Brian Bi
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## Section 11.1. Definition of a Ring

Exercise 11.1.1 Let $$x = 7 + \sqrt[3]{2}$$. Then $$(x - 7)^3 - 2 = 0$$, so $$x$$ is the root of a polynomial. Let $$y = \sqrt{3} + \sqrt{-5}$$. Then $$y^2 = 3 + 2\sqrt{-15} + (-5)$$. So $$y^2 +2 = 2\sqrt{-15}$$, or $$(y^2 + 2)^2 = -60$$. So $$y$$ satisfies the polynomial equation $$(y^2 + 2)^2 + 60 = 0$$. Therefore both $$x$$ and $$y$$ are algebraic.

Exercise 11.1.2 The proof is based on the following well-known result that we will state and prove here:

Lemma: For all $$n \in \mathbb{N}$$, $$\cos(n\theta)$$ can be written as $$P_n(\cos\theta)$$ where $$P_n$$ is a polynomial with integer coefficients.

Proof: By induction. The base cases of $$n = 0, n = 1$$ are obvious. Assume $$n \ge 2$$. Using the identity $$2\cos a \cos b = \cos(a - b) + \cos(a + b)$$, we find $$2\cos((n-1)\theta) \cos \theta = \cos((n-2)\theta) + \cos(n\theta)$$. By the induction hypothesis, $$\cos((n-1)\theta) = P_{n-1}(\cos\theta)$$ and $$\cos((n-2)\theta) = P_{n-2}(\cos\theta)$$. Therefore $$2P_{n-1}(\cos\theta)\cos\theta = P_{n-2}\cos\theta + \cos(n\theta)$$, or $$\cos(n\theta) = 2(\cos\theta) P_{n-1}(\cos\theta) - P_{n-2}(\cos\theta)$$. The RHS is now a polynomial in $$\cos\theta$$.

Now let $$z = \cos(2\pi/n)$$. By the Lemma, $$P_n(z) = \cos(2\pi) = 1$$. So $$z$$ satisfies the polynomial $$P_n(z) - 1 = 0$$ and is algebraic.

Exercise 11.1.3 Certainly $$\mathbb{Q}[\sqrt{2}, \sqrt{3}]$$ contains $$\sqrt{2} + \sqrt{3}$$ and therefore $$\mathbb{Q}[\sqrt{2} + \sqrt{3}] \subseteq \mathbb{Q}[\sqrt{2}, \sqrt{3}]$$. If we can show that $$\sqrt{2}, \sqrt{3} \in \mathbb{Q}[\sqrt{2} + \sqrt{3}]$$, then this will establish that $$\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2}, \sqrt{3}]$$. Write $$\gamma = \sqrt{2} + \sqrt{3}$$ as in the problem statement. Observe that $$\gamma^3 = 11\sqrt{2} + 9\sqrt{3}$$, so $$\mathbb{Q}[\gamma]$$ contains $$\frac{1}{2}\gamma^3 - \frac{9}{2}\gamma = \sqrt{2}$$ and $$-\frac{1}{2}\gamma^3 + \frac{11}{2}\gamma = \sqrt{3}$$. Therefore $$\mathbb{Q}[\gamma] = \mathbb{Q}[\sqrt{2}, \sqrt{3}]$$.

We will show that $$\mathbb{Z}[\gamma]$$ does not contain the numbers $$\sqrt{2}, \sqrt{3}$$, so it is not equal to $$\mathbb{Z}[\sqrt{2}, \sqrt{3}]$$. We will assume that $$\{1, \sqrt{2}, \sqrt{3}, \sqrt{6}\}$$ are linearly independent over $$\mathbb{Q}$$ although we do not prove it here. We will state and prove a Lemma:

Lemma: Let $$n \in \mathbb{N}$$. If $$n$$ is even, then $$\gamma^n = a + b\sqrt{6}$$ where $$a$$ is odd and $$b$$ is even. If $$n$$ is odd, then $$\gamma^n = a\sqrt{2} + b\sqrt{3}$$ where $$a$$ and $$b$$ are both odd.

Proof: By induction on $$n$$. The base cases $$n = 0, n = 1$$ are easily verified. For $$n = 2$$ use $$\gamma^n = (\sqrt{2} + \sqrt{3}) \gamma^{n-1}$$ and expand.

Let $$x \in \mathbb{Z}[\gamma]$$. Then $$x = \sum_{i=0}^n c_i \gamma^i$$ for some integers $$c_i$$. The RHS can be written as $$p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6}$$ for some integers $$p, q, r, s$$, where, by the Lemma, the $$\sqrt{2}$$ and $$\sqrt{3}$$ terms come only from odd powers. Since each $$\gamma^i$$ for odd $$i$$ has odd coefficients for both $$\sqrt{2}$$ and $$\sqrt{3}$$, arbitrary integer combinations of the $$\gamma^i$$ must have both coefficients even or both odd. The number $$\sqrt{2}$$ has one unique representation $$0 + 1\sqrt{2} + 0\sqrt{3} + 0\sqrt{6}$$ by the linear independence property we stated earlier; since 1 and 0 do not have the same parity, $$\sqrt{2} \notin \mathbb{Z}[\gamma]$$. A similar argument applies to $$\sqrt{3}$$.

