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\noindent{\bf Math 431 October 5 homework discussion}

\medskip

\ex
Find odd square-free integers $a$ and $b$ such that $(\sqrt{a}+\sqrt{b})/2$
is an algebraic integer.

\A  The minimal polynomial for $\sqrt{a}+\sqrt{b}$ is
$$x^4-2(a+b)x^2+(a-b)^2.$$
So $(\sqrt{a}+\sqrt{b})/2$ is an algebraic integer if and only if
$a+b$ is even and $a-b$ is divisible by~16.  If $a$ and~$b$ are
both odd the first condition is automatic, and the second done is
true if and only if $a$ and~$b$ are congruent modulo~4.  Of course,
trivial examples exist, e.g., $a = b$.  And this analysis shows that
a random guess has a 50/50 chance of being right.

\ex
Prove that $D(fg) = D(f) D(g) \Res(f,g)^2$ where $D(f)$ denotes
the discriminant of a polynomial~$f$.

\A It is fairly routine to express all quantities in terms of the
roots of the polynomial.  Here is an argument (not particularly
shorter) that only uses properties of the resultant.
\begin{eqnarray*}
D(fg) & = & \pm f_m^{-1} g_n^{-1} R(fg,fg'+f'g) \\
& = & \pm f_m^{-1} g_n^{-1} R(f,fg'+f'g) R(g,fg'+f'g)\\
& = & \pm f_m^{-1} g_n^{-1} R(f,f'g) R(g,fg')\\
& = & \pm f_m^{-1} g_n^{-1} R(f,f')R(f,g) R(g,f)R(g,g')\\
& = & \pm D(f) D(g) R(f,g)^2.
\end{eqnarray*}
Here the definition $D(f) = (-1)^{m(m-1)/2} f_m^{-1} R(f,f')$
has been used several times, the ``quotient/remainder'' property
has been used, and multiplicativity has been used.  The $\pm$ sign
can be specified precisely and one finds that the final sign is
$$(-1)^{mn + m(m-1)/2 + n(n-1)/2 + (m+n)(m+n-1)/2} = 1.$$

\ex
Prove or disprove: If $E$ and $F$ are distinct cubic extensions
of~$\Q$, and $K$ is a number field that contains both, then the
degree of $K$ over~$\Q$ is divisible by~9.

\A The conjugate fields $\Q(\sqrt[3]{2})$ and 
$\Q(e^{2\pi i/3} \sqrt[3]{2})$ both have degree~3 over~$\Q$,
but, as discussed in class, the splitting field 
$\Q(\sqrt[3]{2},e^{2\pi i/3})$ contains both of them and has
degree~6 over~$\Q$.  (In fact, if $E$ and $F$ are ``unrelated''
any field~$K$ will have degree divisible by~9.)

\ex
Prove that the discriminant of the polynomial $f(x) = x^n+ax+b$ is
$$D(f) = (-1)^{n(n-1)/2} n^n b^(n-1) + (-1)^{(n+2)(n+3)/2} (n-1)^{n-1} a^n.$$

\A
Properties of the resultant come to the rescue.
\begin{eqnarray*}
D(x^n+ax+b) & = & (-1)^{n(n-1)/2} R(x^n+ax+b, nx^{n-1}+a) \\
&  = & (-1)^{n(n-1)/2} R(nx^{n-1}+a,x^n+ax+b) \\
& = & (-1)^{n(n-1)/2} n^n R(x^{n-1}+\frac{a}{n},x(x^{n-1}+\frac{a}{n})
+ \frac{(n-1)a}{n}+b) \\
& = & (-1)^{n(n-1)/2} n^n R(x^{n-1}+\frac{a}{n}, \frac{(n-1)a}{n}+b) \\
& = & (-1)^{n(n-1)/2} (-1)^n n^n R(\frac{(n-1)a}{n}+b,x^{n-1}+\frac{a}{n})\\
& = & (-1)^{n(n-1)/2} (-1)^n n^n \left (
(-b)^{n-1} + \frac{a}{n} \left ( \frac{(n-1)a}{n} \right)^{n-1} \right) \\
& = & (-1)^{n(n-1)/2} \left ( n^n b^{n-1} + (-1)^{n-1} (n-1)^{n-1} a^n
\right )\\
& = & (-1)^{n(n-1)/2} n^n b^{n-1} + (-1)^{(n+2)(n+3)/2} (n-1)^{n-1} a^n\\
\end{eqnarray*}

\ex
Show that $(1-\zeta_8)^4$ is an associate of 2 in $\Z[\zeta_8]$
where $\zeta_8 = e^{2\pi i/8}$ is a primitive eighth root of unity.

