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\title{MATH 129, HW 4}
\author{Bryden Cais}

\begin{document}
\maketitle

\section{Units, roots of unity, and the unit circle}
\begin{enumerate}

	\item Suppose that a monic polynomial $P(X)=c_0+c_1X+\cdots+X^N\,\in\,\Z[X]$
	has roots $\alpha_1,\ldots,\alpha_N$ with $|\alpha_j|<A$ for some $A>0$
	and all $j$.  Since the coefficients of $P$ are given as the elementary symmetric functions 
	in the roots of $P$, we may write
	\begin{eqnarray*}
	    c_{N-k-1}\ =\ \sum_{j_0<\cdots<j_k}\alpha_{j_0}\cdots\alpha_{j_k},
	\end{eqnarray*}
	from which we obtain the inequality
	\begin{eqnarray*}
	    |c_{N-k-1}|\leq {N\choose{k+1}}\,|A|^{k+1},  
	\end{eqnarray*}
	which shows that all the coefficients of $P$ are bounded.  Since $P\,\in\,\Z[X]$, there
	can only be finitely many such $P$ of a given degree $N$.

	\item Because of containment properties, if we show that $\mu(\Z^{\mathrm{ab}})=\mu$,
	we have shown that $\mu(\bar{\Z})=\mu(\Q^{\mathrm{ab}})=\mu(\bar{\Q})=\mu$.  To see
	that $\mu(\Z^{\mathrm{ab}})=\mu$, notice that $\mu(\Z^{\mathrm{ab}})\subset\mu$ by
	definition, and that we have $\mu\subset\mu(\Z^{\mathrm{ab}})$ because any root
	of unity $\alpha$ has irreducible polynomial $\Phi_d$ (where $\Phi_d$ is the d th 
	cyclotomic polynomial) for some $d$ and the galois group of $\Phi_d$ is $\left(\Z/d\Z\right)^*$,
	and is hence abelian.  That is, any root of unity is in $\Z^{\mathrm{ab}}$.

	We next show that $V(\bar{\Z})=\mu,$ and therefore also that $V(\Z^{\mathrm{ab}})=\mu$,
	since obviously $V(\bar{\Z})\supset V(\Z^{\mathrm{ab}})$ and $\mu\subset V(\Z^{\mathrm{ab}})$
	since every conjugate of a root of unity is a root of some cyclotomic polynomial and is thus
	a root of unity itself.
	Let $x\,\in\,V(\bar{\Z})$ have monic irreducible polynomial $P$ of degree $N$.
	Consider $x,x^2,x^3,\ldots$.  For each $j$, we have that the minimal polynomial
	of $x^j$ has degree $N$ or less since $x^j\,\in\,\Q(x)$ and $[\Q(x):\Q]=N$.
	Moreover, since all
	the conjugates of $x$ have absolute value $1$, so do all the conjugates of $x^j$.
	Finally, the minimal polynomial of $x^j$ is in $\Z[X]$ since $x$ is integral, and
	thus so is any symmetric function in any of the conjugates of $x^j$.
	We can reinterpret the result of item $1$ in the following useful way: there are at most
	a finite number of distinct algebraic integers of absolute value less than $A>0$ and degree
	less than or equal to $N>0$.  Clearly, an infinite number of such algebraic integers
	would imply an infinite number of monic polynomials of degree at most $N$ with roots
	bounded in absolute value by $A$, and this contradicts item $1$.
	Therefore, there are only a finite number of distinct $x^j$, 
	so that we have $x^i=x^j$ for some $i\not=j$ and therefore that
	$x$ is a root of unity.
	
	Now we show that $V(\Q^{\mathrm{ab}})=U(\Q^{\mathrm{ab}})$ and therefore also that
	$U(\Z^{\mathrm{ab}})=\mu$ since, again, $\Z^{\mathrm{ab}}\subset \Q^{\mathrm{ab}}$.
	Observe that complex conjugation $x\longrightarrow \bar{x}$ is always an automorphism
	of the galois group of $\Q(\alpha)$ for any $\alpha\,\in\,\Q^{\mathrm{ab}}$.  Since
	this galois group is assumed to be abelian, any $\sigma\,\in\,G_{\Q}$ commutes
	with complex conjugation.  Therefore, if $x\bar{x}=1$,
	\begin{eqnarray*}
	    1=\sigma(x\bar{x})=\sigma(x)\sigma(\bar{x})=\sigma(x)\bar{\sigma(x)}=|\sigma x|,
	\end{eqnarray*}
	so that $x\,\in\,V(\Q^{\mathrm{ab}})$.  Since obviously 
	$V(\Q^{\mathrm{ab}})\subset U(\Q^{\mathrm{ab}})$, we have $V(\Q^{\mathrm{ab}})=U(\Q^{\mathrm{ab}})$.
	
