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\title{Algebraic Number Theory, PS 2}
\author{Bryden Cais}
\begin{document}
\maketitle
\begin{enumerate}

    \item Let $K$ be a local field of characteristic 0 with residue field $k$, and let $|*|$ be the normalized valuation of $K$.
    Since $K$ is local, $k$ is finite, say $\#k=q$, and hence has characteristic $p$ for some prime $p$.  The image of $\Z$ in $\O_K$
    then maps to $\Z/(p)$ in the residue field.  An integer $n$ maps to $0$ if and only if $p|n$, moreover, we have $p=\pi^e$ and hence
    the power of
    $\pi$ dividing the image of $n$ in $\O_K$ is equal to $re$, where $r$ is the power of $p$ dividing $n$ in $\Z$.
    Recall that
    $|*|$ is a nonarchimedian valuation, so that the infinite series
    $$\sum_{i=0}^{\infty} \a_i$$
    converges if and only if
    $$\lim_{i\rightarrow \infty} \a_i=0.$$
    We claim that the formal power series
    \begin{align*}
        \log (1+x)=x-x^2/2+x^3/3&-\cdots\\
        \exp(x)=1+x+x^2/2!+x^3/3!&+\cdots
    \end{align*}
    converge precisely for $|x|<1$ provided that $e<p-1$.  Indeed, we have
    \begin{align*}
        \lim_{n\rightarrow \infty} \left|\frac{x^n}{n}\right|&=\lim_{n\rightarrow\infty}q^{-(n\ord_{\pi} (x)-e\ord_p (n))}.
    \end{align*}
    But $e\ord_p(n)\leq e\log_p(n)$.  Observe that
    $$\lim_{n\rightarrow\infty} n\ord_{\pi} (x)-e\ord_p (n)=\infty$$
    if and only if $\ord_{\pi}(x)>0,$ that is, if and only if $|x|<1$.

    Now it is elementary to determine that the power of $p$ dividing $n!$ is
    $$d_p(n)=\sum_{i=0}^{\infty} \left\lfloor\frac{n}{p^i}\right\rfloor< \sum_{i=0}^{\infty}\frac{n}{p^i}=\frac{n}{p-1}.$$
    It follows that
    $$|x^n/n!|< q^{-(n\ord_{\pi}(x)-en/(p-1))}.$$
    But $$n\ord_{\pi}(x)-en/(p-1)=n\left(\ord_{\pi}(x)-\frac{e}{p-1}\right),$$
    which tends to $\infty$ for $\ord_{\pi}(x)>e/(p-1)$.  Supposing $e< (p-1)$, this shows that
    the series for $\exp(x)$ converges for $|x|<1$.

    Next, observe that \textit{as formal power series} we have
    \begin{align}
        \log ((1+x)(1+y))&=\log(1+x)+\log(1+y)\label{loghom}\\
        \exp(x+y)&=\exp(x)\exp(y)\label{exphom}\\
        \log\exp(z)&=z\label{id1}\\
        \exp\log(1+z)&=1+z\label{id2},
    \end{align}
    with equality holding whenever all series in question converge.

    Now let
    $$U=\{\alpha\in\O_K:|\alpha-1|<1\}$$ and
    $$D=\{\alpha\in\O_K:|\alpha|<1\}.$$
    Observe that $U$ is a subgroup of $\O_K^{\times}$ and every $\alpha\in U$ has the form $1+\pi x$ for some $x\in\O_K$.
    Moreover, we see that $D$ is an additive subgroup of $\O_K$ since $|*|$ is nonarchimedian so that
    if $\alpha,\beta\in D$ then $|\alpha+\beta|<\max\{|\a|,|\b|\}<1$.  In addition, it is clear that if $\alpha\in D$
    then $1+\alpha\in U$ and conversely.

    We know that the power series defining $\log(1+x)$ converges for $x\in D$, or equivalently, $\log(x)$ gives a function
    $$\log:U\rightarrow K.$$
    In fact, for $\alpha\in U$ we have
    $$|\log(\alpha)|=|\log(1+(\a-1))|=\left|(\a-1)-\frac{(\a-1)^2}{2}+\cdots\right|\leq \max\left\{\frac{|\a-1|^i}{i}\right\}<1$$
    since $|\a-1|<1$.
    Moreover, by (\ref{loghom}), we see that $\log$ defines a homomorphism
    $$\log:U\rightarrow D.$$

    Similarly, the series defining $\exp(x)$ converges for $x\in D$ and a quick calculation gives
    $$|\exp(x)-1|=\left|\sum_{k=1}^{\infty}\frac{x^k}{k!}\right|\leq \max\left\{\left|\frac{x^k}{k!}\right|\right\}<1$$\
    provides that $\ord_{\pi}(x)>e/(p-1)$, so that the image of $D$ under $\exp$ lies in $U$.  Thus, by (\ref{exphom}) we have a homomorphism
    $$\exp:D\rightarrow U.$$

