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\begin{document}
\thispagestyle{empty}
\noindent \huge{Team Number:} \TeamNumber
\vspace{.5cm}

\noindent \huge{PUMaC 2022* Power Round Cover Sheet}

\vspace{.5cm}
\normalsize
\begin{paragraph}{Notes:}
Remember that this sheet comes first in your solutions. You should submit solutions for the problems in increasing order. Write on one side of the page only. The start of a solution to a problem should start on a new page. Please mark which questions for which you submitted a solution to help us keep track of your solutions.
\end{paragraph}
\begin{center}
\begin{tabular}{|c|c|c|}\hline
Problem Number & Points & Attempted?\\\hline
1.1.1 & 15 & Yes \\\hline
1.1.2 & 5 & Yes \\\hline
1.1.3 & 5 & Yes \\\hline
1.1.4 & 5 & Yes \\\hline
1.1.5 & 10 & Yes \\\hline
1.1.6 & 5 & Yes \\\hline
1.2.1 & 10 & Yes \\\hline
1.2.2 & 5 &  \\\hline
1.2.3 & 10 & Yes \\\hline
1.2.4 & 10 & \\\hline
1.2.5 & 10 & \\\hline
1.3.1 & 10 & Yes \\\hline
1.3.2 & 10 & Yes \\\hline
1.3.3 & 25 & Yes \\\hline
1.3.4 & 10 & Yes \\\hline
2.1.1 & 10 & Yes \\\hline
2.1.2 & 10 & Yes \\\hline
2.2.1 & 10 & Yes \\\hline
2.2.2 & 20 & Yes \\\hline
2.2.3 & 10 & Yes \\\hline
2.2.4 & 10 & Yes\\\hline
2.2.5 & 20 & Yes\\\hline
2.2.6 & 5 & \\\hline
2.2.7 & 10 & Yes\\\hline
2.3.1 & 15 & Yes\\\hline
2.3.2 & 5 & Yes\\\hline
2.3.3 & 15 & \\\hline
2.3.4 & 10 & \\\hline
2.3.5 & 15 & Yes\\\hline
2.3.6 & 10 & \\\hline
2.4.1 & 15 & Yes\\\hline
2.4.2 & 10 & \\\hline
2.4.3 & 15 & \\\hline
\end{tabular}
\hspace{0.5 cm}
\begin{tabular}{|c|c|c|}\hline
Problem Number & Points & Attempted?\\\hline
2.4.4 & 30 & \\\hline
2.4.5 & 10 & Yes\\\hline
2.4.6 & 10 & Yes\\\hline
2.4.7 & 25 & Yes\\\hline
2.4.8 & 20 & \\\hline
3.0.1 & 5 & Yes \\\hline
3.0.2 & 15 & Yes \\\hline
3.0.3 & 10 & Yes \\\hline
3.0.4 & 20 & Yes \\\hline
4.1.1 & 30 & Yes \\\hline
4.1.2 & 25 & Yes \\\hline
4.1.3 & 15 & Yes \\\hline
4.1.4 & 25 & Yes \\\hline
4.1.5 & 40 & Yes \\\hline
4.2.1 & 10 & Yes \\\hline
4.2.2 & 15 & Yes \\\hline
4.2.3 & 20 & Yes \\\hline
4.2.4 & 25 & Yes \\\hline
4.2.5 & 30 & \\\hline
4.3.1 & 10 & \\\hline
4.3.2 & 5 & \\\hline
4.3.3 & 40 & \\\hline
4.3.4 & 30 & \\\hline
5.1.1 & 25 & \\\hline
5.2.1 & 25 & \\\hline
5.2.2 & 10 & \\\hline
5.2.3 & 15 & \\\hline
5.2.4 & 20 & \\\hline
5.2.5 & 15 & \\\hline
5.3.1 & 15 & \\\hline
5.3.2 & 10 & \\\hline
5.3.3 & 30 & \\\hline
5.3.4 & 10 & \\\hline
\end{tabular}
\end{center}
\newpage 
\small
\tableofcontents
\normalsize
\newpage

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% SOLUTIONS START HERE
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Problem 1.1.1}

\begin{proof}

\begin{enumerate}
    \item 
    In $2\Z$, there does not exist multiplicative identity 1.
    \item 
    We can show that $\C[x]$ has all the ring's properties:
    \begin{itemize}
        \item $\C[x]$ is closed under addition and multiplication
        \item \\Addition in $\C[x]$ is adding the coefficients together, which does not change the degree. Suppose $i_1, i_2 \in \C[x]$ such that $i_1=a_1+b_1x$ and $i_2=a_2+b_2x$. We have $i_1+i_2=(a_1+a_2)+(b_1+b_2)x \in \C[x]$, and $i_1i_2=(a_1+b_1x)(a_2+b_2x)=(a_1a_2+x^2b_1b_2)+(a_1b_2+a_2b_1)x$. Both the sum and multiplication of $i_1, i_2$ lie in $\C[x]$, so $\C[x]$ is closed under addition and multiplication.
        \item For each element $i$ in $\C[x]$, we can always find an element $i'$ such that $i+(i')=0$.