Exercise 11.1.4 Suppose $$x = p/2^q$$ where $$p \in \mathbb{Z}, q \in \mathbb{N}$$. We have $$\alpha^{4n} = 2^{-4n},$$ therefore $$\mathbb{Z}[\alpha]$$ contains some number of the form $$1/2^{4n}$$ where $$4n \ge q$$, therefore it also contains $$x = (p 2^{4n-q})(1/2^{4n})$$. That is, $$\mathbb{Z}[\alpha]$$ contains all real dyadic fractions. A similar argument shows that $$\mathbb{Z}[\alpha]$$ contains all imaginary dyadic fractions, that is, numbers of the form $$ip/2^q$$. Since the dyadic fractions are dense in $$\mathbb{R}$$, the desired result follows from bounding the distance along the real and imaginary axes separately.

Exercise 11.1.5 Let $$R$$ be a subring of $$\mathbb{R}$$. Then $$R$$ contains 1, implying that it contains all integers. Now, if $$R$$ contains any nonzero number $$\alpha$$ such that $$0 < |\alpha| < 1$$, then the powers of that number get arbitrarily close to 0, so $$R$$ is not discrete. If $$R$$ contains a positive number $$\beta$$ that is not an integer, then since $$R$$ contains the integer $$\lfloor \beta \rfloor$$, it must contain the number $$\beta - \lfloor \beta \rfloor$$, which is strictly between 0 and 1, and again the powers of that number get arbitrarily close to 0 so $$R$$ is not discrete. If $$R$$ contains a negative number $$\beta$$ that's not an integer, then the previous argument applies to $$-\beta$$. So $$R$$ can only be discrete if all its elements are integers; therefore, the only discrete subring of $$\mathbb{R}$$ is $$\mathbb{Z}$$.

Exercise 11.1.6

1. Presumably the intent was that $$S$$ contains the rationals such in the reduced form $$a/b$$, the denominator $$b$$ is not divisible by 3. Let $$a/b$$, $$c/d$$ be two rationals of this form. Then $$(a/b)(c/d) = (ac)/(bd)$$ has a denominator that isn't divisible by 3, so the same will be true after reducing to lowest terms. Also $$(a/b) + (c/d) = (ad + bc)/(bd)$$ and again the denominator isn't divisible by 3. So $$S$$ is closed under addition and multiplication, and is therefore a subring of $$\mathbb{Q}$$.
2. $$S$$ is not a subring of $$R$$ because it's not closed under multiplication; the function $$\sin t \cos t$$ is not in $$R$$. To see this, observe that this function can be written as $$\frac{1}{2}\sin 2t$$. The functions $$1, \sin t, \cos t, \sin 2t, \cos 2t, \ldots$$ are linearly independent over $$\mathbb{R}$$ (this can be proven using the integration trick used to compute Fourier coefficients), so $$\frac{1}{2}\sin 2t$$ can't be written as any other real linear combination of the basis functions, including any integer combination.

Exercise 11.1.7

1. $$R$$ is a ring. To see this, observe that $$(R, +, \cdot)$$ is isomorphic to $$((\mathbb{Z}/2\mathbb{Z})^U, +, \times)$$ where the entry indexed by $$u \in U$$ is 1 if $$u$$ is in the subset and 0 otherwise, and the operations $$+$$ and $$\times$$ act element-wise; that this is a ring follows from the fact that $$\mathbb{Z}/2\mathbb{Z}$$ is.
2. $$R$$ is not a ring because the $$\circ$$ operation isn't commutative. However, $$\circ$$ is associative, and $$(R, +)$$ does form an abelian group. (In fact even the distributive law holds in the form given in the text, which requires right-distributivity only.)

Exercise 11.1.8 In $$\mathbb{Z}/n\mathbb{Z}$$ the element $$m$$ is a unit iff there is some $$a$$ with $$am \equiv 1 \pmod{n}$$, or in other words $$am - bn = 1$$ for some $$a, b \in \mathbb{Z}$$. This equation has solutions iff $$\gcd(m, n) = 1$$. Therefore the units in $$\mathbb{Z}/n\mathbb{Z}$$ are the equivalence classes of integers that are relatively prime to $$n$$. For $$n = 8$$ these are the equivalence classes of 1, 3, 5, and 7; for $$n = 12$$ these are the equivalence classes of 1, 5, 7, and 11.

Exercise 11.1.9 Let $$a, b \in R$$. Using the distributive property and the commutativity of multiplication, we obtain $$(a + 1)(b + 1) = a(b+1) + 1(b+1) = (b+1)a + (b+1)1 = ba + 1a + b1 + 11 = ba + a + b + 1$$. Similarly we can show $$(b+1)(a+1) = ab + b + a + 1$$. But these are equal by commutativity of multiplication, so $$ba + a + b + 1 = ab + b + a + 1$$. Cancelling $$ab$$ on the left and $$1$$ on the right, we are left with $$a + b = b + a$$.