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\noindent
\parbox{1.95in}{\bf Math 4010}
\hfill {\Large \bf Homework \#4} \hfill
\parbox{1.95in}{\bf \hfill Due: Wed., Sept. $24^\text{th}$, 2024}

\vspace{0.2in}

\begin{description}
\item[\#1 Congruently Yours:] Notice that $f(x)=x-1$, $g(x)=x^2+1$, and $h(x)=x^2-4$ are pairwise relatively prime in $\mathbb{Q}[x]$. Let $I=(f(x))$, $J=(g(x))$, and $K=(h(x))$.

\begin{enumerate}[(a)]
\item Find the set of all solutions $m(x) \in \mathbb{Q}[x]$ such that $m(x)+I=4+I$, $m(x)+J=2x+J$. 

I suggest you use my SAGE interact to do some Euclidean algorithm stuff.

\item {}{\bf [Grad.]} Now solve $m(x)+I=4+I$, $m(x)+J=2x+J$, and $m(x)+K=3x+4+K$.
 \end{enumerate}
  
{\it How?} Use Chinese remaindering to solve $m(x)+I=1+I$ and $m(x)+J=2x+J$. You will get some solution set $\ell(x)+IJ$. Then for the grad part solve $m(x)+IJ=\ell(x)+IJ$ and $m(x)+K=3x+4+K$. The undergrad answer isn't too bad. The grad answer is kind of ugly -- use technology!

\item[\#2 Moving up a Notch:] Consider the field $\mathbb{F}=\mathbb{C}(x)$ (i.e., ratio of polynomials with complex coefficients). This is the field of fractions of $R=\mathbb{C}[x]$.

I want to prove that the polynomial $f(Y) = Y^2-x \in \mathbb{F}[Y]$ is irreducible. Let's do this in a less than efficient manner:
\begin{enumerate}[(a)]
\item Explain why it is enough to show $f(Y)$ is irreducible in $R[Y]$.
\item Explain why $x$ is an irreducible element of $R$ (use a quotient ring argument). 
\item Suppose $f(Y)$ is not irreducible in $R[Y]$. Then $f(Y)=(Y-a)(Y-b)$ for some $a,b \in R$. Why is this impossible?
\item Now we know that $f(Y)$ is irreducible in $\mathbb{C}(x)[Y]$. Thus $\mathbb{E} = \mathbb{C}(x)[Y]/(f(Y))$ is an extension field of $\mathbb{F}=\mathbb{C}(x)$. Essentially, we have tacked some new function onto the field of (complex coefficient) rational functions. What should we call it (i.e., secretly what is $Y$)?
\end{enumerate}

\item[\#3 Irreducibly Annoying:] We can detect irreducibility of low degree polynomials (i.e., degree 2 and 3) by seeing if they have any roots. Once we move up to fourth degree polynomials, things get more complicated -- a fourth degree polynomial can be the product of two irreducible quadratics and thus have no roots!

\begin{enumerate}[(a)]
\item Generate a list of all irreducible quadratic and cubic polynomials in $\mathbb{Z}_2[x]$. 
\item There are 16 quartic (fourth degree) polynomials in $\mathbb{Z}_2[x]$. Find the irreducibles. 

{\it Note:} By plugging in 0 and 1, you can easily eliminate those with roots. Any other reducibles must be products of two irreducible quadratics -- which you already know from part (a).

\item Find the monic irreducible quadratic polynomials in $\mathbb{Z}_3[x]$.

\item Give an example of a monic irreducible cubic in $\mathbb{Z}_3[x]$. 

``Thank you Dr. Cook for not asking us to tediously find all of these too.''
\end{enumerate}

\item[\#4 A Field with Nine:] Build the field of order 9, $\mathbb{F}_9$ -- do this by constructing a quotient of $\mathbb{Z}_3[x]$. Next, let $\alpha$ be the coset represented by $x$. List all nine (distinct) elements and fill out a table as follows:

\begin{tabular}{|r||c|c|c|} \hline
element $z=$ & 0 & 1 & $\cdots$ \Large\strut \\ \hline
additive inverse $-z=$ & 0 & 2 & $\cdots$ \Large\strut \\ \hline
additive order $|z|=$ & 1 & 3 & $\cdots$  \Large\strut \\ \hline
mutiplicative inverse $z^{-1}=$ & DNE & 1 & $\cdots$ \Large\strut \\ \hline
multiplicative order $|z|=$ & DNE & 1 & $\cdots$  \Large\strut \\ \hline
\end{tabular}

{\it Note:} DNE = does not exist.


\end{description}

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