        
        \item 
    \end{itemize}
    
\end{enumerate}

 


\end{proof}

\newpage 

\section{Problem 1.1.2}

\begin{proof}
We can show that $\Z/n\Z$ is a ring because it has all the properties.
\begin{itemize}
    \item $\Z/n\Z$ is closed under addition and multiplication
    \begin{itemize}
        \item Based on the definition of $\Z/n\Z$, $\Z/n\Z$ only has \{0,1,\dots,n-1\} because they are the remainders after divided by n. \forall a,b $\in$ $\Z$, $a+b$ and ab must \in 
        $\{0,1,\dots,n-1\}$ $after divided by n, which they are both in$ $\Z/n\Z$.  $Therefore, $\Z/n\Z$ is closed under addition and multiplication.$           
    \end{itemize}
\item Addition and multiplication are associative in $\Z/n\Z$ 
    \begin{itemize}
        \item $Addition$: \forall a,b \in \mathbb{R}, a+b=b+a. $Therefore, a+b=b+a in $\mathbb{Z}/n\mathbb{Z}$ because they have the same remainder divided by n.$
        \\
        \item Multiplication: \forall a,b \in \mathbb{R}, ab=ba. $Therefore, ab=ba in $\mathbb{Z}/n\mathbb{Z}$ because they have the same remainder divided by n.$
    \item The same reason applies to the commutative property.
    \item 
    \end{itemize}
\end{itemize}
\end{proof}

\newpage 

\section{Problem 1.1.3}

\begin{proof}
	Because $R$ is the ring, it has the multiplicative identity 1. Because 1 is in the ring, the additive inverse -1 must be in the ring. The element $x \in \mathbb{R}$ such that $r+xr=1\cdot r+x \cdot r = (1+x)r = 0$ is the additive inverse of the multiplicative identity = -1.
\end{proof}

\newpage 

\section{Problem 1.1.4}

\begin{proof}
	Suppose that the set of odd integers $I$, as a subset of $\mathbb{Z}$, is an ideal. We know $1 \in I$ and $2 \in \mathbb{Z}$. However, $1 \cdot 2 \notin I$. Therefore, it does not hold the second property of Definition 1.1.3 and is not an ideal.
\end{proof}

\newpage 

\section{Problem 1.1.5}

\begin{proof}
	All the prime ideals are $\langle x-a \rangle, a \in \mathbb{C}$
 \\ Since, according to the theorem that any nonconstant polynomials in $\mathbb{C}$ can be written as a product of linear factors, $ab \in \langle x-a \rangle$ if and only if $a$ or $b$ is divisible by $\langle x-a \rangle$
\end{proof}

\newpage 

\section{Problem 1.1.6}

\begin{proof}
	$1.)$ $S=\{x\in\mathbb{R}|x<0\}$ and $S'=\{x\in\mathbb{R}|x>0\}$
    \\Suppose $f(s)=f(t)$, we have $\mid s \mid= \mid t \mid$, $s<0$, and $t<0$. So, $s=t$
    \\Therefore, the function is $\textbf{injective}$.
    \\For all $a \in \mathbb{R^+}$, There is $-a \in \mathbb{R^-}$ such that $f(-a)=a$
    \\Therefore, the function is $\textbf{surjective}$.
    \\ \\ 2.)
    $f(x)=e^x$ never reach the negative value, therefore, the function is $\textbf{not surjective}$.
    \\Suppose $f(s)=f(t)$, we have $e^s=e^t$, $e^{s-t}=1$, $s-t=0$, $s=t$.
    \\Therefore, the function is $\textbf{injective}$.
    \\ \\ 3.)
    For $f(x)=sin(x)=0$, there are more than solutions in $[0,2\pi]$, which is $0, \pi, 2\pi$.
    \\Therefore, the function is $\textbf{not injective}$.
    \\Because the range of $sin(x)$ for $x\in[0,2 \pi]$ is $[-1,1]$,  the function is $\textbf{surjective}$.
\end{proof}

\newpage 

\section{Problem 1.2.1}

\begin{proof}
	As the example in page 7 already shown that $r=a+b \sqrt{2}$ is a ring. We will further show that it is also a field. For $r=a+b \sqrt{2} \in R$, where $a,b \in \mathbb{Q}$, $\dfrac{1}{r}=\dfrac{a-b \sqrt{2}}{a^2-2b^2}=\dfrac{a}{a^2-2b^2}+\dfrac{b }{2b^2-a^2}\sqrt{2}\in R$
 \\ Therefore, $r=a+b \sqrt{2}$ is a field.
\end{proof}

\newpage 

\section{Problem 1.2.2}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 1.2.3}

\begin{proof}
	consider the smallest positive element in I, say i. Assume that there is an element j that is not divisible by i. We can write $j = ai+b$ when $0<b<i$. Because $-1 \in \mathbb{Z}$, we also have that $-i$ is in $I$. 
 \\ If $a>0$, We have $j+(-i) \in I, ... ,j+a(-i)=b \in I$
 \\ If $a<0$, We have $j+i \in I, ... ,j+(-a)i=b \in I$
 \\ If $a=0$, We have $b \in I$
 \\which contradicts the assumption that i is the smallest positive element in I
 \\ Therefore, there does not exist j that is not divisible by i.
 \\ Because of this, every element in I can be generated by i (the smallest positive element in I).
 \\ $\mathbb{Z}$ is a PID.
\end{proof}

\newpage 

\section{Problem 1.2.4}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 1.2.5}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 1.3.1}

\begin{proof}
If $r_1 + I = r'_1 + I$, $r_2 + I = r'_2 + I$, then $r'_1 = r_1 + i'_1$, $r'_2 = r_2 + i'_2$ for some $i'_1, i'_2 \in I$. Suppose $r_1 + r_2 + i \in (r_1+I) + (r_2 + I) = (r_1+r_2) + I$, where $i \in I$. Then $r_1 + r_2 + i = (r'_1 + r'_2) + (i-i'_1-i'_2)) \in (r'_1+r'_2)+I=(r'_1 + I) + (r'_2 +I)$. Hence, $(r_1+I) + (r_2 + I) \subseteq (r'_1+I) + (r'_2 + I)$. Similarly, $(r'_1+I) + (r'_2 + I)\subseteq (r_1+I) + (r_2 + I)$. Hence, $(r_1+I) + (r_2 + I) = (r'_1+I) + (r'_2 + I)$. \\

