*Algebra*

## 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

- 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 \(\mathbb{Q}\). - \(S\) 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

- \(R\) 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.
- \(R\) 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 \(a + b = b + a\).