    Finally, by identities (\ref{id1}) and (\ref{id2}) we see that $\exp$ and $\log$ are in fact inverses to one another, and hence that
    $$\log: U\rightarrow D\quad\text{and}\quad \exp:D\rightarrow U$$
    are mutually inverse algebraic and topological isomorphisms (since $\exp,\log$ are both continuous functions).
    Since $\O_K\simeq D$ via the mapping $x\rightarrow \pi x$ we see that
    $$\O_K\simeq U\subset \O_K^{\times}$$ both algebraically and topologically via the map
    $$x\rightarrow \exp(\pi x).$$

    Now consider the map
    $$\Z_p^{\times}\rightarrow (\Z/p\Z)^{\times}$$ given by
    $$x\mapsto x\mod p.$$  We have seen that this is a surjection, and the kernel of this map is obviously $U$.
    Moreover, this sequence is \textit{split}, for we have the canonical map
    $$a\mapsto w(a),$$
    where $w$ is the Teichm\"{u}ller character.  In other words, since $x^{p-1}-1$ splits completely over $\F_p$,
    (and since it is a separable polynomial), Hensel's lemma tells us that $\Z_p$ contains the $p-1$ st roots of unity,
    and that modulo $p$, these map bijectively to $\F_p$.  Our splitting homomorphism then maps $a\in \F_p$ to the unique
    $(p-1)^{\text{st}}$ root of unity that is congruent to $a\mod p$.
    It follows that
    $$\Z_p^{\times}\simeq (\Z/p\Z)^{\times}\times U.$$  Now $(\Z/p\Z)^{\times}\simeq \Z/(p-1)$ (by picking a generator for $(\Z/p\Z)^{\times}$),
    while we have shown above that $U\simeq \Z_p$, with $\Z_p$ considered additively.  It follows that
    $$\Z_p^{\times}\simeq \Z/(p-1)\times \Z_p.$$

    \item Let $K$ be a local field of characteristic $p$.  Then we have seen that $K$ is isomorphic to $\F_q((t))$, where
    $q=p^r$ for some $r$ and that $\pi_K=t$ and $\O_k^{\times}$ is the group of formal power series with coefficients
    in $\F_q$ having nonzero constant term.  Recall that $x\rightarrow x^p$ is a homomorphism of $K$.
    We claim that the index of $(\O_K^{\times})^p$ in $\O_K^{\times}$ is infinite.  For we can realize $\O_K^{\times}$
    as the inverse limit
    $$\lim_{\substack{\longleftarrow\\m} }(\F_q[t]/t^m)^{\times}$$
    (with the obvious projection maps $\pi_{ij}:(\F_q[t]/t^i)^{\times}\rightarrow (\F_q[t]/t^j)^{\times}$ for $i>j$ given
    by reduction modulo $t^j$).
    Thus, it is enough to show that the index of
    $$\left((\F_q[t]/t^m)^{\times}\right)^p=(\F_q[t^p]/t^m)^{\times}$$
    in $(\F_q[t]/t^m)^{\times}$ grows without bound as $m\rightarrow \infty.$  However, this is easy:
    The size of $(\F_q[t]/t^m)^{\times}$ is
    $$(q-1)q^{m-1}$$
    (since we are considering polynomials with nonzero constant term of maximum degree $m-1$),
    while the size of $(\F_q[t^p]/t^m)^{\times}$ is clearly
    $$(q-1)q^{\lfloor (m-1)/p\rfloor}.$$
    It follows that the index of $(\F_q[t^p]/t^m)^{\times}$ in $(\F_q[t]/t^m)^{\times}$ is
    $$q^{m-1-\lfloor (m-1)/p\rfloor}\rightarrow \infty$$
    as $m\rightarrow \infty$.  Our claim follows.

    \item The proof given in class that a local field of characteristic 0 has only finitely many
    extensions breaks down in characteristic $p>0$ for the following reason: \textbf{in characteristic $p$,
    not every irreducible polynomial is separable} (consider, for example, the polynomial
    $X^p-t\in \F_p((t))[X]$).  Let us explore where this fundamental fact causes problems in the proof.
    We set up a correspondence $C$ between the points of
    $$M=\m_K\times\m_K\times\cdots\times\m_K\times(\m_K\setminus \m_K^2)$$
    and the Eisenstein polynomials of degree $n$ (here, $n$ is the number of copies of $\m_K$),
    by mapping a point $a=(a_0,\ldots,a_n)\in M$ to the polynomial $C(a):=a_0x^n+\cdots + a_n$.
    Clearly $M$ is compact since it is a finite product of compact sets.  We then used Krasner's lemma
    to construct an open set $U_a$ about each point $a\in M$ such that for every $x\in U_a$,
    the fields cut out by the roots of $C(x)$ are the \textit{same} as the fields cut out by the roots of
    $C(a)$.  However, Krasner's lemma crucially requires that the polynomial $C(a)$ be \textit{separable}
    in order to construct the open set $U_a$ as above.  In characteristic $0$, this separability is
    a consequence of irreducibility.  As noted above, this is false in characteristic $p$.  Thus, in characteristic
    $p$ we in general cannot construct an open cover of $M$.  Put differently, the set $N\subset M$ of Eisenstein
    polynomials about which there is an open set $V$ such that every polynomial in $V$ generates the same field extensions
    is \textit{not} compact since it is not even closed in $M$.  Observe that $N$ is just the set of separable Eisenstein
    polynomials, which is the complement of the set $N_s$ of non-separable Eisenstein polynomials in $M$.  The set $N_s$
    is clearly closed since $f\in N_s$ if and only if the resultant of $C(f)$ vanishes.  (So we can view $N_s$ as a subvariety
    of $M$).