Moreover, if $r_1r_2 + i \in (r_1+I) + (r_2 + I) = (r_1r_2) + I$, then $r_1r_2 + i = (r'_1-i'_1)(r'_2-i'_2)+i = r'_1r'_2 +(i - i'_1r'_2 - i'_2(r'_1 + i'_1)) \in (r'_1r'_2)+I=(r'_1 + I) \cdot (r'_2 +I)$. Hence $(r_1+I) \cdot (r_2 + I) \subseteq (r'_1+I) \cdot (r'_2 + I)$. Similarly, $(r'_1+I) \cdot (r'_2 + I)\subseteq (r_1+I) \cdot (r_2 + I)$. Hence, $(r_1+I) \cdot (r_2 + I) = (r'_1+I) \cdot (r'_2 + I)$. 
\end{proof}

\newpage 

\section{Problem 1.3.2}

\begin{proof}
	If $r_1+I, r_2+I \in R/I$, then $(r_1+I)+(r_2+I) = (r_1+r_2)+I$, $(r_1+I)(r_2+I) = (r_1r_2)+I$. Hence, all the properties of addition and multiplication of $R$ carries down over to $R/I$. Define $0$ in $R/I$ to be $0 + I = I$ and $1$ to be $1 + I$. Then $R/I$ is a ring.
\end{proof}

\newpage 

\section{Problem 1.3.3}

\begin{proof}
\begin{enumerate}
    \item If $i, i' \in I_1 \cap I_2$, then $i+i'\in I_1$, $i+i' \in I_2$, so $i+i' \in I_1 \cap I_2$. If $i \in I_1 \cap I_2$, $r \in R$, then $ri \in I_1$, $ri \in I_2$, so $ri \in I_1 \cap I_2$. Hence, $I_1 \cap I_2$ is an ideal.
    \item Let $\phi: R/(I_1 \cap I_2) \longrightarrow (R/I_1) \times (R/I_2)$, $r+ I_1 \cap I_2 \rightsquigarrow (r + I_1, r+ I_2)$. Let $(r + I_1, r + I_2) = (r' + I_1, r' + I_2)$ if and only if $r+I_1 = r'+I_1$ and $r + I_2 = r' + I_2$ if $r + I_1 \cap I_2 = r' + I_1 \cap I_2$. Hence, $\phi$ is well-defined. It is a homomorphism since $\phi((r_1+ I_1 \cap I_2) + (r_2 +I_1 \cap I_2))=\phi(r_1+I_1 \cap I_2) + \phi(r_2+I_1 \cap I_2)$ and $\phi(r(r_1+I_1 \cap I_2))=r\phi(r_1+I_1 \cap I_2)$.
    \item $(r + I_1, r + I_2)=0$ if and only if $r \in I_1 \cap I_2$, which implies that $r+I_1\cap I_2 = I_1 \cap I_2$. Therefore, the kernal of $\phi$ contains only the zero element, so $\phi$ is injective.
    \item Since $I_1 + I_2 = R$, $1 = i_1 + i_2$ for some $i_1 \in I_1, i_2 \in I_2$. $(i_1+i_2)r_1=r_1=i_1r_1+i_2r_1$, but since $i_1r_1 \in I_1$, we also have $i_2r_1 \in r_1 + I_1$. Similarly, $i_1r_2 \in r_2 + I_2$. Consider $r = i_2r_1 + i_1r_2$. Since $i_1r_2\in I_1$, $i_2r_1 \in r_1 + I_1$, we have $r \in r_1 + I_1$. Similarly, $r \in r_2 + I_2$. Therefore, $\phi(r+ (I_1 \cap I_2)) = (r+I_1, r+I_2) = (r_1 + I_1, r_2 + I_2)$. This shows that $\phi$ is surjective.
\end{enumerate}
	
\end{proof}

\newpage 

\section{Problem 1.3.4}

\begin{proof}
	Let $I_1 = \langle m \rangle$, $I_2 = \langle n \rangle$. If $m, n$ are relatively prime, then $I_1 + I_2 = \Z$ and $I_1 \cap I_2 = \langle mn \rangle$. By the previous problem, there is a bijection $\phi$ from $\Z/(I_1 \cap I_2)$ to $(R/I_1) \times (R/I_2)$ that takes $r + I_1 \cap I_2$ to $(r+I_1, r+I_2)$. Hence, there is a unique $r$ such that $\phi(r + I_1 \cap I_2) = (r_1 + I_1, r_2 + I_2)$. This $r$ is the unique residue (mod $mn$), as desired.
\end{proof}

