Posted on June 1, 2019

Here are my exercise solutions and notes for Neukirch’s Algebraic Number Theory.

1.1. Recall that (ab)(\overline{ab})=(a\overline{a})(b\overline{b}) for a,b\in\mathbb{C}. Therefore N(ab)=N(a)N(b) for a,b\in\mathbb{Z}[i]. Moreover, N(a)\in\mathbb{Z}_{\ge 0} for all a\in\mathbb{Z}[i]. Thus if ab=1 we must have N(a)N(b)=1 and therefore N(a)=N(b)=1.

1.2. Suppose that \alpha\beta=\epsilon\gamma^n for \alpha,\beta,\gamma,\epsilon\in\mathbb{Z}[i] with \epsilon a unit and \alpha,\beta relatively prime. Since \mathbb{Z}[i] is a UFD we may write out the prime factorizations \alpha=\epsilon&39;p_1^{a_1}\ldots p_r^{a_r} and \beta=\epsilon&39;&39;q_1^{b_1} \ldots q_{\ell}^{b_\ell} with \epsilon&39;,\epsilon&39;&39; units and the p_i’s and q_j’s prime. Since \alpha,\beta are relatively prime, the p_i’s and q_j’s are all distinct primes. Therefore, since \alpha\beta=\epsilon\gamma^n, we must have n\mid a_i and n\mid b_j for all i,j. I.e., we have \alpha=\epsilon&39;\xi^n and \beta=\epsilon&39;&39;\eta where \xi=p_1^{a_1/n}\ldots p_r^{a_r/n} and \eta=q_1^{b_1/n}\ldots q_\ell^{b_\ell/n}.

1.3. Observe that if x^2+y^2=z^2 for x,y,z\in\mathbb{Z}, then \alpha\overline{\alpha}=z^2 for \alpha=x+iy. Therefore by Exercise 1.2 we have x+iy=\epsilon\xi^2 with \epsilon a unit and \xi=u+iv. Therefore x+iy=\epsilon(u^2-v^2+2uvi). So x=\epsilon(u^2-v^2) and y=\epsilon 2uv. From Exercise 1 we see that the only units in \mathbb{Z}[i] are \pm 1,\pm i. We cannot have \epsilon=\pm i as x,y\in\mathbb{Z}. Thus we may ignore the \epsilon as it just swaps the sign.

Observe that if (u,v)>1 then we we would have (x,y)>1, a contradiction. If u,v are both odd, then u^2-v^2\equiv 0(\text{mod } 3) and 2uv\equiv 2(\text{mod } 3). Thus z^2\equiv 2 (\text{mod } 3), a contradiction.

1.4. Recall that an ordering on a ring satisfies a\le b implies a+c\le b+c for all a,b,c\in R, as well as 0\le a and 0\le b implies 0\le ab for all a,b\in R. The first implication shows that -a\le 0\le a for all a\ge 0; in particular -1\le 0\le 1. If i\ge 0, then i^2=-1\le 0, a contradiction. However, if i\le 0 then -i\ge 0 and (-i)^2=-1\le 0.

1.5. Observe that we may define a norm on \mathbb{Z}[\sqrt{-d}] by N(\alpha)=\alpha\overline{\alpha} where if \alpha=a+b\sqrt{-d} we have \overline{\alpha}=a-b\sqrt{-d}. This gives N(a+b\sqrt{-d})=a^2+b^2d. Again we see that N(\alpha\beta)=N(\alpha)N(\beta) and thus an element \alpha is a unit if and only if N(\alpha)=1. The only elements of \mathbb{Z}[\sqrt{-d}] with norm 1 are \pm 1.

1.6. After struggling for a while and digging a bit on math stack exchange, it appears that this question is fairly difficult for its place in the book. Perhaps I will revisit this question after progressing further.