    \item Let $|*|$ be a nontrivial valuation on $\Q$.  We distinguish two cases:
    \begin{enumerate}[{Case} 1:]
        \item There exists some positive integer $n$ such that $|n|>1$.  Let $n_0$ be the least such positive integer.
        Then since $|n_0|>1,$ there exists some positive real number $r$ such that $|n_0|=n_0^r$.  Let
        $n\in \N$.  Then we claim that $|n|=n^r$.  We proceed as follows: writing
        \begin{align}
            n&=a_0+a_1 n_0+\cdots +a_k n_0^k \label{rep}
        \end{align}
        with $0\leq a_i< n_0$ we have
        \begin{align*}
            |n|&\leq |a_0|+|a_1| |n_0|+\cdots +|a_k| |n_0|^k\\
            &\leq 1+n_0^r+\ldots +n_0^{rk}\\
            &\leq n_0^{rk}\sum_{j=0}^{\infty} n_0^{-j}\\
            &\leq C n^{r}
        \end{align*}
        for some positive constant $C$ {\em independent of $n$}, where we have used the fact that $|a_i|\leq 1$ since $a_i < n_0$ and the obvious
        fact that $n_0>1$.
        It follows that for every integer $N>0$ we have
        $$|n^N|=|n|^N\leq C n^{rN},$$
        so that
        $$|n|\leq C^{1/N} n^r.$$  Taking $N$ arbitrarily large shows that
        \begin{align}
        |n|\leq n^r\label{ineq1}
        \end{align}
        for all positive integers $n$.
        On the other hand, we obviously have $n_0^{k+1}>n\geq n_0^{k}$ by (\ref{rep}), so that
        \begin{align*}
            |n|&=|n+n_0^{k+1}-n_0^{k+1}|\\
            &\geq |n_0^{k+1}|-|n_0^{k+1}-n|\\
            &=n_0^{(k+1)r}-|n_0^{k+1}-n|\\
            &\geq n_0^{(k+1)r}-(n_0^{k+1}-n)^r\\
            &\geq n_0^{(k+1)r}\left(1-\left(1-\frac{n}{n_0^{k+1}}\right)^r\right)\\
            &\geq n_0^{(k+1)r}\left(1-\left(1-\frac{1}{n_0}\right)^r\right),
        \end{align*}
        where we have used the inequality (\ref{ineq1}) on the third line and the fact that $n>n_0^k$ on the penultimate line.
        Setting
        $$D=1-\left(1-\frac{1}{n_0}\right)^r,$$ we see that
        $$|n|\geq Dn^r$$ where $D$ is independent of $n$.  Using the same trick as above, we have
        $$|n|^N\geq D n^{Nr}$$ for all positive integers $N$ and hence that
        \begin{align}
        |n|\geq n^r.\label{ineq2}
        \end{align}
        Combining inequalities (\ref{ineq1}) and (\ref{ineq2}) we find that
        \begin{align}
        |n|= n^r\label{ineq3}
        \end{align}
        for all positive integers $n$.  It follows that $|x|=|x|_{0}^r$ for all $x$ in $\Q$, where $|*|_0$ is the usual absolute value
        on $\R$.  Since $\Q$ is dense in $\R$, we then have
        $$|*|=|*|_0^r,$$ so that $|*|$ is equivalent to $|*|_0$.