\newpage 

\section{Problem 2.1.1}

\begin{proof}
	1.)
 \\ Let $v_1$, $v_2$, and $v_3$ be polynomials with complex coefficients and $V$ is the set of polynomials with complex coefficients.
 \\ Let $v_1 = a_n \cdot x^n + ... + a_2 \cdot x^2 + a_1 \cdot x + a_0$, $v_2 = b_n \cdot x^n + ... + b_2 \cdot x^2 + b_1 \cdot x + b_0$, and $v_3 = c_n \cdot x^n + ... + c_2 \cdot x^2 + c_1 \cdot x + c_0$.
 \\
 \\$\textbf {property 1: }$
 \\$v_1 + v_2 = (a_n + b_n) \cdot x^n + ... +(a_2 + b_2) \cdot x^2 +(a_1 + b_1) \cdot x + (a_0 + b_0) $
 \\ For $s \in \mathbb{C}$, $s \cdot v_1 = (s \cdot a_n) \cdot x^n + ... + (s \cdot a_2) \cdot x^2 + (s \cdot a_1) \cdot x + (s \cdot a_0)$
 \\Because $\mathbb{C}$ is closed under addition and scalar multiplication, $v_1+v_2 \in V$ and $s \cdot v_1 \in V$.
 \\
 \\$\textbf {property 2: }$
 $(v_1+v_2)+v_3 = v_1+(v_2+v_3) = (a_n+b_n+c_n) \cdot x^n + ... +(a_2+b_2+c_3) \cdot x^2+(a_1+b_1+c_1) \cdot x + (a_0+b_0+c_0) $
 \\ $s_1\cdot(s_2\cdot v_1)=(s_1 \cdot s_2)\cdot v_1=s_1 \cdot s_2 \cdot a_n \cdot x^n+...+s_1 \cdot s_2 \cdot a_2 \cdot x^2 + s_1 \cdot s_2 \cdot a_1 \cdot x + s_1 \cdot s_2 \cdot a_3$
 \\
 \\$\textbf {property 3: }$
 $v_1+v_2 = v_2+v_1$ because the coefficients are in the set of complex numbers which have commutative property.
 \\
 \\$\textbf {property 4: }$
 $0$ is the identity element for $+$
 \\
 \\$\textbf {property 5: }$
 The additive inverse is the polynomial with all the coefficients being additive inverses of those of initial polynomial.
 \\
 \\$\textbf {property 6: }$
 $1$ is the identity element for $\cdot$
 \\
 \\$\textbf {property 7: }$
 Because the coefficients are in $\mathbb{C}$, the polynomials also follow the distributive law.
 \\
 \\2.)
 \\$\textbf {property 1: }$
 \\$v_1 + v_2 = (a_n + b_n) \cdot x^n + ... +(a_2 + b_2) \cdot x^2 +(a_1 + b_1) \cdot x + (a_0 + b_0) $
 \\ For $s \in \mathbb{Q}$, $s \cdot v_1 = (s \cdot a_n) \cdot x^n + ... + (s \cdot a_2) \cdot x^2 + (s \cdot a_1) \cdot x + (s \cdot a_0)$
 \\Because $\mathbb{R} \cdot \mathbb{Q} \in \mathbb{R}$ and $\mathbb{R}$ is closed under addition, $v_1+v_2 \in V$ and $s \cdot v_1 \in V$.
 \\
 \\$\textbf {property 2: }$
 $(v_1+v_2)+v_3 = v_1+(v_2+v_3) = (a_n+b_n+c_n) \cdot x^n + ... +(a_2+b_2+c_3) \cdot x^2+(a_1+b_1+c_1) \cdot x + (a_0+b_0+c_0) $
 \\ $s_1\cdot(s_2\cdot v_1)=(s_1 \cdot s_2)\cdot v_1=s_1 \cdot s_2 \cdot a_n \cdot x^n+...+s_1 \cdot s_2 \cdot a_2 \cdot x^2 + s_1 \cdot s_2 \cdot a_1 \cdot x + s_1 \cdot s_2 \cdot a_3$
 \\
 \\$\textbf {property 3: }$
 $v_1+v_2 = v_2+v_1$ because the coefficients are in the set of real numbers which have commutative property.
 \\
 \\$\textbf {property 4: }$
 $0$ is the identity element for $+$
 \\
 \\$\textbf {property 5: }$
 The additive inverse is the polynomial with all the coefficients being additive inverses of those of initial polynomial.
 \\
 \\$\textbf {property 6: }$
 $1$ is the identity element for $\cdot$
 \\
 \\$\textbf {property 7: }$
 Because the coefficients are in $\mathbb{R}$, the polynomials also follow the distributive law.
\end{proof}

\newpage 

\section{Problem 2.1.2}

\begin{proof}
\begin{enumerate}
    \item If $v=0$, then $1 \cdot v = 0$ but $1 \neq 0$, implying that $S \cup \{v\}$ is linearly dependent. Suppose $v \neq 0$. Since $S$ is a spanning set, $v = c_1s_1 + c_2s_2 + \cdots + c_ns_n$, where $c_1, c_2, \ldots, c_n \in k$ and $s_1, s_2, \ldots, s_n \in S$. Since $v \neq 0$, not all of $c_1, c_2, \ldots, c_n$ is $0$. But $c_1s_1 + c_2s_2 + \cdots + c_ns_n - 1 \cdot v = 0$ is a linear combination that equals $0$. Hence, $S \cup \{v\}$ is linearly dependent.
    \item Let $v = c_1s_1 + c_2s_2 + \cdots + c_ns_m$, where $c_1, \ldots, c_m$ and $s_1, \ldots, s_m \in S$. Let $w = s_m$. Then, $w = v - c_1(c_m)^{-1}s_1 - c_2(c_m)^{-1}s_2 - \cdots - c_{m-1}(c_m)^{-1}s_{m-1}$. Suppose $v_0 = k_1s'_1 + k_2s'_2 + \cdots + k_ns'_n$, where $k_1, \ldots, k_n \in k$ and $s'_1, \ldots, s'_n \in S$. If $w \not\in \{s'_1, \ldots, s'nm\}$, then $v_0$ is a linear combination of elements in $(S-\{w\}) \cup \{v\}$. If $w \in \{s'_1, \ldots, s'_n\}$, without loss of generality, assume $s'_n = w$. Then, $v_0 = k_1s'_1 + \cdots + k_{n-1}s'_{n-1} + k_n (v - c_1(c_m)^{-1}s_1 - \cdots - c_{m-1}(c_m)^{-1}s_{m-1})$. This is a linear combination of elements in $(S-\{w\}) \cup \{v\}$. Hence, $(S-\{w\}) \cup \{v\}$ is a spanning set.
\end{enumerate}
\end{proof}

