Tag Archives: roots of unity

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.

There are m distinct pᵏmth roots of unity over a field of characteristic p, for p a prime not dividing m

Let n = p^km, where p is a prime not dividing m, and let F be a field of characteristic p. Prove that there are m distinct nth roots of unity over F.

Note that x^n-1 = x^{p^km}-1 = (x^m)^{p^k}-1 = (x^m-1)^{p^k}. That is, the distinct nth roots of unity over F are precisely the distinct roots of x^m-1 over F.

If m=1, then certainly there is only 1 root of x-1.

Suppose m > 1. Now D(x^m-1) = mx^{m-1}, which has only the root 0 with multiplicity m-1. Clearly 0 is not a root of x^m-1, so that x^m-1 and its derivative are relatively prime, and thus x^m-1 is separable. Hence there are m distinct mth roots of unity over F, and so m distinct nth roots of unity over F.

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.

If ζ is a primitive nth root of unity and d divides n, then ζᵈ is a primitive (n/d)th root of unity

Let \zeta be a primitive nth root of unity and let d|n. Prove that \zeta^d is a primitive n/dth root of unity.

Certainly (\zeta^d)^{n/d} = 1. Now if (\zeta^d)^t = 1, we have n|dt, and so n/d divides t. So n/d is the order of \zeta^d, and thus \zeta^d is a primitive n/dth root of unity.

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 mth and nth roots of unity, respectively. Show that \zeta_m\zeta_n is a primitive mnth 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 mnth 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 mnth root of unity.

Find the irreducible polynomials of degree 1, 2, and 4 over ZZ/(2)

Find all the irreducible polynomials of degree 1, 2, or 4 over \mathbb{F}_2, and verify that their product is x^{16}-x.

Note that there are 2^k (monic) polynomials of degree k, as each non-leading coefficient can be either 0 or 1.

The polynomials of degree 1, p_1(x) = x and p_2(x) = x+1, are both irreducible.

There are 4 polynomials of degree 2, one of which is irreducible.

  1. x^2 = xx is reducible
  2. x^2+1 = (x+1)^2 is reducible
  3. x^2+x = x(x+1) is reducible
  4. p_3(x) = x^2+x+1 has no roots, and so is irreducible.

Before we address the degree 4 polynomials, we prove a lemma.

Lemma 1: If p(x) is a degree 4 polynomial over \mathbb{F}_2 with constant term 1 and p factors as a product of quadratics, then the linear and cubic terms of p are equal. Proof: We have (as we assume) p(x) = (x^2+ax+b)(x^2+cx+d) = x^4 + (a+c)x^3 + (b+d+ac)x^2 + (ad+bc)x+bd. Now bd = 1, so that b = d = 1. So p(x) = x^4 + (a+c)x^3 + acx^2 + (a+c)x + 1, as desired. \square

Lemma 2: \Phi_5(x) = x^4+x^3+x^2+x+1 is irreducible over \mathbb{F}_2. Proof: Clearly this has no roots. By Lemma 1, if \Phi_5(x) factors into two quadratics, then the factors’ linear terms, a and c, satisfy a+c=ac=1, which is impossible over \mathbb{F}_2. \square

There are 16 polynomials of degree 4.

  1. x^4 = xx^3 is reducible
  2. x^4+1 = (x^2+1)^2 is reducible
  3. x^4+x = x(x^3+1) is reducible
  4. x^4+x^2 = x^2(x^2+1) is reducible
  5. x^4+x^3 = x^3(x+1) is reducible
  6. p_4(x) = x^4+x+1 clearly has no roots, and by the lemma, has no quadratic factors. So p_4 is irreducible.
  7. x^4+x^2+1 = (x^2+x+1)^2 is reducible
  8. p_5(x) = x^4+x^3+1 clearly has no roots, and by Lemma 1, has no quadratic factors. So p_5 is irreducible.
  9. x^4+x^2+x = x(x^3+x+1) is reducible
  10. x^4+x^3+x = x(x^3+x^2+1) is reducible
  11. x^4+x^3+x^2 = x^2(x^2+x+1) is reducible
  12. x^4+x^2+x+1 has 1 as a root, so is reducible
  13. x^4+x^3+x^2+1 = (x+1)(x^3+1) is reducible
  14. x^4+x^3+x^2+1 has 1 as a root, so is reducible
  15. x^4+x^3+x^2+x = x(x^3+x^2+x+1) is reducible
  16. p_6(x) = x^4+x^3+x^2+x+1 = \Phi_5(x) is irreducible