        \item For every positive integer $n$, one has $|n|\leq 1$.  Since $|*|$ is nontrivial, there exists some positive integer $n$
        with $|n|<1$. Let $n_0$ be the least positive integer such that $|n_0|<1$.  Then $n_0$ must be prime.  For if not, say $n_0=ab$
        with $a, b>1$, then we have $|n_0|=|a||b|=1$ since $a,b<n_0$.  Now let $p\neq n_0$ be any prime.  Then
        for any positive integers $j,k$ there exist integers $u,v$ (depending, of course on $j,k$)
        such that
        $$p^ju+n_0^kv=1.$$
        It follows that
        \begin{align*}
            1&=|p^ju+n_0^kv|\\
            &\leq |p|^j|u|+|n_0|^k|v|\\
            &\leq |p|^j+|n_0|^k.
        \end{align*}
        If $|p|<1$ then we may choose $j,k$ sufficiently large so that $|p|^j+|n_0|^k <1,$ a contradiction.  It follows that
        $|p|=1$ for all primes $p\neq n_0$.  Therefore, for any integer $n$ we have
        \begin{align*}
            |n|=|n_0|^{\ord_p(n)},
        \end{align*}
        where $\ord_{n_0}(n)$ denoted the highest power of $n_0$ dividing $n$.  As before, this means that
        \begin{align}
            |x|=|x|^{\ord_{n_0}(x)},
        \end{align}
        for all $x\in\Q$, where we have extended $\ord_{n_0}$ to $\Q$ in the usual way.  Let $r=\log |n_0|/\log n_0$.  Then we have
        $$|x|=n_0^{r\ord_{n_0}(x)}=|x|^r_{n_0},$$
        where $|*|_{n_0}$ is the usual $n_0$-adic valuation on $\Q$.  In short, this shows that $|*|$ is equivalent to $|*|_{n_0}$ and we are done.
    \end{enumerate}

    \item Let $K$ be a local field equipped with a valuation $v$ and $\pi$ a uniformizer of $K$.  Let $\v$ be an extension of
    $v$ to $K^{sep}$.  Suppose that
    $$f(X)=X^n+a_1X^{n-1}+\cdots + a_n$$ with $a_i\in K^{sep}$ is a monic polynomial with roots
    $$\a_1,\a_2,\ldots,\a_n\in K^{sep}$$ numbered so that we have
    \begin{alignat*}{5}
        \v(\a_1)&=\v(\a_2)&=\ldots& = \v(\a_{l_1})&=m_1\\
        \v(\a_{l_1+1})&=\v(\a_{l_1+2})&=\ldots &= \v(\a_{l_2})&=m_2\\
        \vdots & &&& \vdots\\
        \v(\a_{l_{r}+1})&=\v(\a_{l_{r}+2})&=\ldots &= \v(\a_{n})&=m_{r+1}\\
    \end{alignat*}
    and $$m_1< m_2 < m_3<\ldots < m_{r+1}.$$
    Recall that we have
    $$a_i=s_i(\a_1,\ldots,\a_n),$$
    where $s_i$ is the $i^{\text{th}}$ symmetric polynomial in the $\a_j$.  We then have
    \begin{alignat*}{3}
        v(a_1)&=\v(s_1(\a_1,\ldots,\a_n))  \geq \min \{\v(\a_{i_1})\}  &=m_1\\
        v(a_2)&=\v(s_2(\a_1,\ldots, \a_n))  \geq \min \{\v(\a_{i_1}\a_{i_2})\}  &=2m_1\\
        \vdots &  & \vdots\\
        v(a_{l_1-1})&=\v(s_{l_1-1}(\a_1,\ldots, \a_n))  \geq \min \{\v(\a_{i_1}\cdots\a_{i_{l_1-1}})\}  &=(l_1-1)m_1\\
    \end{alignat*}
    while since $\v(\a_1\cdots \a_{l_1})$ is \textit{strictly} less than $\v(\a_{i_1}\cdots \a_{i_l})$ for any other set
    $\{i_1,\ldots,i_l\}$, it follows that in fact
    $$v(a_{l_1})=\min \{\v(\a_{i_1}\cdots\a_{i_{l_1}})\} =l_1m_1$$
    Continuing in this manner, we have
     \begin{alignat*}{3}
        v(a_{l_1+1})&=\v(s_{l_1+1}(\a_1,\ldots,\a_n))  \geq \min \{\v(\a_{i_1}\cdots \a_{i_{l_1+1}})\}  &=l_1m_1+m_2\\
                v(a_{l_1+2})&=\v(s_{l_1+2}(\a_1,\ldots,\a_n))  \geq \min \{\v(\a_{i_1}\cdots \a_{i_{l_1+2}})\}  &=l_1m_1+2m_2\\
        \vdots &  & \vdots\\
        v(a_{l_2-1})&=\v(s_{l_2-1}(\a_1,\ldots, \a_n))  \geq \min \{\v(\a_{i_1}\cdots\a_{i_{l_2-1}})\}  &=l_1m_1+(l_2-l_1-1)m_1\\
    \end{alignat*}
    while again, since $\v(\a_1\cdots \a_{l_2})$ is \textit{strictly} less than $\v(\a_{i_1}\cdots \a_{i_2})$
    for any other set
    $\{i_1,\ldots,i_2\}$, it follows that we have
    $$v(a_{l_2})=\min \{\v(\a_{i_1}\cdots\a_{i_{l_2}})\} =l_1m_1+(l_2-l_1)m_2.$$

    We therefore see that the Newton polygon of $f$ has precisely the vertices
    $$(0,0),\ (l_1,l_1m_1),\ (l_2,l_1m_1+(l_2-l_1)m_2),\ldots,(l_{r+1},l_1m_1+(l_2-l_1)m_2+\cdots+(l_{r+1}-l_r)m_{r+1}).$$
    Observe that the $j^{\text{th}}$ line segment in the Newton polygon is
    $$\frac{(l_1m_1+(l_2-l_1)m_2+\cdots+(l_{j}-l_{j-1})m_{j})-(l_1m_1+(l_2-l_1)m_2+\cdots+(l_{j-1}-l_{j-2})m_{j-1})}{l_j-l_{j-1}}=m_j$$
    and that the $x$ length of this segment is $l_j-l_{j-1}$.  Thus we see that the Newton polygon of $f$ contains a line segment of $x$-length $m$
    and slope $s$ if and only if $f$ has precisely $m$ roots with valuation $s$.