\newpage

\section{Problem 2.2.1}

\begin{proof}
\textbf{Example 1:} $ 2, x, x^2$
\\ If $a \cdot 2 + b \cdot x + c \cdot x^2 = 0$ as polynomials, where $a,b,c \in \mathbb{C}$, then $a=b=c=0$. Furthermore, every polynomial of degree at most 2, by definition, can be written in the form $\frac{a}{2} \cdot 2 + b \cdot x + c \cdot x^2$, and so these elements $2,x,x^2$ are a basis.
\\
\textbf{Example 2:} $ 3, x, x^2$
\\ If $a \cdot 3 + b \cdot x + c \cdot x^2 = 0$ as polynomials, where $a,b,c \in \mathbb{C}$, then $a=b=c=0$. Furthermore, every polynomial of degree at most 2, by definition, can be written in the form $\frac{a}{3} \cdot 3 + b \cdot x + c \cdot x^2$, and so these elements $3,x,x^2$ are a basis.
\end{proof}

\newpage 

\section{Problem 2.2.2}

\begin{proof}
	1.)
 \\ Let S be a spanning set in field $k$. Because $S$ is a spanning set, $v$ can be written as $v=w_1 \cdot c_1 + w_2 \cdot c_2 + ... + w_x \cdot c_x$ where $w_i \in S$ and $c_i \in k$
 \\ We have a linear combination, $w_1 \cdot c_1 + w_2 \cdot c_2 + ... + w_x \cdot c_x + v \cdot (-1) = 0$. Therefore, $S \cup \{v\}$ is linearly dependent.
 \\
 2.) Because $v=w_1 \cdot c_1 + w_2 \cdot c_2 + ... + w_x \cdot c_x$ and $v \neq 0$, there must be $w_i \in \{w_1, w_2, ...,w_x \}$ such that $w_i \neq 0$
 \\$w_i = \dfrac{v-w_1\cdot c_1-...-w_{i-1}\cdot c_{i-1}-w_{i+1}\cdot c_{i+1}-...-w_x\cdot c_x}{c_i}$
 \\ We will prove that $w_i$ is the vector such that $(S-\{w\}) \cup \{v\}$ is a spanning set.
 \\ Let $V$ be the vector space. Suppose that $(S-\{w_i\}) \cup \{v\}$ is NOT a spanning set. \\There must exist a vector $a \in V$, such that there is no expression of $a$ as a linear combination of vectors in $(S-\{w\}) \cup \{v\}$.
 \\ However, because $S$ is a spanning set, $a$ can be written as a linear combination of vectors in $S$.
 \\ If there exists a linear combination that does not include $w_i$, then $(S-\{w_i\}) \cup \{v\}$ is a spanning set which leads to CONTRADICTION.
 \\ If all the linear combinations include $w_i$, then substitute $w_i$ with
 \\$\dfrac{v-w_1\cdot c_1-...-w_{i-1}\cdot c_{i-1}-w_{i+1}\cdot c_{i+1}-...-w_x\cdot c_x}{c_i}$
 \\Because this linear combination exists, $(S-\{w_i\}) \cup \{v\}$ is a spanning set which leads to CONTRADICTION.
 \\
 \\We can conclude that there exists a vector $w_i$ in $S$ such that $(S-\{w_i\}) \cup \{v\}$ is a spanning set
\end{proof}

\newpage 

\section{Problem 2.2.3}

\begin{proof}
	According to the condition $L \not\subset S$, $|L|>0$ and $|L-S|>0$.
 \\ Because $L$ is a linearly independent set, all elements in $L$ are non-zero.
 \\ Now, we have that $S$ is a spanning set, and Let $v$ be a vector that doesn't lie in $S$, but in $L-S$.
 \\According to Problem 2.2.2(2), there must exists a vector $w_i$ in $S$ such that $(S-\{w_i\}) \cup \{v\}$ is a spanning set.
 \\We want to show that this vector $w_i$ exists in $S-L$, not just $S$.
 \\
 
\end{proof}

\newpage 

\section{Problem 2.2.4}

\begin{proof}
	Continuing the process in Problem 2.2.3, we can see that $|L-S|$ is decreasing until it reaches $0$ when $L \subset S$. Because $S$ is finite, $L$ must be finite. Moreover, $|L|\leq|S|$
 \\We can conclude that the size of every linearly independent set is at most the size of every spanning set.
\end{proof}

\newpage 

\section{Problem 2.2.5}

\begin{proof}
1.) If the spanning set is not a basis yet, there exists a linear combination between some vectors in the set. From those vectors, remove one. This will not affect its spanning property. Repeat the process continuously will result in a spanning set that has no linear combination, which is a basis.
\\ 2.) If the linearly independent set is not a basis yet, the span of all the vectors in the set does not cover the vector space. Suppose there is a vector $a$ outside of the span of the current set, we may add vector $a$ to our linearly independent set. This will not affect its linearly independent property because $a$ is outside of the span of all the vectors in the set. Repeat the process continuously will result in a linearly independent set that span the vector space, which is a basis.
\end{proof}

\newpage 

\section{Problem 2.2.6}

\begin{proof}
	
\end{proof}

\newpage 

\section{Problem 2.2.7}

\begin{proof}
	Suppose $W$ is a subspace of $V$, but $dimW > dimV$.
 
