Question 6

(a)

Let \(\alpha = a + bi + cj + dk\). Note that \[\begin{align*} N(\overline{\alpha}) &= a^2 + (-b)^2 + (-c)^2 + (-d)^2 \\ &= a^2 + b^2 + c^2 + d^2 \\ &= N(\alpha) \end{align*}\] and \[\begin{align*} \alpha\overline{\alpha}&= (a + bi + cj + dk)(a - bi - cj - dk) \\ &= a^2 - abi - acj - adk \\ &\quad + bia - b^2i^2 - bicj - bidk \\ &\quad + cja - cjbi - c^2j^2 - cjdk \\ &\quad + dka - dkbi - dkcj - d^2k^2 \\ &= a^2 + b^2 + c^2 + d^2 \\ &= N(\alpha) \end{align*}\] also, as \(\overline{\overline{\alpha}} = \alpha\), we have \(N(\overline{\alpha}) = \overline{\alpha}\overline{\overline{\alpha}} = \overline{\alpha}\alpha\).

(b)

Let \(\beta = p + qi + rj + sk\), then \[\begin{align*} \alpha\beta &= (a + bi + cj + dk)(p + qi + rj + sk) \\ &= ap + aqi + arj + ask \\ &\quad + pbi - bq + brk - bsj \\ &\quad + pcj - cqk - cr + csi \\ &\quad + pdk + dqj - dri - ds \\ &= ap - bq - cr - ds \\ &\quad + (aq + pb + cs - rd)i \\ &\quad + (ar + pc - bs + qd)j \\ &\quad + (as + pd + br - qc)k \\ \end{align*}\]

Define \(h: H_\mathbb{R} \to M_2(\mathbb{C})\) as \[ a + bi + cj + dk \mapsto \begin{pmatrix} a + bi & -c - di \\ c - di & a - bi \end{pmatrix} \] Let \(w = a + bi, x = c + di, y = p + qi, z = r + si\) and \(w^*\) denote its complex conjugate, we have \[\begin{align*} \begin{pmatrix} w & -x \\ x^* & w^* \end{pmatrix} \begin{pmatrix} y & -z \\ z^* & y^* \end{pmatrix} &= \begin{pmatrix} wy - xz* & -wz - xy^* \\ x^*y + w^*z^* & -x^*z + w^*y^* \end{pmatrix} \\ &= \begin{pmatrix} wy - xz* & -wz - xy^* \\ (wz + xy^*)^* & (wy - xz^*)^* \end{pmatrix} \\ wy - xz^* &= (a+bi)(p+qi) - (c+di)(r-si) \\ &= ap - bq - cr - ds + (aq + bp + cs - rd)i \\ wz + xy* &= (a+bi)(r+si) + (c+di)(p-qi) \\ &= ar - bs + pc + dq + (as + br + pd - qc)i \end{align*}\] therefore \(h(\alpha\beta) = h(\alpha)h(\beta)\) (\(h\) preserving addition is obvious).

Now observe that \(N(\alpha) = \det(h(\alpha))\) and result follows.

(c)

Suppose \(\alpha \in H_\mathbb{Z}\) is a unit, let \(\beta\) such that \(\alpha\beta = 1\), then \(N(\alpha)N(\beta) = 1\) and since \(N(H_\mathbb{Z}) \subseteq \left\{0,1,2,\dots\right\}\) we have \(N(\alpha) = N(\beta) = 1\). Conversely suppose \(N(\alpha) = 1\), we know that \[ \alpha^{-1} = \frac{\overline{\alpha}}{N(\alpha)} \in H_\mathbb{Q} \] and as \(N(\alpha) = 1\), it turns out that \(\alpha^{-1} = \overline{\alpha}\in H_\mathbb{Z}\), so \(\alpha\) is a unit. Observe that the only possible elements in \(H_\mathbb{Z}\) with norm 1 are precisely \(\left\{\pm 1,\pm i,\pm j,\pm k\right\}\), so the set of units is the quaternion group.