    \item Let $L/K$ be a finite separable extension of local fields with corresponding residue field extension $l/k$.
    Recall that
    $$\phi:I(L/K)\rightarrow l^{\times}$$
    defined by
    \begin{align}
    \phi(\s)&=a_{\s}\label{phidef}\\
    \intertext{where}
    \s(\pi_L)&=(a_{\s}+a_1\pi_L+\cdots)\pi_L.\label{asig}
    \end{align}
    induces an injection
    $$I^t(L/K)\hookrightarrow l^{\times}.$$
    We claim that this can be an isomorphism.  Consider, for example, the extension
    $$L/K=\Q_p(\z_{p^{m}})/\Q_p,$$
    where $\z_{p^m}$ is a primitive root of unity of order $p^m$ and $m>1$.  We have seen that this is a totally
    ramified extension of degree
    $$\varphi(p^m)=(p-1)p^{m-1}$$ with Galois group $(\Z/p^m\Z)^{\times}$.  It is not difficult to see from this that the maximal tamely ramified extension
    of $\Q_p$ in $\Q_p(\z_{p^m})$ is $K^{t}=\Q_p(\z_p)$; it is a totally tamely ramified extension of degree $p-1$ with Galois group
    $(\Z/p\Z)^{\times}$.  It follows that $L/K^t$ is a totally wildly ramified extension of degree $p^{m-1}$.  We therefore have
    \begin{alignat*}{4}
        \#I(L/K)&= \#\gal(L/K^{ur})&=\#\gal(L/K)&=(p-1)p^{m-1}\\
        \#I_p(L/K)&=\#\gal(L/K^t)&=[\Q_p(\z_{p^{m}}):\Q_p(\z_p)]&=p^{m-1}\\
        \# I^t(L/K)&=\# I(L/K)/I_p(L/K)&= (p-1)p^{m-1}/p^{m-1}&=(p-1).
    \end{alignat*}
    Since $L/K$ it totally ramified, we have $[l:k]=1$ and hence $l=k=\F_p$.  Now $I^t(L/K)$ injects into $\F_p$, but since both
    have cardinality $p-1$, we have a surjection.  It follows that
    $$\phi:I^t(\Q_p(\z_{p^m})/\Q_p)\longrightarrow \F_p$$
    is an isomorphism (since we have seen in class that it is a homomorphism).

    Next, using (\ref{phidef}) and (\ref{asig}) we find
    \begin{align*}
        \s\t(\pi_L)&=(a_{\s\t}+\cdots)\pi_L\\
        &=\s((a_{\t}+\cdots)\pi_L)\\
        &=(\s(a_{\t})+cdots)\s(\pi_L)\\
        &=(\s(a_{\t})+\cdots)(a_{\s}+\cdots)\pi_L\\
    \end{align*}
    so that
    \begin{align}
        a_{\s\t}=\s (a_{\t}) a_{\s}\label{cocy}.
    \end{align}
    Using (\ref{cocy}) we find
    \begin{align*}
        a_{\t\s\t^{-1}}&=\t\s(a_{\t^{-1}})a_{\t\s}\\
        &=\t\s(a_{\t^{-1}})\t(a_{\s})a_{\t},
    \end{align*}
    but we are now viewing $\t,\s$ as elements of $\gal(l/k)$ since $a_{\s},a_{\t}$ etc. are elements of $l$.  Since
    $\gal(l/k)$ is abelian ($l,k$ are finite fields) we have
    \begin{align}
        a_{\t\s\t^{-1}}&=\s\t(a_{\t^{-1}})\t(a_{\s})a_{\t}.\label{almost}
    \end{align}
    Now observe that by (\ref{cocy}) we have
    \begin{align*}
        a_{\t\t^{-1}}&=\t(a_{\t^{-1}})a_{\t}\\
        &=a_{\id}\\
        &=1,
    \end{align*}
    so that
    $$\t a_{\t^{-1}}=\frac{1}{a_{\t}}.$$
    Substituting this relation in (\ref{almost}), we have
    \begin{align}
        a_{\t\s\t^{-1}}&=\frac{a_{\t}}{\s a_{\t}}\t(a_{\s}).\label{coho}
    \end{align}
    Thus, if $\s\in I(L/K)=\gal(L/K^{ur})$ then $\s$ fixes $K^{ur}$, and since
    $$\gal(K^{ur}/K)\simeq \gal(l/k),$$ we see that $\s$ projects to the identity element of $\gal(l/k)$, and hence fixes $a_{\t}\in l$.
    Thus, if $\s\in I(L/K)$ and if $\tau\in \gal(L/K)$ projects to a Frobenius element, then we have
    \begin{align}
        a_{\t\s\t^{-1}}&=\frac{a_{\t}}{\s a_{\t}}\t(a_{\s})\\
        &=\t(a_{\s})\\
        &=a_{\s}^{\# k},
    \end{align}
    as required.