\end{proof}

\newpage 

\section{Problem 2.3.1}

\begin{proof}
	Because $f(v_1+v_2)=f(v_1)+f(v_2)$ for all $v_1, v_2 \in V$
 \\ Substitute $v1$ with $0$ and $v2$ with $0$: 
 \\ $f(0)=2 \cdot f(0)$
 \\ $f(0)=0$
 \\ Substitute $v1$ with $v$ and $v2$ with $-v$: 
 \\ $0=f(0)=f(v)+f(-v)$
 \\ We will prove that $f(sv)=sf(v)$ for all $s \in \mathbb{N}$
 \\ Base case: $s=1: $ $f(1\cdot v)=1 \cdot f(v)$ which is trivial
 \\ Inductive case: Assume $f(nv)=nf(v)$, will prove that $f((n+1)v)=(n+1)f(v)$
 \\Substitute $v1$ with $nv$ and $v2$ with $v$: 
 \\$f(nv+v)=f(nv)+f(v)$
 \\$f((n+1)v)=nf(v)+f(v)=(n+1)f(v)$
 \\Therefore, $f(sv)=sf(v)$ for all $s \in \mathbb{N}$
 \\Substitute $v$ with $\cfrac{v}{s}$ in the latest equation
 \\ $f(\cfrac{v}{s})=\cfrac{f(v)}{s}$ for all $s \in \mathbb{N}$
 \\Combining these two equations and $-f(v)=f(-v)$, \\we have that $T:V \rightarrow W$ satisfies $T(sv)=sT(v)$ for all $s \in \mathbb{Q}$
 \\ $f$ is a linear transformation.
\end{proof}

\newpage 

\section{Problem 2.3.2}

\begin{proof}
	Because $kerT$ is the set of elements $v\in V$such that $T(v)=0_W$, $kerT$ is obviously a subset of $V$. 
 \\ Let $T(v_1)=0$ and $T(v_2)=0$, we have $T(v_1 + v_2)=0$ and $T(s \cdot v_1)=0$
 \\Let $T(v_1)=0$, we know $T(0)=0$, so $T(-v_1)=0$
 \\Therefore, $kerT$ is a subspace of $V$
\end{proof}

\newpage 

\section{Problem 2.3.3}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 2.3.4}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 2.3.5}

\begin{proof}
     1.) The basis of $kerT$ consists of $w_1,w_2,...,w_n$
 \\According to Problem 2.3.2 , $kerT$ is a subspace of $V$. According to Problem 2.2.5(2), the basis of $kerT$, which is linearly independent, can be extended to a basis of $V$.
 \\ 2.) \\$imT=T(c_1w_1+c_2w_2+...+c_nw_n+c_{n+1} w_{n+1}+...+c_mw_m)$
 \\$=0+c_{n+1}T(w_{n+1})+...+c_mT(w_m)$
 \\Now, we know that $T(w_{n+1}),...,T(w_m)$ span $imT$
 \\Suppose there exists a linear combination between them,
 \\$a_{n+1}T(w_{n+1})+...+a_mT(w_m)=0$
 \\We will have $T(a_{n+1} w_{n+1}+...+a_mw_m)=0$
 \\Therefore, $a_{n+1} w_{n+1}+...+a_mw_m$ is in $kerT$ which spans $w_1,w_2,...,w_n$
 \\We will have that $w_1, w_2,...,w_m$ is not a basis of $V$, which is a CONTRADICTION.
 \\So, $T(w_{n+1}),...,T(w_m)$ form a basis for $imT$
 
 
\end{proof}

\newpage 

\section{Problem 2.3.6}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 2.4.1}

\begin{proof}
	For each $v$, because $w_i$ are bases (linearly independent), there exists a unique $\begin{pmatrix}
  a_{1} \\
  a_{2} \\
  \vdots \\
  a_{m} \\
\end{pmatrix}$
\\ So, the map is well-defined.
\\ $f: V \rightarrow k^n$ is injective because the same
$\begin{pmatrix}
  a_{1} \\
  a_{2} \\
  \vdots \\
  a_{m} \\
\end{pmatrix}$
would give the same $v$.
\\Suppose $f: V \rightarrow k^n$ is not surjective
\\There is no $v$ that yields $\begin{pmatrix}
  b_{1} \\
  b_{2} \\
  \vdots \\
  b_{m} \\
\end{pmatrix}$
, then there will exist $v=b_1w_1+...+b_nw_n \notin V$, which is a CONTRADICTION.
\\Therefore, there is an isomorphism between $V$ and $k^n$
\\
\end{proof}

\newpage 

\section{Problem 2.4.2}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 2.4.3}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 2.4.4}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 2.4.5}

\begin{proof}
	1.)
 $\begin{pmatrix}
1 & 0 & -1\\
0 & 1 & 2
\end{pmatrix}$
\\ 2.)
$\begin{pmatrix}
1 & 0 & 0 & -\dfrac{37}{19}\\
0 & 1 & 0 & 2\\
0 & 0 & 1 & -\dfrac{15}{19}
\end{pmatrix}$
\end{proof}

\newpage 

\section{Problem 2.4.6}

\begin{proof}
	For any column, if there exists more than one nonzero value, do the row operation such that there is only one nonzero value left, then divide that row by that last value to create 1.
\end{proof}