So there are six irreducible polynomials of degree 1, 2, or 4 over \mathbb{F}_2:

  1. p_1(x) = x
  2. p_2(x) = x+1
  3. p_3(x) = x^2+x+1
  4. p_4(x) = x^4+x+1
  5. p_5(x) = x^4+x^3+1
  6. p_6(x) = x^4+x^3+x^2+x+1

It is easy (if tedious) to verify that the product of these polynomials is x^{16}-x. (WolframAlpha agrees.)

If the roots of a monic irreducible polynomial over ZZ all have modulus at most 1, then the roots are roots of unity

Let p(x) \in \mathbb{Z}[x] be a monic irreducible polynomial and suppose that for each root \zeta of p(x), we have ||\zeta || \leq 1, where || \cdot || denotes complex modulus. Show that the roots of p(x) are roots of unity.

Note that p(x) factors as p(x) = \prod (x - \zeta_i), where \zeta_i ranges over the roots of p. Now the constant coefficient of p(x) is \prod \zeta_i, and thus ||\prod \zeta_i|| = \prod ||\zeta_i|| \leq 1. On the other hand, the constant coefficient is an integer, and so is either 0 or 1. If the constant coefficient is 0, then x divides p(x). So the constant coefficient of p is 1, and we have ||\zeta_i|| = 1 for all i.

By Lemma 11.8, the \zeta_i are all roots of unity.

The conjugates of a root of unity are roots of unity

Let \zeta be a root of unity. Show that the conjugates of \zeta are also roots of unity.

If \zeta is a root of 1, then \zeta is a root of p(x) = x^n - 1 for some n. Thus the minimal polynomial of \zeta over \mathbb{Q} is also a root of p(x), so the conjugates of \zeta are roots of unity.

Basic properties of the p-adic integers

Let p be a prime. Let I = \mathbb{N}, let A_k = \mathbb{Z}/(p^k), and let \mu_{j,i} : \mathbb{Z}/(p^j) \rightarrow \mathbb{Z}/(p^i) be the natural projection maps. The inverse limit \varprojlim \mathbb{Z}/(p^k) is called the ring of p-adic integers and is denoted \mathbb{Z}_p.

  1. Show that every element of \mathbb{Z}_p can be written uniquely as a formal sum \sum_{\mathbb{N}^+} b_k p^k where b \in \{0, \ldots, p-1\}. Describe the algorithms for addition and multiplication of such formal sums corresponding to addition and multiplication in \mathbb{Z}_p. [Hint: Write a least residue in each \mathbb{Z}/(p^k) in its base p expansion and then describe the maps \mu_{j,i}.] (Note that \mathbb{Z}_p is uncountable.)
  2. Prove that \mathbb{Z}_p is an integral domain and contains an isomorphic copy of \mathbb{Z}.
  3. Prove that the formal sum \sum b_kp^k as in part (a) is a unit in \mathbb{Z}_p if and only if b_0 \neq 0.
  4. Prove that p\mathbb{Z}_p is the unique maximal ideal of \mathbb{Z}_p and that \mathbb{Z}_p/(p) \cong \mathbb{Z}/(p). Prove further that every ideal of \mathbb{Z}_p is of the form (p^n) for some n \geq 0. (Here p = \sum b_kp^k, where b_1 = 1 and b_k = 0 otherwise.)
  5. Show that if a_1 \not\equiv 0 mod p then there is an element \alpha = (a_i) in \mathbb{Z}_p such that a_k^p \equiv 1 mod p^k and \mu_{k,0}(a_k) = a_1 for all k. Deduce that \mathbb{Z}_p contains p-1 distinct p-1th roots of 1. (That is, p-1 distinct elements a such that a^p = 1.)