    \item Let $K$ be a local field of characteristic $0$ and let $T$ be the maximal abelian, tamely ramified
    extension of $K$.  Then certainly $T\supset K^{ur}$ since $K^{ur}/K$ as abelian.
    Since $T/K^{ur}$ is a tamely ramified extension, we have
    $$I_p(T/K^{ur})=\{1\},$$
    so that
    $$I(T/K^{ur})=I^t(T/K^{ur})$$ and we have (as in problem 6) an injection
    $$\phi:I(T/K^{ur})\hookrightarrow (k^{sep})^{\times},$$
    where $k^{sep}/k$ is the associated residue field extension of $T/K^{ur}$.  Now there is some $\tau\in \gal(T/K^{ur})$
    that projects to the Frobenius automorphism of $\gal(k^{sep}/k)$.  Since $\gal(T/K^{ur})$ is abelian, problem
    6 tells us that we have
    $$\phi(\t\s\t^{-1})=\phi(\s)=a_{\s}^{\# k}=\a_{\s}$$
    for any $\s\in I(T/K^{ur})$.  We conclude that in fact
    $$\phi:I(T/K^{ur})\hookrightarrow k^{\times}.$$ This at any rate gives
    $$\deg(T/K^{ur})\leq \# k.$$  How to obtain equality?

    Well, in the special case that $k=\F_p$, we know that $K^{ur}(\z_p)/K^{ur}$ is a totally tamely ramified abelian extension of degree
    $p-1$, where $\z_p$ is a primitive $p^{\text{th}}$ root of unity.  This shows that $\deg(T/K^{ur})\geq p-1$.  I have been trying to
    find a similarly explicit argument for the general case, but to no avail yet.



    \item Let $K$ be a local field of characteristic 0 and $\bar{K}$ an algebraic closure of $K$.  Suppose further
    that the residue field $k$ of $K$ has characteristic $p$.  Let $\z_n$ be a primitive
    $n^{\text{th}}$ root of unity and let $m_n=n!p^{-\ord_p(n!)}$; that is, $m_n$ is $n!$ with the highest power of $p$
    removed.  Set
    \begin{align}
    a_N&=\sum_{n=1}^N \z_{m_n}\pi_K^{m_n}.\label{cauch}
    \end{align}
    Obviously $\{a_i\}$ forms a Cauchy sequence.  Suppose that this sequence converges to some
    $$\a=\sum_{n=1}^{\infty} \z_{m_n}\pi_K^{m_n}\in \bar{K}.$$  We will obtain a contradiction.

    Let $\s\in\gal(\bar{K}/K)$.  We claim that for $(i,p)=1$ we have $|\z_i-\s(\z_i)|=1$.  For
        $$\b=\prod_{\s(z_i)\neq z_i}(\z_i-\s(\z_i))$$
        divides
        $$\prod_{\substack{j|i\\\s(\z_i)\neq \z_i}}(\z_i-\s(\z_j))=\frac{\mathrm{d}}{\mathrm{d}x}(x^i-1)|_{x=\z_i}=\frac{i}{\z_i},$$
    and since $(p,i)=1$ we see that $\left|\frac{i}{\z_i}\right|=1$ and since $\b\in \O_K$, we must have $|\b|=1$.  It follows that
    $$|\s(\z_i)-\z_i|=1.$$

    Since $$\s(a_N)-a_N=\sum_{n=1}^N (\s(\z_{m_n})-\z_{m_n})\pi_K^{m_n}$$
    for any $\s\in\gal(\bar{K}/K)$, we see that since $|(\s(\z_{m_n})-\z_{m_n})\pi_K^{m_n}|=|\pi_K|^{-m_n}$ if $\s$ does not fix $\z_{m_n}$
    and since $|\a+\b|=\min\{|\a|,|\b|\}$ if $|\a|\neq |\b|$, we must have
    $$|\s(a_N)-a_N|=|\pi_K|^{-m_n},$$
    where $n$ is the smallest positive integer $j$ such that $\s(\z_{m_j})\neq \z_{m_j}$.  In particular, if
    $\s(\z_{m_N})\neq \z_{m_N}$ then $n\leq N$ whence $|\s(a_N)-a_N|\neq 0$ so that $\s(a_N)\neq a_N$.  Since
    $K(a_N)\subseteq K(\z_{m_N})$ by our construction of $a_N$, and since any $\s\in \gal(K(\z_{m_N})/K)$ acts nontrivially
    on $a_n$ we in fact have equality.  We have seen that the degree of $K(\z_{m_N})$ is $f$, where $f$ is the smallest
    positive integer such that $q^f\equiv 1\mod m_N$.  It is clear that as $m_N\rightarrow \infty$ we also have $f\rightarrow \infty$.

