by the uniqueness part of the division algorithm in $\mathbb{Q}[x]$) $x^r-1 = 0$, so $r = 0$ and $d|n$. If $q > 1$, then $x^n - 1 = x^r(x^d-1)(\sum_{i=0}^{q-1} (x^d)^i) + (x^r-1)$, and again we have $r = 0$, so $d|n$.