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\begin{flashcard}[Frobenius-homomorphism-prop]
\begin{proposition}[Frobenius ring homomorphism]
  % Proposition 1.4
  \label{frob_hom_prop}
  \glsdefn{frob_hom}{Frobenius homomorphism}{N/A}
  \cloze{
  Let $R$ be a ring of characteristic $p$ ($p$ a
  prime). Then the \emph{Frobenius} $\phi : R \to R$, $x
  \mapsto x^p$ is a
  ring homomorphism.
  }
\end{proposition}

\begin{proof}
  \cloze{
  Clearly $\phi(1) = 1$ and $\phi(xy) = \phi(x)
  \phi(y)$. Also,
  \[ (X + y)^p = x^p + y^p + \sum_{r = 1}^{p - 1}
  {p \choose r} x^{p - r} y^r \]
  For $1 \le r \le p - 1$ the binomial coefficient
  ${p \choose r} = \frac{p!}{r!(p - r)!}$ is
  divisible by $p$ (since $p$ is prime). Therefore
  $\phi(x + y) = (x + y)^p = x^p + y^p = \phi(x) +
  \phi(y)$.
  }
\end{proof}
\end{flashcard}

\begin{remark*}
  We have $\phi(a) = a$ for all $a \in \FF_p
  \subset R$ (proof by induction on $a$), which
  implies $a^p \equiv a \pmod{p}$ for all integers
  $a$ (Fermat's Little
  Theorem).
\end{remark*}

\begin{flashcard}[tower-law-thm]
\begin{theorem}[Tower Law]
  \label{tower_law}
  \cloze{
  Let $M \extendover L$ and $L \extendover K$ be
  \glspl{field_ex}. Then $M \extendover K$ is
  \gls{finite_ex} if and
  only if $M \extendover L$ and $L \extendover K$
  are \gls{finite_ex}. In
  this case,
  \[ \degover{M}{K} = \degover{M}{L}\degover{L}{K} \]
  }
\end{theorem}

\begin{proof}
  \cloze{
  The forwards direction holds since any $K$-basis
  for $M$ spans $M$ as an $L$-vector space, and
  $L$ is a $K$-vector subspace of $M$.

  We now suppose that $M \extendover L$ and $L
  \extendover K$
  are \gls{finite_ex}, say $v_1, \ldots, v_n$ is a
  $K$-basis for $L$, $w_1, \ldots, w_m$ is a
  $L$-basis for $M$. We claim that $\{v_i
  w_j\}_{\substack{1 \le i \le n \\ 1 \le j \le
  m}}$ is a $K$-basis for $M$.
  \begin{enumerate}[(spanning)]
    \item[(spanning)] If $x \in M$ then $x =
      \sum_j \lambda_j w_j$ for some $\lambda_j
      \in L$ and $\lambda_j = \sum_i \mu_{ij} v_i$
      for some $\mu_{ij} \in K$. Therefore $x =
      \sum_{i, j} \mu_{ij} v_i w_j$.
    \item[(independent)] Suppose $\sum_{i, j}
      \mu_{ij} v_i w_i = 0$ for some $\mu_{ij} \in
      K$. Then
      \[ \sum_j \ub{\left( \sum_i \mu_{ij} v_i
      \right)}_{\in L} w_j = 0 \]
      $w_1, \ldots, w_m$ are linearly independent
      over $L$ so $\sum_i \mu_{ij} v_i = 0$ for
      all $j$. Also, $v_1, \ldots, v_n$ are
      linearly independent over $K$ so $\mu_{ij} =
      0$ for all $i, j$.
  \end{enumerate}
  This $K$-basis for $M$ then easily implies the
  desired result.
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[bracket-extension-defn]
\begin{definition}
  \glssymboldefn{field_adj}{smallest field
  containing}{$K(\alpha)$}
  \glssymboldefn{ring_adj}{smallest ring
  containing}{$K[\alpha]$}
  \label{bracket_extension_defn}
  Let $L \extendover K$ be a \gls{field_ex}. Let
  $\alpha_1, \ldots, \alpha_n \in L$.\cloze{
  \[ K[\alpha_1, \ldots, \alpha_n] = \{f(\alpha_1,
  \ldots, \alpha_n) \mid f \in K[X_1, \ldots,
  X_n]\} \]
  This is the smallest \fcemph{subring} of $L$ to contain
  $K$ and $\alpha_1, \ldots, \alpha_n$.
  \[ K(\alpha_1, \ldots, \alpha_n) = \left\{
  \frac{f(\alpha_1, \ldots, \alpha_n)}{g(\alpha_1,
  \ldots, \alpha_n)} : f, g \in K[X_1, \ldots, X_n],
  g(\alpha_1, \ldots, \alpha_n) \neq 0 \right\}  \]
  This is the smallest \fcemph{subfield} of $L$ to
  contain $K$ and $\alpha_1, \ldots, \alpha_n$.
  }
\end{definition}
\end{flashcard}

\begin{example*}
  $\fieldadj{\QQ}{\sqrt{2}} = \{a + b \sqrt{2} \mid a, b \in
  \QQ\} \subset \RR$ (use that $\frac{1}{a +
  b\sqrt{2}} = \frac{a - b\sqrt{2}}{a^2 - 2b^2}$).