\newpage 

\section{Problem 2.4.7}

\begin{proof}
	Start at the first column, if there exists more than one nonzero value, do the row operation such that there is only one nonzero value left,
 \\then divide that row by itself to create 1,
 \\then switch the pivot to the first row.
 \\If there exists only one nonzero, only divide and switch its row to the first row.
 \\
 \\ At the second column, if there exists at least one nonzero value on the different row than the existed pivot, do the row operation such that the value on the same row with the existed pivot become 0 (this will not affect the column to the left because those row without existed pivot only have 0 to the left of the current column), then continue until there is only one nonzero value left, divide to make it 1, then switch it to the second row.
 \\If all the nonzero values are on the same row with the existed pivot, we leave it as it was.
 \\
 \\ For all the other column, we do the same process did in the second column.
 \\ This satisfies the properties of the reduced row echelon form.
 \\ 1) Every row with at least one nonzero entry has their leftmost nonzero entry as a 1 because the process always divide the whole row to make 1.
 \\ 2) Each pivot is the only nonzero entry in its column was guaranteed through the process.
 \\ 3) The pivot of the ith row is left of the pivot of the jth row if i < j is guaranteed by switching the row with new pivot to the row below the earlier pivot.
 \\ 4) The rows with all zeros are on the bottom of the matrix. This is also guaranteed by switching the row.
\end{proof}

\newpage 

\section{Problem 2.4.8}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 3.0.1}

\begin{proof}
	If $m_1, m_2 \in I$, then by the definition of an ideal, $m_1 + m_2 \in I$, and for all $r \in R, m \in M$, we have $r \cdot m \in M$. Hence, $M$ is closed under addition and scalar multiplication. Since the operations of $+$ and $\cdot$ are those of the ring $R$, these operations follow the commutative, associative, and distributive laws. Since $0 \in I$, and for all $m \in I \subseteq R$, $0 + m = m$, the operation $+$ has an identity element. Since $1_R \cdot m = m$ for all $m \in I \subseteq R$, property 6 also suffices. Lastly, for all $m \in I$, we have $-m \in I$ with $m + (-m) = 0$. Hence, $I$ is an $R$-module under the addition and multiplication operations of the ring $R$.
\end{proof}

\newpage 

\section{Problem 3.0.2}

\begin{proof}
	Let $M$ be a finitely generated free module. Let $\{m_1, m_2, \ldots, m_n\}$ be a free basis of $M$. Then every element $m \in M$ can be uniquely expressed as $m = r_1 m_1 + r_2 m_2 + \cdots + r_n m_n$ for some $r_1, r_2, \ldots, r_n \in R$. Consider the map $\phi: M \longrightarrow R^n$ defined by
    $$\phi(m) = \phi(r_1 m_1 + r_2 m_2 + \cdots + r_n m_n) = (r_1, r_2, \ldots, r_n)$$
    This mapping is well-defined, since for each $m$ the corresponding $\{r_1, r_2, \ldots, r_n\}$ is unique. Furthermore, if $m = r_1 m_1 + r_2 m_2 + \cdots + r_n m_n$, $m' = r'_1 m_1 + r'_2 m_2 + \cdots + r'_n m_n$, and $r \in R$, then
    $$\phi(m+m') = \phi((r_1+r'_1)m_1 + (r_2+r'_2)m_2 + \cdots + (r_n+r'_n)m_n) $$
    $$= (r_1+r'_1, r_2+r'_2, \ldots, r_n+r'_n) = (r_1, r_2, \ldots, r_n) + (r'_1, r'_2, \ldots, r'_n) = \phi(m) + \phi(m')$$
    $$\phi(rm) = \phi(rr_1m_1 + rr_2m_2 + \cdots rr_nm_n) = (rr_1, rr_2, \ldots, rr_n) = r (r_1, r_2, \ldots, r_n) = r\phi(m)$$
    This shows that $\phi$ is a homomorphism. For all $(r_1, r_2, \ldots, r_n) \in R^n$, we have $\phi(r_1m_1 + r_2m_2 + \cdots + r_nm_n) = (r_1, r_2, \ldots, r_n)$, so $\phi$ is surjective. If $\phi(m) = \phi(r_1m_1 + r_2m_2 + \cdots + r_nm_n) = 0$, then $(r_1, r_2, \ldots, r_n) = 0$, which implies that $m = r_1m_1 + r_2m_2 + \cdots r_nm_n = 0$. Hence, $\phi$ is injective. We conclude that $\phi$ is an isomorphism from $M$ to $R_n$. Therefore, $M$ is isomorphic to $R^n$.
\end{proof}

\newpage 

\section{Problem 3.0.3}

\begin{proof}
	Notice that for all $m, m' \in M$ and $r\in R$, 
    $$\kappa_{M, N}(m+m') = (m+m') + N = (m+N) + (m'+N) = \kappa_{M, N}(m) + \kappa_{M, N}(m')$$
    $$\kappa_{M, N}(r\cdot m) = (r\cdot m) + N = r\cdot (m+N) = r\kappa_{M, N}(m)$$
    Hence, $\kappa_{M, N}$ is a homomorphism. For all $m+N \in M/N$, we have $m + N = \kappa_{M, N}(m)$, so $\kappa_{M, N}$ is a surjective homomorphism. Note that $\kappa_{M, N}(m) = m+N = 0+N$ if and only if $m \in N$. Therefore, the kernal of $\kappa_{M, N}$ is $N$.
\end{proof}

