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\noindent Math 463/563 - Homework 4\bigskip
{\bf 6.2} Define $\phi\colon\Z/6\Z\to\Z/2\Z\times\Z/3\Z$ by $\phi(1)=(1,1)$. It's straightforward to check that $\phi$ is injective and surjective, so it's an isomorphism. However $\Z/8\Z$ is not isomorphic to $\Z/2\Z\times\Z/4\Z$ since the former has an element of order 8 while the latter does not.\bigskip
{\bf 6.4} (a) The elements consists of $0,1,\alpha,\alpha+1$. Note that $\alpha(\alpha+1)=1$ so every nonzero element is invertible. Hence this ring is a field with four elements, $\F_4$. \bigskip
(b) Note $(\alpha+1)^2=\alpha^2+1=0$. So letting $x=\alpha+1$, we get that the ring is isomorphic to $\F_2[x]/(x^2)$. \bigskip
(c) Since $\alpha$ and $\alpha+1$ are both idempotents, we have $\F_2[\alpha]\cong\alpha\F_2[\alpha]\times(\alpha+1)\F_2[\alpha]$. Listing out the elements gives $\alpha\F_2[\alpha]\cong\F_2$ and $(\alpha+1)\F_2[\alpha]\cong\F_2$. Hence $\F_2[\alpha]\cong \F_2\times\F_2$.\bigskip
{\bf 6.8} (a) We showed in 3.13 that $IJ\subseteq I\cap J$. To show $I\cap J\subseteq IJ$, suppose $x\in I\cap J$. Write $x=a+b$ where $a\in I$ and $b\in J$. Since $x,b\in J$, we have $a\in J$ also, and similarly $b\in I$. Let $1=i+j$ for some $i\in I,j\in J$. Then $x=i(a+b)+(a+b)j\in IJ$.\bigskip
(b) Let $a=a_I+a_J$ and $b=b_I+b_J$ where $a_I,b_I\in I$ and $a_J,b_J\in J$. Let $x=a_J+b_I$. It's straightforward to check that $x-a\in I$ and $x-b\in J$. \bigskip
(c) Define $\varphi\colon R\to R/I\times R/J$ by the usual quotient maps. The kernel is $I\cap J=IJ=0$ by hypothesis and (b) shows that $\varphi$ is surjective.\bigskip
(d) Let $i\in I,j\in J$ be the unique elements such that $i+j=1$. The idempotents are $i$ and $j$.\bigskip
{\bf 7.1} Suppose $R$ is an integral domain of order $n$. Let $x\in R$ be a nonzero element and consider the sequence $x,x^2,x^3,\ldots$. Since $R$ is finite, there is some distinct positive integers $i\langle j$ such that $x^i=x^j$. Hence $x^i(x^{j-i}-1)=0$. Since $R$ is a domain and $x$ is nonzero, we must have $x^{j-i}=1$. Thus $x$ is invertible and so $R$ is a field.\bigskip
{\bf 7.5} Addition: If $a/b,c/d\in S^{-1}R$ define $a/b+c/d=(ad+bc)/bd$ which is an element of $S^{-1}R$ since $S$ is multiplicatively closed. To show it's well defined, suppose $a'/b'=a/b$ and $c/d=c'/d'$. Then $a'/b'+c'/d'=(a'd'+b'c')/b'd'$. But then $0=(ab'-a'b)dd'+(cd'-c'd)bb'=(ad+bc)(b'd')-(a'd'+b'c')(bd)$ implies $(ad+bc)/bd=(a'd'+b'c')/b'd'$. Commutativity and associativity follow from that of $R$. Identity is $0/a$ for any $a\in S$. Inverse of $a/b$ is $(-a)/b$.
Multiplication: Define $a/b\cdot c/d=ac/bd$. To show this is well defined, again suppose $a'/b'=a/b,c'/d'=c/d$. Then $ac/bd=a'c'/b'd'$ since $acb'd'-a'c'bd=ab'cd'-a'bc'd=0$. It is associative since $R$ is and $a/a$ for any $a\in S$ is the multiplicative identity.
Distributive law: Follows directly from the distributive law of $R$.\bigskip
{\bf 8.1} Let $f\in\Z[x]$ and $(f)$ be a principal ideal. Suppose $\deg f=0$, so $f\in\Z$. Then $(f,x)$ is a proper ideal which properly contains $(f)$. Hence $(f)$ is not maximal. Suppose $\deg f\geq 1$. Let $p\in\Z$ be a prime which does not divide any of the coefficients of $f$. Clearly $(f)\neq (f,p)$ since $p\notin (f)$. Moreover $(f,p)\neq\Z[x]$ since $\Z[x]/(f,p)\cong \F_p[x]/(f)$ which is a vector space over $\F_p$ of dimension $\deg f$. Hence $(f)$ is not maximal. There are no maximal principal ideals.\bigskip
{\bf 8.3} Since $\F_2$ is a field, it suffices to check that $x^3+x+1$ is irreducible. This can be done by listing out all degree 1 and 2 polynomials and observing that none of them multiply out to equal $x^3+x+1$. On the otherhand, over $\F_3$, it factors as $x^3+x+1=(x+2)(x^2+x+2)$. Hence $\F_3[x]/(x^3+x+1)$ is not a field, e.g., $x+2$ is not invertible.\bigskip
{\bf 8.4} Maximal ideals in $\mathbb{R}[x]$ correspond to irreducible polynomials. Irreducible polynomials over $\mathbb{R}$ are either linear or quadratic with two complex conjugate roots. If we consier the upper half plane as $\C$ minus complex numbers with negative imaginary component, then we get a bijection as follows: Send any linear polynomial $x-a$ to $a$ lying on the real axis. Send any quadratic polynomial to the root with positive imaginary component. It's clear this gives a bijection between maximal ideals and the upper half plan.\bigskip
{\bf Part B:} 563 Students Only\bigskip
{\bf 2.2} Since $s$ sends each element $x\in X$ to $sx$, $\rho_s$ sends $e_x$ to $e_{sx}$ for basis elements $(e_x)$ in $V$. Thus $\operatorname{Tr}(\rho_s)=\#\{e_x\mid e_x=e_{sx}\}=\#\{x\mid x=sx\}$. So $\chi_{X(s)}$ is the number of elements in $X$ fixed by $s$.\bigskip
{\bf 2.3} Existence: Set $\rho_s'=(\rho_s^t)^{-1}$. It suffices to show that $\langle \rho_s e_j,\rho'_se_i\rangle =\langle e_j,e_i\rangle $ for basis elements $e_j,e_i'$. Let $\rho_s=r_{ij}(s)$ and $\rho_s'=r_{ij}'(s)$. Then
\begin{align*}
\langle \rho_s e_j,\rho'_se_i'\rangle &=\sum_{l,k}r_{lj}(s)r_{ki}'(s)\langle e_l,e_k'\rangle \\
&=\sum_{l,k}r_{lj}(s)r_{ki}'(s)\delta_{kl}\\
&=\sum_k r_{kj}(s)r_{ki}'(s)\\
&=\sum_k r_{jk}^t(s)r_{kj}'(s)\\
&=(\rho^t_s\rho'_s)_{ji}=\delta_{ij}.
\end{align*}
Uniqueness: The previous calculation shows that $\rho'$ must satisfy $(\rho^t_s\rho'_s)= \operatorname{Id}$ so there is only one solution.
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