\A The minimal polynomial of $\zeta_8$ is $f(x) = x^4+1 $.  The norm
of $x-\zeta_8$ is $f(x)$ (see a later problem), so
$N(1-\zeta_8) = 2$.  It is easy to verify the equation
$$(1-\zeta_8)^4 = -4\zeta_8 + 6 \zeta_8^2 - 4 \zeta_8^3 = 2u$$
where $u = -2\zeta_8+3 \zeta_8^2 - 2 \zeta_8^3$; taking the
norm of both sides shows that $N(u) = 1$.  Thus $u$ is a unit
and $2$ and $(1-\zeta_8)^4$ are indeed associates.

\ex
Show that if $\m_\alpha(x) = x^3-x-2x^2-8$ and $F = \Q(\alpha)$
then 
$$(\alpha^2+\alpha)/2 \in \Z_F \backslash \Z[\alpha].$$

\A
There are several ways to find the equation of $\beta = \alpha^2+\alpha$.
One is to find $\beta^2$ and $\beta^3$ explicitly and find a linear
dependence on $1,\beta,\beta^2,\beta^3$.  Another is to find the
matrix of multiplication by $\beta$ (by multiplying $\beta$ by each
of $1, \alpha, \alpha^2$ and expressing the results in terms of the
same basis) and taking the characteristic polynomial of the matrix.  
Still another
is to do the symmetric function exercise implicit in finding the
coefficients of
$$ f(x) = \prod_{i=1}^3 (x - \alpha_i^2 - \alpha_i)$$
(where the $\alpha_i$ are the conjugates of $\alpha$).  Finally, one
can compute the resultant
$$\Res_x(y-x^2-x,x^3-x^2-2x-8)$$
to get a cubic in~$y$.  (The intuitive explanation is that in
eliminating $x$ between those two expressions we find an equation
satisfied by~$y$; the more precise explanation is that in taking
the product of the expression $y-x^2-x$ as $x$ takes on the three
different roots of $\m_\alpha$ we are exactly computing $f(x)$ above.)
We find (keeping in mind that signs are irrelevant)
$$\Res(x^2+x-y,x^3-x^2-2x-8)  = 
\Res(x^2+x-y,(x-2)(x^2+x-y)+yx-2y-8)$$
$$ = (2y+8)^2+y(2y-8)-y\cdot y^2 = -(y^3-6y^2+40y-64)$$
from which it follows that $\beta/2$ is an algebraic integer.
Since $1,\alpha,\alpha^2$ is a basis of $\Z[\alpha]$ it is obvious
that $\beta$ is not in $\Z[\alpha]$.







\ex
Let $p$ be an odd prime, $\zeta = e^{2\pi i/p}$, and $F = \Q(\zeta)$.
Find $N_{F/\Q}(\zeta)$.  Find $N_{F/\Q}(\zeta-1)$.

\A If $f(x) = \m_\alpha(x)$ for any algebraic $\alpha$ and $r$ is 
any integer then 
$$N_{\Q(\alpha)/\Q} (\alpha-r) = \prod_{i=1}^n (\alpha_i-r) =
(-1)^n \prod_{i=1}^n (r-\alpha_i) = (-1)^n f(r).$$
The minimal polynomial of $\zeta$ is $1+x+ x^2 + \cdots + x^{p-1}$ and
so we find that $N(\zeta) = 1$ and $N(\zeta-1) = p$.

\ex
Let $I$ be a nontrivial ideal in the ring of integers $\Z_F$ of
a number field~$F$.  Prove that there is a nonzero element in
$I \cap \Z$.

\A If $\alpha$ is a nonzero element of $I$ then $N(\alpha)$ is
nonzero, and $N(\alpha) = \alpha \beta = m \in \Z$, where $\beta$
is the product of the other conjugates of $\alpha$.  Since the
other conjugates are algebraic integers, $\beta$ is an algebraic
integer, and the equation $\alpha\beta = m$ shows that $\beta$ is
in~$F$.  Thus $\beta \in \Z_F$ and $m \in I$.

Alternatively, the constant term of the minimal polynomial of $\alpha$
is easily seen to be in $I \cap \Z$.  (This is no surprise: the
constant term is equal to the norm, up to sign.)

\ex
Find an example of nonunique factorization in $\Q(\sqrt{-15}).$
I.e., find an element that factors into irreducibles in two genuinely
different ways.

\A
The ring of integers is $\Z[\delta]$, where $\delta = (1+\sqrt{-15})/2$;
the norm form is $N(a+b\delta) = a^2+ab+4b^2$.  Almost anything works as
a counterexample; for instance
$$3 \cdot 5 = \sqrt{-15} \cdot \sqrt{15}.$$
To show that each element on both sides is irreducible, it suffices
to show that no element has norm 3 or 5 (e.g., if $3$ factors as
$e = uv$, where $u$ and $v$ aren't units, then $9 = N(3) = N(u)N(v)$,
so that $u$ has norm~3), and this is easy from the
norm form.
Almost any 

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