  1. We will denote by [0,p) the set of integers \{0,1,\ldots,p-1\}. Given (\overline{a_i}) \in \varprojlim \mathbb{Z}/(p^k) with 0 \leq a_i < p^i, define \Phi((\overline{a_i})) by \Phi((\overline{a_i}))_1 = a_1 and \Phi((\overline{a_i}))_{k+1} = (a_{k+1} - a_k)/p^k. I claim that \Phi((\overline{a_i}))_k \in [0,p) for all k. This is certainly true for k=0. Since \mu_{k+1,k}(\overline{a_{k+1}}) = \overline{a_k}, we have a_{k+1} \equiv a_k mod p^k. Thus \Phi((\overline{a_i}))_{k+1} = (a_{k+1} - a_k)/p^k is an integer in [0,p). Thus we see that in fact \Phi((\overline{a_i})) \in \prod_{\mathbb{N}^+} [0,p), and thus \Phi : \varprojlim_{\mathbb{N}^+} \mathbb{Z}/(p^k) \rightarrow \prod_{\mathbb{N}^+} [0,p).

    I claim now that \Phi is a bijection. To see injectivity, suppose \Phi((\overline{a_i})) = \Phi((\overline{b_i})), where 0 \leq a_i,b_i < p^i for each i. We proceed by induction on the index k. For the base case k = 1, we have a_1 = \Phi((\overline{a_i}))_1 = \Phi((\overline{b_i}))_1 = b_1. For the inductive step, suppose that a_k = b_k for some k \geq 1. Then (a_{k+1} - a_k)/p^k = \Phi((\overline{a_i}))_{k+1} = \Phi((\overline{b_i}))_{k+1} = (b_{k+1} - b_k)/p^k. Thus a_{k+1} = b_{k+1}. So (\overline{a_i}) = (\overline{b_i}), and so \Phi is injective. To see surjectivity, let (b_i) \in \prod_{\mathbb{N}^+} [0,p). Now define a_k = \sum_{i=0}^{k-1} b_{i+1}p^i for all k \in \mathbb{N}^+, and consider (\overline{a_i}) \in \prod_{\mathbb{N}^+} \mathbb{Z}/(p^k). Note first that for all s \geq t, \mu_{s,t}(\overline{a_s}) \equiv \overline{\sum_{i=0}^{s-1} b_{i+1}p^i} mod p^t, so that \mu_{s,t}(\overline{a_s}) \equiv \overline{\sum_{i=0}^{t-1} b_{i+1}p^i} \equiv \overline{a_t}. Thus (\overline{a_i}) \in \varprojlim \mathbb{Z}/(p^k). Next, note that 0 \leq a_k < p^k for all k since 0 \leq a_k = \sum_{i=0}^{k-1} b_{i+1}p^i \leq \sum_{i=0}^{k-1} (p-1)p^i = p^k - 1 < p^k. Finally, note that \Phi((\overline{a_i}))_1 = a_1 = \sum_{i=0}^0 b_1p^0 = b_1, and that \Phi((\overline{a_i}))_{k+1} = (a_{k+1} - a_k)/p^k = (\sum_{i=0}^k b_{i+1}p^i = \sum_{i=0}^{k-1} b_{i+1}p^i)/p^k = b_{k+1}p^k/p^k = b_{k+1}. Thus \Phi((\overline{a_i})) = (b_i), and hence \Phi is surjective.