Question 7

(a)

As addition is defined pointwise the properties easily carry from \(R\). We still need to check the properties of multiplication.

Cauchy product commutative when \(R\) is

it follows from commutativity of \(\cdot\) in \(R\) that \[ \sum_{i=0}^n a_ib_{n-i} = \sum_{j=0}^n b_ja_{n-j} \]

Cauchy product associative

applying commutativity of \(+\) and associativity of \(\cdot\) and distributivity in \(R\) to coefficients, we get

\[\begin{align*} \left(\left(\sum a_nx^n\right)\left(\sum b_nx^n\right)\right) \left(\sum c_nx^n\right) &= \sum_n \left(\sum_{j=0}^n \left( \sum_{i=0}^j a_ib_{j-i} \right) c_{n-j}\right) x^n \\ &= \sum_n \left(\sum_{j=0}^n \sum_{i=0}^j a_ib_{j-i}c_{n-j}\right) x^n \\ &= \sum_n \left(\sum_{i=0}^n \sum_{j=i}^n a_ib_{j-i}c_{n-j}\right) x^n \\ % k = j-i &= \sum_n \left(\sum_{i=0}^n \sum_{k=0}^{n-i} a_ib_{k}c_{n-i-k}\right) x^n \\ &= \sum_n \left(\sum_{i=0}^n a_i \left(\sum_{k=0}^{n-i} b_{k}c_{n-i-k}\right)\right) x^n \\ &= \left(\sum a_nx^n\right)\left(\left(\sum b_nx^n\right) \left(\sum c_nx^n\right)\right) \end{align*}\]

Cauchy product distributes

By commutativity we just show \((A+B)C = AC + BC\)

\[\begin{align*} \left(\sum (a_n+b_n)x^n\right) \sum c_nx^n &= \sum_n \left(\sum_{i=0}^n (a_i+b_i)c_{n-i} \right) x^n \\ &= \sum_n \left(\sum_{i=0}^n a_ic_{n-i} + b_ic_{n-i} \right) x^n \\ &= \left(\sum a_nx^n\right) \left(\sum c_nx^n\right) + \left(\sum b_nx^n\right) \left(\sum c_nx^n\right) \end{align*}\]

\(1_R\) is \(1_{R[[x]]}\)

easy to check that \[\left(1 + 0x + 0x^2 + \dots\right) \sum_n a_nx^n = \sum_n \left(\sum_{i=0}^n \delta_{i0}a_{n-i}\right) x^n = \sum_n a_nx^n \] where \(\delta_{ij}\) is \(1\) if \(i=j\) else \(0\)

(b)

We just multiply out \(1-x\) and \(\sum_{j=0}^\infty x^j\). \[\begin{align*} (1-x)\sum_{j=0}^\infty x^j &= \sum_{j=0}^\infty x^j - x\sum_{j=0}^\infty x^j \\ &= \sum_{j=0}^\infty x^j - \sum_{j=1}^\infty x^j \\ &= 1 \end{align*}\]

(c)

Suppose \(f=\sum a_nx^n\) is a unit in \(R[[x]]\), let \(g=\sum b_nx^n\) such that \(1 = fg\), \[ 1 = fg = \sum_n \left(\sum_{j=0}^n a_{j}b_{n-j}\right) x^n \] by comparing coefficients, \(1 = a_0b_0\) and for all \(n>1\), \[ \sum_{j=0}^n a_jb_{n-j} = 0 \] so in particular \(a_0\) is a unit in \(R\).

Conversely suppose \(a_0\) is a unit, let \(a_0b_0 = 1 = b_0a_0\), recursively define for all \(n>1\) \[ b_n = -b_0\sum_{j=1}^n a_jb_{n-j}. \] Now we have \(a_0b_0 = 1\) and for all \(n>1\), \[\begin{align*} a_0b_n &= -\sum_{j=1}^n a_jb_{n-j} \\ 0 &= \sum_{j=0}^n a_jb_{n-j} \end{align*}\] which shows that \(fg = 1\).