1_30_17.tex

\section{January 30, 2017}
\begin{prop} \label{Prop 1, Jan 30}
    Let $L / K$ be a field extension, $a \in L$ algebraic over $K$. Then,
    \begin{itemize}
        \item $M_{a,K} \in k[T]$ is irreducible
        \item $K[T] / (M_{a/K})$ is isomorphic to $K(a) \subset L$
        \item The images of $1, T, \dots T^{n-1}$ in $K[T] / (M_{a/K})$ are a
        basis of this $K$-vector space.
        \item $[K(a) : K] = \deg(a/K) = n$
    \end{itemize}
\end{prop}
\begin{proof}
    By definition of $M_{a/K}$ the evaluation $ev: k[T] \rightarrow L$ gives an
    injection from $K[T] / (M_{a/K}) \rightarrow K(a)$. Since $K(a)$ is a domain,
    so is $K[T] / (M_{a/K})$. By \ref{Thm 8, Jan 6}, this means $(M_{a/K})$ is prime.
    this means that $M_{a/K}$ is prime, so $M_{a/K}$ is irreducible by \ref{Fact 8, Jan 18} \\
    Now we show the images of $1, \dots T^{n-1}$ in $K[T] / (M_{a/K})$ generate the space.
    (Pranav Exclusive) It is enough to show that they generate $T^n$, as for any higher
    dimensional coset, by division algorithm we can reduce to a coset of $(M_{a/K})$ with
    degree less than $M(a/K)$, which is then definitionally generated by $1, \dots T^{n-1}$.
    If $M_{a,K}(T) = T^n + a_{n-1}T^{n-1} + \dots a_0$, Then $T^n = -a_{n-1}T^{n-1} \dots -a_0$
    They are also linearly independent. If not, let $b_{n-1}T^{n-1} + \dots b_0 = 0$
    be a non-trivial relation. Then $Q(T) = b_{n-1}T^{n-1} + \dots b_0 \in K[T]$ lies
    in $(M_{a,k})$ as it is in the kernel of the quotient map. We thus have that
    $M_{a,k} \mid Q$. This contradicts that $\deg(Q) < \deg(M_{a,k})$ \\
    Now we know that $K[T] / (M_{a,K})$ is a domain, containing $K$ and is finite
    dimensional over $K$, so by \ref{Prop 4, Jan 27}, $K[T] / (M_{a,k})$ is a field.
    This implies that $K[T] / (M_{a/K}) \rightarrow K(a)$ is surjective as the image
    in $K(a)$ is now a field that contains $K$ and $a$ and because $K(a)$ is the smallest
    such field.
\end{proof}

\begin{prop}\label{Prop 2, Jan 30}
    Let $L / K$ be an extension, $a \in L$. The following are equivalent.
    \begin{enumerate}[i)]
        \item $a$ is algebraic over $K$
        \item $[K(a) : K] < \infty$
        \item There is an intermediate field $L/E/K$, finite over $K$, containing $a$
    \end{enumerate}
\end{prop}
\begin{proof}
    i) $\Rightarrow$ ii): This comes from the above proposition (\ref{Prop 1, Jan 30}), final condition. \\
    ii) $\Rightarrow$ iii): Just let $E$ be $K(a)$. \\
    iii) $\Rightarrow$ i): Since $E / K$ is finite, the set of powers $1,a,a^2 \dots$ cannot
    be linearly independent over $K$. Thus, there are $n \in \NN$ and $b_0, b_{1}, \dots b_{n} \in K$
    such that $b_na^n + \dots b_0 = 0$ Put $P(T) = b_nT^n + \dots b_0 \in K[T]$. Then $P(a) = 0$,
    so $a$ is algebraic over $K$, as desired.
\end{proof}

\begin{cor} \label{Cor 3, Jan 30}
    Let $L / K$ be an extension, $a,b \in L$ algebraic over $K$. Then
    \begin{itemize}
        \item $a + b$, $a \cdot b$ are algebraic over $K$. $\deg(a + b), (a \cdot b) \leq \deg(a)
        \cdot \deg(b)$
        \item $a \neq 0$ $a^{-1}$ is algebraic over $K$ and $\deg(a^{-1}) = \deg(a)$
    \end{itemize}
\end{cor}

\begin{proof}
    By above proposition (\ref{Prop 2, Jan 30}), $K(a)$ and $K(b)$ are finite over $K$. \\
    Now, $a + b, a \cdot b \in K(a)\cdot K(b)$ and $[K(a) \cot K(b) : K] \leq [K(a) : K]
    \cdot [K(b): K]$. \\
    By \ref{Prop 6, Jan 27} $a + b$, and $a \cdot b$ are algebraic over $K$. Note that $a^{-1} \in K(a)$
    so $K(a^{-1}) = K(a)$
\end{proof}
\begin{defn} \label{Defn 5, Jan 30}
    An extension $L / K$ is called \textbf{algebraic}, if all $a \in L$ are algebraic
    over $K$.
\end{defn}
\begin{cor} \label{Cor 6, Jan 30}
    If $L / K$ is finite, then it is algebraic, and $\deg(a,K) \mid [L:K]$, for
    any $a \in L$.
\end{cor}
\begin{proof}
    \ref{Prop 2, Jan 30} and \ref{Cor 8, Jan 27}
\end{proof}

\begin{rmk}
    $a \in L$ is algebraic over $K \iff K(a) / K$ is algebraic.
\end{rmk}
\begin{rmk}
    There exist infinite algebraic extensions.
\end{rmk}