    Next we prove a technical lemma. Let n \geq 1 be an integer and suppose a and b are least residues mod n; that is, 0 \leq a,b < n. Then the least residue of a+b mod n is a+b - \left\lfloor \frac{a+b}{p} \right\rfloor p. Proof: This follows easily by considering the cases 0 \leq a+b < p and p \leq a+b < 2p. \square

    Suppose now we have (x_i),(y_i) \in \prod_{\mathbb{N}^+} [0,p). Define c(x,y) = (c(x,y)_i) as follows: c_0 = 0 and c_{t+1} = \left\lfloor \dfrac{x_{t+1} + y_{t+1} + c_t}{p} \right\rfloor.

    Next we prove another technical lemma. Suppose (\overline{a_i}), (\overline{b_i}) \in \varprojlim \mathbb{Z}/(p^k), where 0 \leq a_i,b_i < p^i for all i. Then c(\Phi((\overline{a_i})),\Phi((\overline{b_i})))_k = \left\lfloor \dfrac{a_k + b_k}{p^k} \right\rfloor for all k \geq 1. Proof: We proceed by induction on k. For the base case k = 1, we see that c_1 = \left\lfloor \dfrac{\Phi((\overline{a_i}))_1 + \Phi((\overline{b_i}))_1 + c_0}{p} \right\rfloor = \left\lfloor \dfrac{a_1 + b_1}{p} \right\rfloor. For the inductive step, suppose that the conclusion holds for some k \geq 1. Note that 0 \leq a_k + b_k < 2p^k and 0 \leq a_{k+1}+b_{k+1} < 2p^{k+1}, and that a_{k+1} = \Phi((a_i))_{k+1}p^k + a_k and b_{k+1} = \Phi((b_i))_{k+1}p^k + b_k.

    1. Suppose 0 \leq a_{k+1} + b_{k+1} < p^{k+1}. Then \left\lfloor \dfrac{a_{k+1} + b_{k+1}}{p^{k+1}} \right\rfloor = 0. Then 0 \leq \Phi((a_i))_{k+1}p^k + a_k + \Phi((b_i))_{k+1}p^k + b_k < p^{k+1}, and thus 0 \leq \Phi((a_i))_{k+1} + \Phi((b_i))_{k+1} + \dfrac{a_k + b_k}{p^k} < p. Then 0 \leq \Phi((a_i))_{k+1} + \Phi((b_i))_{k+1} + \left\lfloor \dfrac{a_k + b_k}{p^k} \right\rfloor < p, and hence 0 \leq \frac{1}{p}\left(\Phi((a_i))_{k+1} + \Phi((b_i))_{k+1} + \left\lfloor \dfrac{a_k + b_k}{p^k} \right\rfloor \right) < 1. So c_{k+1} = 0 as desired.
    2. Suppose p^{k+1} \leq a_{k+1} + b_{k+1} ^lt; 2p^{k+1}. Then \left\lfloor \dfrac{a_{k+1} + b_{k+1}}{p^{k+1}} \right\rfloor = 1. Similarly to the previous case, we get p \leq \Phi((a_i))_{k+1} + \Phi((b_i))_{k+1} + \dfrac{a_k + b_k}{p^k} < 2p. Since p, \Phi((a_i))_{k+1}, and \Phi((b_i))_{k+1} are integers, we can take floors while preserving the inequality; so p \leq \Phi((a_i))_{k+1} + \Phi((b_i))_{k+1} + \left\lfloor \dfrac{a_k + b_k}{p^k} \right\rfloor < 2p, whence c_{k+1} = 1 as desired.

