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\hfill \textbf{January 29, 2015}

\vspace{0.5 cm} \centerline{
\textbf{\large{Homework 1\hspace{1 cm} Math
6303}}}

\vspace {1 cm}
\begin{enumerate}
\item Let $\mathcal{F}$ be the class of fields of characteristic zero. Prove that for $M\ne \emptyset$ a free field, freely generated by $M$, cannot exist. What happens if $M=\emptyset$.
\item
Let $p(x)$ be an irreducible polynomial in $\mbb{F}[x]$ where $\mbb{F}$ is a field.
\begin{enumerate}
\item
Show that $\mbb{F} \to \mbb{F}[x]/(p(x)), a \mapsto [a]$ is
an injective homomorphism (embedding) bet\-ween fields.
Thus, we can think of  $\mbb{F}[x]/(p(x))$ as an extension
$\mbb{K}$ of $\mbb{F}$.
\item
Show that $p(x)$ has a root in $\mbb{K}$.
\item
Show that for every polynomial  $p(x)$ of degree $n$ with coefficients of a field $\mbb{F}$
admits a field extension $\mbb{K}$ where $p(x)$ has $n$ roots.
\end{enumerate}
\item
Explain why $\mbb{R}[x]/(x^2+1)$ is isomorphic to the complex numbers.

\item
Let the  field  $\mbb{E}$ be an extension  of the field
$\mbb{F}$. Then $\mbb{E}$ can be perceived as  a vector space over
$\mbb{F}$. The dimension of $\mbb{E}$ over $\mbb{F}$ is called
the\textit{ degree}$[\mbb{E}:\mbb{F}]$ of the extension
$\mbb{E}|\mbb{F}$.  An extension $\mbb{E}$ over $\mbb{F}$ is called
\textit{finite} if the degree $[\mbb{E}:\mbb{F}]$ is finite. An
element $\alpha$ of $\mbb{E}$ is called \textit{algebraic} if it is
a root of a polynomial in $\mbb{F}[x]$. Otherwise it is called
\textit{transcendental}.



Let $M$ be a subset of $\mbb{E}$. Then $\mbb{F}(M)$ denotes  the
smallest subfield of $\mbb{E}$ which contains $M$ and $\mbb{F}$.



 An extension $\mbb{E}$ of $\mbb{F}$ is \textit{simple} if
$\mbb{E}=\mbb{F}(\alpha)$
\begin{enumerate}
\item
 Any $\alpha$ in $\mbb{E}$ gives rise to a homomorphism $\mbb{F}[x] \to \mbb{E}, f(x)\mapsto f(\alpha)$.
If $\alpha$ is algebraic then one has that $\mbb{F}(\alpha)\cong
\mbb{F}[x]/(p(x))$ where $p(x)$ is a uniquely determined  monic
irreducible polynomial. The  degree of $p(x)$ is the degree of the
extension. If $\alpha$ is transcendental then $\mbb{F}(\alpha)\cong
\mbb{F}[x]$
\item
Let  $\mbb{E}=\mbb{F}(\alpha)$ where $\alpha$ is algebraic.  Then
for every element $\beta$ there is a polynomial $r(x)$ of degree
less than $n$ where $n$ is the degree of $\mbb{E}$ over $\mbb{F}$
such that $\beta=r(\alpha)$.
\item
Assume that $\mbb{K}$ over $\mbb{E}$ and $\mbb{E}$  over $\mbb{F}$
are finite. Then $\mbb{K}$ over $\mbb{F}$ is finite and one has the
\textit{degree rule}:   \[ [\mbb{K}:\mbb{F}]=[\mbb{K}:\mbb{E}]\cdot
[\mbb{E}:\mbb{F}]
    \]
\item
Prove that the algebraic elements of $\mbb{E}$ over $\mbb{F}$ form a
subfield of $\mbb{E}$ which contains $\mbb{F}$.








\end{enumerate}
\end{enumerate}
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