	Next, we show that $V(\bar{\Z})\not=U(\bar{\Z})$ and therefore that $V(\bar{\Q})\not=U(\bar{\Q})$
	because of containment.  To do this, we must produce some element of $\bar{\Z}$ of norm $1$
	that has a conjugate not of norm $1$.  We claim that
	\begin{eqnarray*}
	    \gamma=(2-\sqrt{3})(\sqrt{7}+2i\sqrt[4]{3})
	\end{eqnarray*}
	is such an algebraic integer.  To see this, first observe that $\gamma\,\in\,\bar{\Z}$, clearly,
	and that the square of the absolute value of $\gamma$ is 
	$(2-\sqrt{3})^2(7+4\sqrt{3})=(2-\sqrt{3})^2(2+\sqrt{3})^2=1$.  Moreover, the automorphism
	$\sigma\,\in\,\mathrm{Gal}(\bar{\Q}/\Q)$ which sends $i\sqrt[4]{3}$ to $\sqrt[4]{3}$ sends
	$\gamma$ to $(2+\sqrt{3})(\sqrt{7}+2\sqrt[4]{3}),$ which evidently does not have norm $1$.

	It is not difficult to show that $U$ and $U(\bar{\Q})$ are distinct.  For example, the algebraic
	numbers are countable (by a standard argument of Cantor), but the set of points on the unit 
	circle is not countable, so that there exist
	uncountably many complex numbers of absolute value $1$ that are not algebraic.

	Now we show that $U(\bar{\Z})$ is distinct from $U(\bar{\Q}),\ V(\Q^{\mathrm{ab}})$, and that
	$\mu$ is distinct from $V(\Q^{\mathrm{ab}})$.  To do this, we consider
	\begin{eqnarray*}
	    \beta\ =\ \frac{3+i\sqrt{7}}{4}.
	\end{eqnarray*}
	Notice that $|\beta|=1$ and that $\beta\,\in\,\bar{\Q}\setminus\bar{\Z}$ since
	its minimal polynomial is $X^2-(3/2)X+1$.  Moreover, we in fact
	have $\beta\,\in\,\Q^{\mathrm{ab}}$ since $\beta$ lies in a degree two extension of $\Q$.  For
	this same reason, all of the conjugates of $\beta$ have norm $1$, so that 
	$\beta\,\in\,V(\Q^{\mathrm{ab}})\setminus\bar{\Z}$.  Finally, $\beta$ is not a root of unity
	since it is not in $\bar{\Z}$.

	It remains to show that $V(\bar{\Q})\not=V(\Q^{\mathrm{ab}})$.

	\item First, every $\alpha\,\in\,K$ satisfies a minimial polynomial of degree less than or equal
	to $N=[K:\Q]$.  Thus, $K\cap \mu$ is finite by item 1, since every root of unity has minimal 
	polynomial of degree less than or equal to $N$ with all roots of absolute value $1$.  Second,
	$K\cap \mu$ is a subgroup of $\mathcal{O}_K^*$ since every root of unity is invertable
	and the product of two roots of unity is again a root of unity.  Finally, this is in fact
	a cyclic subgroup since if $K\cap \mu$ contains roots of unity of order $a,\ b$, then it
	contains a root of unity of order $\mathrm{lcm}(a,b)$.  (This is a standard fact from
	elementary number theory).  Thus, there is some root of unity
	$\rho\,\in\,K\cap \mu$ with order a multiple of the order of any other root of unity, so that
	$K\cap \mu=<\rho>$ is cyclic.

	\item Let $\alpha\,\in\,\bar{\Z}$ and suppose that all of the conjugates of $\alpha$ are
	in a disc centered at $\alpha$ of radius $1/4$.  Then we have $|\alpha-\sigma(\alpha)|<1/4$
	for any $\sigma\,\in\,\mathrm{Gal}(\bar{\Q}/\Q)$.  Now let $\sigma_1,\ldots,\sigma_n$ denote
	the distinct conjugates of $\alpha$.  Then for any $i,\ j$, we have 
	$|\sigma_i(\alpha)-\sigma_j(\alpha)|<1/2$, and therefore, by our formula for
	$\disc(1,\alpha,\ldots,\alpha^{n-1})$, we have 
	\begin{eqnarray*}
	|\disc(\alpha)|=\left|\prod_{i\not=j}(\sigma_i(\alpha)-\sigma_j(\alpha))\right|<1.
	\end{eqnarray*}
	Since $\alpha\,\in\,\bar{\Z}$, we have $\disc(\alpha)\,\in\,\Z$, and therefore
	$\disc(\alpha)=0$.  This implies that $1,\alpha,\ldots,\alpha^{n-1}$ are not linearly
	independent over $\Q$, and therefore that $n=1$, that is, $\alpha\,\in\,\Q$.  Since
	$\alpha$ is rational algebraic integer, it is in $\Z$.
\end{enumerate}