    Thus the claim is proved. We are now prepared to characterize addition in \prod_{\mathbb{N}^+} [0,p). First, note that \Phi((a_i) + (b_i))_1 = a_1 + b_1 - \left\lfloor \dfrac{a_1 + b_1}{p}\right\rfloor p = \Phi((a_i))_1 + \Phi((b_i))_1 - c_1p. Next, for k \geq 1 we have \Phi((a_i) + (b_i))_{k+1} = \frac{1}{p^k} \left( a_{k+1} + b_{k+1} - \left\lfloor \dfrac{a_{k+1} + b_{k+1}}{p^{k+1}} \right\rfloor p^{k+1} - a_k - b_k + \left\lfloor \dfrac{a_k + b_k}{p^k} \right\rfloor p^k \right) = \dfrac{a_{k+1} - a_k}{p^k} + \dfrac{b_{k+1} - b_k}{p^k} + \left\lfloor \dfrac{a_k+b_k}{p^k} \right\rfloor - \left\lfloor \dfrac{a_{k+1} + b_{k+1}}{p^{k+1}} \right\rfloor p = \Phi((a_i))_{k+1} + \Phi((b_i))_{k+1} + c_k - c_{k+1}p. Note that this gives the “digits” of \Phi((a_i) + (b_i)) purely in terms of the digits of \Phi((a_i)) and \Phi((b_i)).

    Now we will address multiplication. First, given elements (a_i), (b_i) \in \prod_{\mathbb{N}^+} [0,p), define (c_i) by c_0 = 0 and c_{t+1} = \left\lfloor \dfrac{c_t + \sum_{i+j=t+1} a_ib_j}{p} \right\rfloor. Note from our proof that \Phi is surjective that if a = (\overline{a_i}) \in \varprojlim \mathbb{Z}/(p^k) such that 0 \leq a_i < p^i, then a_k = \sum_{i=0}^{k-1} \Phi(a)_{i+1} p^i. Now let a = (\overline{a_i}) and b = (\overline{b_i}) be two such elements. Consider now a_kb_k.

    a_kb_k  =  \left( \displaystyle\sum_{i=0}^{k-1} \Phi(a)_{i+1}p^i \right) \left( \displaystyle\sum_{j=0}^{k-1} \Phi(b)_{j+1} p^j \right)
     =  \displaystyle\sum_{t=0}^{2k-2} \left( \displaystyle\sum_{i+j = t+2} \Phi(a)_i \Phi(b)_j \right) p^t
     =  \displaystyle\sum_{t=0}^{2k-2} \left( c_{t+1}p - c_{t+1}p + \displaystyle\sum_{i+j=t+2} \Phi(a)_i \Phi(b)_j \right) p^t
     =  \displaystyle\sum_{t=0}^{2k-2} c_{t+1}p^{t+1} + \displaystyle\sum_{t=0}^{2k-2} \left[ \left( \displaystyle\sum_{i+j=t+2} \Phi(a)_i \Phi(b)_j \right) - c_{t+1}p \right] p^t
     =  \displaystyle\sum_{t=1}^{2k-1} c_{t}p^{t} + \displaystyle\sum_{t=0}^{2k-2} \left[ \left( \displaystyle\sum_{i+j=t+2} \Phi(a)_i \Phi(b)_j \right) - c_{t+1}p \right] p^t
     =  \displaystyle\sum_{t=0}^{2k-1} c_{t}p^{t} + \displaystyle\sum_{t=0}^{2k-2} \left[ \left( \displaystyle\sum_{i+j=t+2} \Phi(a)_i \Phi(b)_j \right) - c_{t+1}p \right] p^t
     =  c_{2k-1}p^{2k-1} + \displaystyle\sum_{t=0}^{2k-2} \left[ \left( \displaystyle\sum_{i+j=t+2} \Phi(a)_i \Phi(b)_j \right) + c_t - c_{t+1}p \right] p^t
    \equiv \displaystyle\sum_{t=0}^{k-1} \left[ \left( \displaystyle\sum_{i+j=t+2} \Phi(a)_i \Phi(b)_j \right) + c_t - c_{t+1}p \right] p^t mod p^k

    We can see that the coefficients in this expression are in [0,p), so that in fact, \Phi(ab)_k = \left( \sum_{i+j=k+2} \Phi(a)_i \Phi(b)_j \right) + c_k - c_{k+1}p. Again, we have the digits of ab in terms of the digits of a and b.