    Now we use Krasner's lemma to show that $K(a_N)\subseteq K(\a)$.  Indeed, by our discussion above, we clearly have
    $$|a_N-\a|=|\pi_K|^{-m_{N+1}}.$$  On the other hand, if $\s\in\gal(\bar{K}/K)$ does not fix $a_N$ then since $K(a_N)=K(\z_{m_N})$
    it cannot fix $\z_{m_N}$, whence $|a_N-\s(a_N)|\geq |\pi_K|^{-m_N}$.  It follows that $K(a_N)\subseteq K(\a)$ so that
    $K(\a)/K$ has degree at least as large as the degree of $K(a_N)/K$.  But we have shown above that as $N\rightarrow \infty,$
    the degree of $K(a_N)/K$ grows without bound.  Hence, $\alpha\not\in\bar{K}$ since otherwise the degree of $K(\a)/K$
    would be finite.  It follows that $\bar{K}$ is not complete.

    \item Let $K$ be a local field with valuation $v$ and $L$ an extension of $K$ with valuation $w$ extending $v$.  Suppose that
    $L/K$ has ramification index $e$ so that $|\pi_L|^e=|\pi_K|$.  It follows that the index of the valuation ring
    $w(L)$ in the ring $v(K)$ is $e$.  Thus, if $L/K$ is any \textit{unramified} extension, we see that
    $$w(L)=v(K).$$
    It follows that the valuation ring of $K^{ur}$ is that of $K$ since the valuation ring of any finite subextension $L/K$
    of $K^{ur}/K$ is $v(K)$.  Since $v(K)$ is discrete, it follows that the extension of $v$ to $K^{ur}$ is discrete.  Observe
    that we are critically using the fact that $K^{ur}/K$ is unramified.  For example, even though any \textit{finite}
    extension $L/K$ has discrete valuation ring (if $v(K)$ is discrete) the extension $\bar{K}/K$, for example, does not have discrete valuation ring.

    Now we claim that $K^{ur}$ is not complete.  Indeed, if $(n,p)=1$ (where $p$ is the characteristic of $K/\pi_K$)
    we see that $x^n-1$ is separable over the residue field of $K$ so that by Hensel's lemma, the irreducible polynomial
    of $\z_n$ (a primitive $n^{\text{th}}$ root of unity), which divides $x^n-1$ and is hence separable,
    remains irreducible over the residue field.  It follows that the extension $l/k$ of the residue fields of the extension
    $K(\z_n)/K$ has the same degree as that of $K(\z_n)/K$ so that $K(\z_n)/K$ is unramified.  (Indeed, we can say more:
    since the residue field of $K^{ur}/K$ is the separable closure of $k$, which is generated by adjoining the roots
    of $x^n-1$ for all $(n,p)=1$ to $k$---this follows directly from the theory of finite fields---we see that
    $K^{ur}/K$ is generated by adjoining all roots of unity of order prime to $p$).
    Thus, the cauchy sequence (\ref{cauch}) is actually a cauchy sequence in $K^{ur}$.  Since $K^{ur}\subset \bar{K}$
    and the sequence (\ref{cauch}) does not even converge to a limit in $\bar{K}$ (c.f. problem 8), we see that
    $K^{ur}$ is not complete.

    \item Let $K=\Q(\sqrt{n},\sqrt{m})$ for $m,n$ distinct and square free (and not equal to $1$).  A rational prime $p$
    cannot be totally inert in $K$.  Recall from the previous problem set that a prime $p$ splits in $\Q(\sqrt{n})$
    if and only if $x^2-n$ splits modulo $p$, in other words, if and only if $n$ is a square modulo $p$.  However,
    $K$ has exactly three quadratic subfields---namely $\Q(\sqrt{n}),\ \Q(\sqrt{m}),\ \Q(\sqrt{mn})$---
    and if $m,n$ are non-squared modulo $p$ then $mn$ is a square modulo $p$ (similarly, if $m,mn$ are non-squares, then
    $m$ is a square modulo $p$ and so forth).  This is elementary.  It follows that if $p$ is inert in any two of the quadratic
    subfields of $K$ then it splits (or ramifies) in the third.  Thus no rational prime can be totally inert in $K$ since
    it must already split in a subfield.