\section{A computation}

	We already know from class that the ring of integers of $K_1=\Q[\zeta_5]$ is $\Z[\zeta_5]$,
	and that $\disc(K_1)=125$.  We now show that the ring of integers of $K_2=\Q[\sqrt[5]{2}]$
	is $\Z[\sqrt[5]{2}]$.  We have the inclusions
	\begin{eqnarray*}
	    \Z[\sqrt[5]{2}]\subset \mathcal{O}_{K_2}\subset\frac{1}{N(P^{\prime}
	    (\sqrt[5]{2}))}\,\Z[\sqrt[5]{2}],
	\end{eqnarray*}
	where $P(x)=X^5-2$.  Using the known imbeddings of $K_2$ in $\C$, we easily compute 
	\begin{eqnarray*}
	    N(P^{\prime}(\sqrt[5]{2})=N(10/\sqrt[5]{2})=5\cdot10^4=\disc(K_2).
	\end{eqnarray*}
	Now, it is obvious that $\Z[\sqrt[5]{2}]=\Z[\sqrt[5]{2}-2]$.  Moreover, since the
	minimal polynomial of $\sqrt[5]{2}$ is $P(X)=X^5-2$, the minimal polynomial of 
	$\sqrt[5]{2}-2$ is $P(X+2)=X^5+10X^4+40X^3+80X^2+80X+30$.  Thus, since 
	$P(X)$ is $2$-Eisenstein and $P(X+2)$ is $5$-Eisenstein, we see that multiplication
	by the primes $2,\ 5$ is injective on $\mathcal{O}_{K_2}/\Z[\sqrt[5]{2}]$.  But since 
	the index of $\Z[\sqrt[5]{2}]$ in $K_2$ divided $\disc(K_2)=5\cdot10^4$, we see
	that in fact $\mathcal{O}_{K_2}/\Z[\sqrt[5]{2}]=0$, so that $\mathcal{O}_{K_2}$ is
	as claimed.  Now, since $K_2$ is of degree $5$ over $\Q$ and $K_1$ is of degree $4$,
	we find that $K_1K_2=\Q(\sqrt[5]{2},\zeta_5)$ is of degree $20$.  Hence, we
	have the following inclusions (since $\gcd(\disc(K_1),\disc(K_2))=125$)
	\begin{eqnarray*}
	    \Z[\sqrt[5]{2},\zeta_5]=\Z[\sqrt[5]{2}][\zeta_5]\subset T\subset \frac{1}{125}
	    \Z[\sqrt[5]{2},\zeta_5],
	\end{eqnarray*}
	where $T$ is the ring of integers of $K_1K_2$.


\section{The sign of the discriminant}

	We know that
	\begin{eqnarray*}
	    \disc(K)\ =\ |\sigma_i(\alpha_j)|^2,
	\end{eqnarray*}
	where $\sigma_1,\ldots,\sigma_n$ denote the $n$ distinct embeddings of
	$K$ in $\C$ and $\alpha_1,\ldots,\alpha_n$ is an integral basis for $K$.  Consider
	\begin{eqnarray*}
	    x\ =\ |\sigma_i(\alpha_j)|.
	\end{eqnarray*}
	We have 
	\begin{eqnarray*}
	    \bar{x}\ =\ |\bar{\sigma_i}(\alpha_j)|,
	\end{eqnarray*}
	where $\bar{\sigma}$ denotes the automorphism $\tau\sigma$ where $\tau$ is complex conjugation.
	If $\sigma$ is a real embedding of $K$ then $\sigma=\bar{\sigma}$, otherwise
	$\bar{\sigma}$ and $\sigma$ are distinct.  Since the sign of the determinant is
	changed when we swap columns or rows, it is not difficult to see that
	\begin{eqnarray*}
	    \bar{x}\ =\ (-1)^k x,
	\end{eqnarray*}
	where $k$ is the number of pairs of complex embeddings (since complex conjugation of the matrix
	$(\sigma_i(\alpha_j))$ swaps precisely $k$ rows).
	Therefore, since 
	\begin{eqnarray*}
	    [K:\Q]=2k+r
	\end{eqnarray*}
	where $r$ is the number of real embeddings, we see that $k=([K:\Q]-r)/2$.  Now if
	$k$ is even, then $\bar{x}=x$ so that the determinant $|\sigma_i(\alpha_j)|$ is real
	and hence its square is positive.  On the other hand, if $k$ is odd,
	$\bar{x}=-x$ so that the determinant $|\sigma_i(\alpha_j)|$ is pure imaginary and
	its square is negative.  Thus, we see that the sign of the discriminant is
	\begin{eqnarray*}
	    (-1)^k\ =\ (-1)^{\frac{[K:\Q]-r}{2}}.
	\end{eqnarray*}

\end{document}