  2. First we’ll show that \mathbb{Z}_p is an integral domain. Let a = (a_i), b = (b_i) \in \mathbb{Z}_p be p-adic expansions with ab = 0 and a \neq 0. We claim that b = 0 and that the carry digits c_k in the product ab are all zero, and proceed by induction on the digit index k. For the base case k = 1, we have a_1b_1 - \left\lfloor \dfrac{a_1b_1}{p} \right\rfloor p = 0. Mod p, we have a_1b_1 \equiv 0. Since a_1 \neq 0, we have b_1 \equiv 0 mod p, and in fact b_1 = 0. Moreover, c_1 = \left\lfloor \dfrac{a_1b_1}{p} \right\rfloor = 0. For the inductive step, suppose that for some k, we have b_i = c_i = 0 whenever 1 \leq i \leq k. Comparing the k+1th coefficients of 0 = ab, we have 0 = \sum_{i+j=k+2} a_ib_j + c_k - c_{k+1}p. Then 0 = a_1b_{k+1} - \left\lfloor \dfrac{a_1b_{k+1}}{p} \right\rfloor p. Mod p, we have a_1b_{k+1} \equiv 0, so that b_{k+1} = 0. Thus b = 0, and so \mathbb{Z}_p is an integral domain.

    Now we demonstrate an embedding of \mathbb{Z} in \mathbb{Z}_p. Let \pi_k : \mathbb{Z} \rightarrow \mathbb{Z}/(p^k) be the natural projection. Certainly we have \pi_i = \mu_{j,i} \circ \pi_i for all i \leq j. By the universal property of direct limits, we have a unique group homomorphism \pi : \mathbb{Z} \rightarrow \varprojlim \mathbb{Z}/(p^k) such that \mu_i \circ \pi = \pi_i for all i. \pi is a ring homomorphism since \pi(ab) = (\overline{ab}) = (\overline{a}\overline{b}) = (\overline{a})(\overline{b}) = \pi(a)\pi(b). Moreover, \pi is injective, as we show. Suppose \pi(a) = \pi(b). Then a \equiv b mod p^k for all k. Then a-b is divisible by p^k for all positive integers k, and thus a-b = 0. So \pi is injective, and \mathbb{Z}_p contains a copy of \mathbb{Z}.

  3. Suppose first that a = (\overline{a_i}), b = (\overline{b_i}) \in \mathbb{Z}_p with ab = 1. Comparing the first entries of ab and 1, we have a_1b_1 \equiv 1 mod p, so that b_1 \neq 0. Suppose now that (\overline{b_i}) \in \mathbb{Z}_p with b_1 \neq 0. Now b_1 is invertible mod p. Since \mu_{i,1}(b_i) = b_1, b_i is invertible mod p^i. Certainly then (\overline{b_i})(\overline{b_i}^{-1}) = 1.
  4. Let M denote the set of all nonunits in \mathbb{Z}_p. By part (c), a \in M if and only if the first digit of a is nonzero. We claim that M = p \mathbb{Z}_p. First, we claim that p(a_i) = (b_i), where (a_i) and (b_i) are p-adic expansions and b_1 = 0 and b_{i+1} = a_i, and that the carries (c_i) in the product p(a_i) are all zero. We proceed by induction on i. For the base case i = 1, we have b_1 = 0 \cdot a_1 + 0 - \left\lfloor \dfrac{0}{p} \right\rfloor = 0 and c_1 = \left\lfloor \dfrac{a_1 \cdot 0 + 0}{p} \right\rfloor = 0. For the inductive step, c_{t+1} = \left\lfloor \dfrac{\sum_{i+j = t+2} p_ia_j + c_t}{p} \right\rfloor = \left\lfloor \dfrac{a_{t}}{p} \right\rfloor = 0 and b_{t+1} = \sum_{i+j=t+2} p_i a_j + c_t - c_{t+1}p = a_t, as desired. Thus we have M = p \mathbb{Z}_p. By §7.4 #37, since the set of nonunits is an ideal, p \mathbb{Z}_p is the unique maximal ideal of \mathbb{Z}_p.

