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sheet2-2024-solutionsABC.pdf
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6. (i) Suppose that $c_0, c_1, \ldots, c_{p-2}$ are integers and that
\[
\frac{1}{p}\left(c_0+c_1(\zeta-1)+\cdots+c_{p-2}(\zeta-1)^{p-2}\right) \in \mathcal{O}_K
\]
Show that all the $c_i$ are divisible by $p$. (Hint: suppose not, and let $r$ be minimal such that $p \nmid c_r$. You may wish to recall Questions 3 and 4.)
(ii) Show that $1, \zeta, \ldots, \zeta^{p-2}$ is an integral basis for $\mathcal{O}_K$.
Solution: (i) Suppose not, and that $r$ is minimal such that $p \nmid c_r$. Subtracting off elements of $\mathbf{Z}[\zeta-1]$, we have
\[\tag4
\alpha:=\frac{1}{p}\left(c_r(\zeta-1)^r+\cdots+c_{p-1}(\zeta-1)^{p-1}\right) \in \mathcal{O}_K
\]
Now use the result of Question 4, that is to say
\[
\frac{1}{p}(\zeta-1)^{p-1} \in \mathcal{O}_K
\]
Thus, multiplying (4) through by $(\zeta-1)^{p-2-r}$, we see that
\[
\frac{1}{p} c_r(\zeta-1)^{p-2} \in \mathcal{O}_K
\]
However by Question 3 the norm of the left-hand side is $c_r^{p-1} / p$. This is not an integer, and so we get a contradiction.
(ii) A slick way to proceed here is notice that $m_{\zeta-1}(X)=m_\zeta(X+1)$, and so $m_{\zeta-1}^{\prime}(\zeta-1)=$ $m_\zeta^{\prime}(\zeta)$, and so by Questions 1 and 5,
\begin{align*}
\operatorname{disc}_{K / \mathbf{Q}}\left(1,(\zeta-1),(\zeta-1)^2, \ldots,(\zeta-1)^{p-2}\right) & =\operatorname{disc}_{K / \mathbf{Q}}\left(1, \zeta, \ldots, \zeta^{p-2}\right) \\
& =(-1)^{(p-1) / 2} p^{p-2}\tag5
\end{align*}
Since $p$ is the only prime dividing this discriminant, a result from lectures shows that any element of $\mathcal{O}_K$ is of the form $\frac{1}{p}\left(c_r(\zeta-1)^r+\cdots+c_{p-1}(\zeta-1)^{p-1}\right)$, and hence by part (i) of the question lies in $\mathbf{Z}[\zeta-1]$, which is contained in $\mathbf{Z}[\zeta]$.
An alternative way to proceed (the one I originally had in mind) is to note that
\[\tag6
\mathbf{Z}[\zeta-1]=\mathbf{Z}[\zeta]
\]
This is true because, for any algebraic integer $t$, $\mathbf{Z}[t \pm 1] \subseteq \mathbf{Z}[t]$, by binomial expansion of each power $(t \pm 1)^i$. Applying this with $t=\zeta-1$ and the $+$ sign gives $\mathbf{Z}[\zeta] \subseteq \mathbf{Z}[\zeta-1]$, and applying it with $t=\zeta$ and the $-$ sign gives the opposite inclusion $\mathbf{Z}[\zeta-1] \subseteq \mathbf{Z}[\zeta]$.
Now, by lectures and Question 5, any element $x \in \mathcal{O}_K$ lies in $\frac{1}{p} \mathbf{Z}[\zeta]$, and hence by (6) lies in $\frac{1}{p} \mathbf{Z}[\zeta-1]$. By part (i) of the question, $x$ therefore lies in $\mathbf{Z}[\zeta-1]$, and so finally by (6) again, we have $x \in \mathbf{Z}[\zeta]$. |
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abababa
发表于 2024-12-28 17:14
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