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

### 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 $n$th 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 $n$th 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 $m$th roots of unity over $F$, and so $m$ distinct $n$th 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 $n$th root of unity and let $d|n$. Prove that $\zeta^d$ is a primitive $n/d$th 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 $n/d$th 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 $m$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.

### 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 $p-1$th 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 $k+1$th 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 $x = -1$.