  Note that $\fieldadj{\QQ}{\sqrt{2}} =
  \fieldadj{\QQ}{1 + \sqrt{2}} =
  \glsref[field_adj_notation]{\QQ\left( \frac{17}{3 - \sqrt{2}}
  \right)}$
  etc.
\end{example*}

\begin{remark*}
  In \cref{bracket_extension_defn}, another way to
  see the ``smallest'' subring / subfield exists
  is to take the intersection of all such subrings
  / subfields.
\end{remark*}
\vspace{-1em}

\textbf{Exercise:} Check that
\[ \fieldadj{K}{\alpha_1} =
\fieldadj{\fieldadj{K}{\alpha_1, \ldots, \alpha_{n -
1}}}{\alpha_n} = \fieldadj{\fieldadj{K}{\alpha_1}}{\alpha_2, \ldots,
\alpha_n} \]

\begin{flashcard}[simple-extension-defn]
\begin{definition*}[Simple extension]
  \glsdefn{simple_ex}{simple}{N/A}
  \cloze{
  A \gls{field_ex} $L \extendover K$ is a \emph{simple
  \gls{ex}} if $L = \fieldadj{K}{\alpha}$ for some $\alpha
  \in L$.
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[algebraic-element-and-min-poly-defn]
\begin{definition}[Minimal polynomial]
  \glsdefn{min_poly}{minimal polynomial}{minimal
  polynomials}
  \cloze{
  Let $L \extendover K$ be a \gls{field_ex} and $\alpha
  \in L$. Then there is a unique ring homomorphism
  $\phi : \ringadj{K}{X} \to L$ such that $\phi(c) = c$ for
  all $c \in K$ and $\phi(X) = \alpha$. Indeed,
  \[ \phi \left( \sum c_i X^i \right) = \sum c_i
  \alpha^i \]
  (i.e. $\phi$ is ``evaluation at $\alpha$''). Since
  $\ringadj{K}{X}$ is a PID, we have that $\ker(\phi) = (f)$
  for some $f \in \ringadj{K}{X}$.

  \glsdefn{alg_el}{algebraic}{N/A}
  We say $\alpha$ is \emph{algebraic} over $K$ if
  $f \neq 0$. In this case $f$ is irreducible and
  unique up to multiplication by elements of
  $\mult{K}$. We scale $f$ so that it is monic, and
  call it the \emph{minimal polynomial} of
  $\alpha$ over $K$. It is the monic polynomial in
  $\ringadj{K}{X}$ of least degree with $\alpha$ as a root.
  }
\end{definition}
\end{flashcard}
\vspace{-1em}

By the first isomorphism theorem for rings,
\[ \frac{\ringadj{K}{X}}{(f)} \cong
\ringadj{K}{\alpha} \]
Note that by \es[2(ii)]{1} we
have that the left side is a \gls{field}. Hence
$\fieldadj{K}{\alpha} = \ringadj{K}{\alpha}$. For this reason, we
usually write $\fieldadj{K}{\alpha}$ instead of
$\ringadj{K}{\alpha}$
in these cases.

Moreover, $\degover{K(\alpha)}{K} = \deg f$, since if
$\deg f = n$ then $1, \alpha, \alpha^2, \ldots,
\alpha^{n - 1}$ is a $K$-basis for $\fieldadj{K}{\alpha}$.

\begin{example*}
  $K = \QQ$, $L = \RR$, $\alpha = \sqrt[d]{2}$.
  $\alpha$ is a root of $f(X) = X^d - 2$, and $f$
  is irreducible in $\ringadj{\ZZ}{X}$ by Eisenstein's
  criterion (with $p = 2$). Therefore by Gauss'
  Lemma it is irreducible in $\ringadj{\QQ}{X}$, so $f$ is
  the \gls{min_poly} of $\alpha$. So
  $\degover{\fieldadj{\QQ}{\sqrt[d]{2}}}{\QQ} = d$.
\end{example*}

\begin{remark*}[A method for computing inverse]
  Let $\alpha \in L$ be \gls{alg_el} over $K$ with
  \gls{min_poly} $f$. Let $0 \neq \beta \in
  \ringadj{K}{\alpha}$, say $\beta = g(\alpha)$ for some $g
  \in \ringadj{K}{X}$. Since $f$ is irreducible and $\beta
  \neq 0$, we see that $f$ and $g$ are coprime.
  Running Euclid's algorithm gives $r, s \in
  \ringadj{K}{X}$
  such that $r(X)f(X) + s(X) g(X) = 1$. Plugging
  in $X = \alpha$, we get
  \[ r(\alpha)\ub{f(\alpha)}_{= 0} +
  s(\alpha) \ub{g(\alpha)}_{=\beta} = 1 ,\]
  therefore $\frac{1}{\beta} = s(\alpha)$.
\end{remark*}