    Observe that $K/\Q$ is a Galois extension with Galois group $\Z/(2)\times \Z/(2)$.  We thus have the fundamental identity
    $$efd=4,$$
    where $e=e(\wp|p),\ f=f(\wp|p),$ and $d$ is the number of primes lying over $p$ (here $\wp|(p)$ in $\O_K$).  Moreover, how
    $p$ splits in the three quadratic subfields of $K$ completely determines how $p$ splits in $K$.  We have seen that a rational
    prime $p$ cannot be totally inert in $K$.  However, our argument also shows that $p$ either splits in
    exactly one of the three quadratic subfields or it splits in all three (in terms of the Legendre symbol, the options are
    $-1\cdot -1=1$ or $1\cdot 1=1$ or $-1\cdot 1= -1$; in any case there are either one or three 1's).  Moreover, a prime $p\neq 2$
    ramifies in $\Q(\sqrt{n})$ if and only if $p|n$.  Since $p$ divides
    one of $m,n$ if and only if $p|mn,$ we see that a prime $p$ ramifies in any quadratic subfield, it
    ramifies in two of them.  Moreover, since $m,n$ are square free, $p$ cannot ramify in all three subextensions since
    if $m,n$ contain $p$ as a factor then $mn$ contains $p$ as a square factor and hence $p$ does \textit{not} divide
    the discriminant of $\Q(\sqrt{mn})$ (similarly if $m,mn$ have a square-free factor of $p$ then $p\not| n$ etc.).
    Finally, if $p\neq 2$ ramifies in any extension, it ramifies in two, and hence must split in the third.

    If $p=2$, the behavior is different: for example, it is possible for $p$ to ramify in all three extensions.
    Similar criteria can be worked out for $p=2$, but this is routine so we omit it here.

    Let $K_1,K_2,K_3$ be the three
    quadratic subfields of $K$.  We thus have the following
    table of data and examples (up to permutation of the labelling of $K_i$):
    \begin{center}
    \begin{tabular}{|c|c|c|c|c|c|c|}
    \hline
    $K_1$ & $K_2$ & $K_3$ & K & m & n & p\\
    \hline
    splits & inert & inert & $p=\p_1\p_2$ & 2 & -1 & 5\\
    ramifies & ramifies & splits & $p=\p_1^2\p_2^2$ & 5 & -1 & 5\\
    splits & splits & splits & $p=\p_1\p_2\p_3\p_4$ & 3 & -1 & 13 \\
    \hline
    \end{tabular}
    \end{center}

    \item We claim that it is possible to find $f\in \Z[x]$ that is irreducible in $\Q[x]$ but reducible in
    $\Q_p[x]$ for every $p$.  First observe that problem $10$ gives us an easy way to construct an $f\in \Z[x]$
    that is irreducible over $\Q$ but reducible over $\F_p$ for every $p$.  In particular, let $f$ be the field polynomial
    of the extension $\Q(\sqrt{m},\sqrt{n}).$  Certainly $f$ is irreducible of degree $4$ over $\Q$ (since $m,n$ are distinct
    and square free, not equal to 1).  However, as we showed in problem 10, no rational prime $p$ is inert in $\Q(\sqrt{m},\sqrt{n})$,
    from which it follows that $f$ is reducible mod $p$ for every $p$ (since
    $$\Z[X]/(f,p)=\F_p[x]/(f)$$
    is not an integral domain since $p$ is reducible in $\Z[x]/f$, so that $f$ is reducible in $\F_p[x]$.)

    To construct $f\in\Z[x]$ that is irreducible over $\Q$ but reducible over $\Q_p$ for every $p$ we proceed as follows.
    Recall that for $p$ odd, $\Q_p$ has three distinct quadratic extensions, while $\Q_2$ has seven.  In particular,
    the seven quadratic extensions of $\Q_2$ are precisely the seven quadratic subfields of
    $$K_2:=\Q_2(\sqrt{-1},\sqrt{2},\sqrt{5}).$$  It follows that $\sqrt{3}\in K_2$ and hence that $K_2(\sqrt{3})/\Q_2$
    has degree $8$.  Moreover, since $\Q_p$ for $p>3$ has only 3 distinct quadratic extensions, the degree of
    $$\Q_p(\sqrt{-1},\sqrt{2},\sqrt{3},\sqrt{5})/\Q_p$$
    is at most 4.

    Now Let $f$ be the field polynomial of $\Q(\sqrt{-1},\sqrt{2},\sqrt{3},\sqrt{5})$.  A short
    computation with PARI tells us that
    $$f(x)=x^{16} - 72x^{14} + 1932x^{12} - 22552x^{10} + 154038x^8 - 582456x^6 + 1440748x^4 - 1486824x^2 + 3721041.$$
    Since $f$ is the field polynomial of a degree $16$ extension (and of degree 16), it is certainly irreducible over $\Q$.
    However, if $f$ were irreducible over $\Q_p$ for some $p$ then the degree of the field $\Q_p(\alpha)/\Q_p$
    would be $16$; but we have just seen that all the roots of $f$ are contained in a field extension of $\Q_p$
    of degree at most $8$.  Thus, $f(x)$ is reducible in $\Q_p[x]$ for every $p$ but irreducible in $\Q[x]$.

\end{enumerate}

\end{document}