    Now consider the surjection \mu_1 : \mathbb{Z}_p \rightarrow \mathbb{Z}/(p). Clearly \mathsf{ker}\ \mu_1 = p\mathbb{Z}_p. By the First Isomorphism Theorem for rings, we have \mathbb{Z}_p/(p) \cong \mathbb{Z}/(p).

    Now let I \subseteq \mathbb{Z}_p be a nonzero ideal. Choose k maximal such that p^k divides every element of I; we have k \geq 1 since I \subseteq p \mathbb{Z}_p. Then I \subseteq (p^k). Now choose a \in I such that p^k divides a but p^{k+1} does not. Then we can write a = p^kb where p does not divide b. Then b is a unit, and we have p^k = p^kbb^{-1} = ab^{-1} \in I; thus (p^k) \subseteq I and so I = (p^k).

  5. Let \overline{b_1} \in \mathbb{Z}/(p) be nonzero. We will define an element b \in \varprojlim \mathbb{Z}/(p^k) such that (1) each \overline{b_i} is a unit in \mathbb{Z}/(p^i), (2) \overline{b_i}^{p-1} \equiv 1 mod p^i, and (3) \mu_{i,1}(\overline{b_i}) = b_1 inductively as follows. The first coordinate is \overline{b_1}. Certainly \overline{b_1} is a unit mod p, \overline{b_1}^{p-1} \equiv 1 mod p by Fermat’s Little Theorem, and \mu_{1,1}(\overline{b_1}) = b_1.

    Suppose now that \overline{b_k} is defined and has the desired properties. By §7.6 #7, the surjection \mu_{k+1,k} : \mathbb{Z}/(p^{k+1}) \rightarrow \mathbb{Z}/(p^k) is surjective on the units; thus there is a unit \overline{a_{k+1}} such that \mu_{k+1,k}(\overline{a_{k+1}}) = b_k. Let \overline{b_{k+1}} = \overline{a_{k+1}}^p; certainly \overline{b_{k+1}} is a unit. Moreover, \mu_{k+1,1}(\overline{b_{k+1}}) = (\mu_{k,1} \circ \mu_{k+1,k})(\overline{b_{k+1}}) = \mu_{k,1}(\mu_{k+1,k}(\overline{a_{k+1})^p}) = \mu_{k,1}(\overline{b_k}^p) = \mu_{k,1}(\overline{b_k}) = b_1. Since \mu_{k+1,k}(\overline{a_{k+1}}) = \overline{b_k}, we have a_{k+1} = qp^k + b_k for some q. Consider now that b_{k+1}^{p-1} \equiv (a_{k+1}^p)^{p-1} \equiv ((qp^k + b_k)^p)^{p-1}. Since \mathbb{Z}/(p^{k+1}) is an integral domain of characteristic p, we have b_{k+1}^p \equiv ((qp^k)^p + b_k^p)^{p-1}. Note that k+1 \leq kp, so that p^{kp} \equiv 0 mod p^{k+1}, so we have b_{k+1}^{p-1} \equiv (b_k^p)^{p-1} \equiv (b_k^{p-1})^p mod p^{k+1}. By the induction hypothesis, b_k^{p-1} = tp^k + 1 for some t. Thus b_{k+1}^{p-1} \equiv (tp^k + 1)^p \equiv (tp^k)^p + 1^p \equiv 1 mod p^{k+1}, as desired.

    Thus we may construct such a b = (\overline{b_k}) for each unit in \mathbb{Z}/(p), and these are clearly distinct. Moreover, b^{p-1} = (\overline{b_i}^{p-1}) = (\overline{1}) = 1. So \mathbb{Z}_p contains (at least) p-1 distinct roots of 1.

In an integral domain, there are at most two square roots of 1

Prove that if R is an integral domain and x^2 = 1 for some x \in R, then x = 1 or x = -1.

If x^2 = 1, then x^2 - 1 = 0. Evidently, then, (x-1)(x+1) = 0. Since R is an integral domain, we must have x-1 = 0 or x+1 = 0; thus x = 1 or x = -1.