\newpage 

\section{Problem 3.0.4}

\begin{proof}
\begin{enumerate}
    \item Given $m + \ker \phi \in M/\ker \phi$, let $\bar{\phi}(m + \ker \phi) = \bar{\phi}(\kappa_{M, N}(m)) = \phi(m)$. We first prove that this mapping is well-defined. \\
    If $m, m' \in M$, $m + \ker \phi = m' + \ker \phi$, then $m' = m + k$, where $k \in \ker \phi$. Since $\phi$ is a homomorphism, 
    $$\phi(m') = \phi(m+k) = \phi(m)+\phi(k)=\phi(m)$$
    Therefore, $\bar{\phi} \kappa_{M, N}(m) = \bar{\phi} \kappa_{M, N}(m')$., so the mapping is well defined. Also notice that if $m, m' \in M$, $r\in R$ then 
    $$\bar{\phi}((m+m')+\ker \phi) = \phi(m+m')=\phi(m)+\phi(m')=\bar{\phi}(m+\ker \phi)+\bar{\phi}(m'+\ker \phi)$$
    $$\bar{\phi}(rm+\ker \phi) = \phi(rm) = r\phi(m) = r \bar{\phi}(m+\ker \phi)$$
    Therefore, $\bar{\phi}$ is the homomorphism we want.
    \item We will prove that $\bar{\phi}$ is an isomorphism from $M/\ker \phi$ to $\im \phi$. \\
    If $\bar{\phi}(m+\ker \phi) = 0$, then $\phi(m) = 0$, which implies that $m \in \ker \phi$. Therefore, $m + \ker \phi = 0 + \ker \phi$. Hence, $\ker \bar{\phi} = \{0+\ker\phi\}$. \\
    If $\bar{\phi}(m+\ker \phi) = \bar{\phi}(m'+\ker \phi)$, then $$\bar{\phi}((m+\ker \phi) - (m' + \ker\phi)) = \bar{\phi}((m-m')+\ker \phi) = \bar{\phi}(m+\ker \phi) - \bar{\phi}(m'+\ker \phi) = 0$$
    Which implies that $(m+\ker\phi)-(m'+\ker\phi) \in \ker \bar{\phi}$. But $\ker \bar{\phi}$ contains only one element, namely, $0 + \ker \phi$. Hence, $(m+\ker\phi)-(m'+\ker\phi)=0+\ker\phi$. So $m+\ker\phi=m'+\ker\phi$. This shows that $\bar{\phi}$ is injective.\\
    For each $m\in M$, $\bar{\phi}(m+\ker\phi)=\phi(m)$. Therefore, $\im \bar{\phi} = \im \phi$. This implies that $\bar{\phi}$ is a bijective map from $M/\ker\phi$ to $\im \phi$, which means that $M/\ker\phi$ is isomorphic to $\im \phi$.
\end{enumerate}
	
\end{proof}

\newpage 

\section{Problem 4.1.1}

\begin{proof}
\begin{enumerate}
    \item Let $R$ be a Noetherian ring. Suppose for the sake of contradiction that there is an ideal $I$ of $R$ that was not generated by finitely many elements. Let $i_1 \in I$, and $I_1 = <i_1>$. Suppose $I_n = <i_1, i_2, \ldots, i_n>$. Since $I$ is not finitely generated, $I_n \subset I$ but $I_n \neq I$. Hence, there is an element $i_{n+1}$ such that $i_{n+1} \notin I_n$. Let $I_{n+1} = <i_1, i_2, \ldots, i_n, i_{n+1}>$. In this way, we have created an ascending chain $I_1 \subset I_2 \subset I_3 \subset \cdots $ such that for each $n \in \N$, $I_{n+1} \neq I_n$. Hence, $R$ is not a Noetherian ring, contradiction. Therefore, every ideal can be generated by a finite set of elements in $R$.
    \item Let $I = \bigcup_{i=1}^\infty I_i$. If $i_1, i_2 \in I$, then $i_1 \in I_{n_1}$, $i_2 \in I_{n_2}$ for some $n_1, n_2 \in \N$. Without loss of generality, assume $n_1 \leq n_2$. Then $I_{n_1} \subseteq I_{n_2}$, so $i_1 \in I_{n_2}$. By the definition of ideal, $i_1 + i_2 \in I_{n_2} \subset I$. Hence, $I$ is closed under addition. If $i \in I$, then there exists $n \in \N$ such that $i \in I_n$. If $r \in R$, then by definition, $r \cdot i \in I_n \subset I$. Hence, $I$ is an ideal of $R$.
    \item Suppose $I_1 \subseteq I_2 \subseteq I_3 \subseteq \ldots$ be a chain of ideals. Let $I = \bigcup_{i=1}^\infty I_i$. By part 2, $I$ is an ideal. From the condition we know that $I$ can be generated by a finite set of elements $\{i_1, i_2, \ldots, i_m\}$. Since $i_1, i_2, \ldots, i_m \in I$, there is $n_1, n_2, \ldots, n_m \in \N$ such that $i_1 \in I_{n_1}, i_2 \in I_{n_2}, \ldots, i_{n_m} \in I_m$. Choose $n = \max \{n_1, n_2, \ldots, n_m\}$. Then, $i_1, i_2, \ldots, i_m \in I_n$, so $<i_1, i_2, \ldots, i_m> \subseteq I_n$. But $I_n \subseteq I = <i_1, i_2, \ldots, i_m>$. Hence, $I_n = <i_1, i_2, \ldots, i_m>$. For all $n' > n $, $I_n = <i_1, i_2, \ldots, i_m> \subseteq I_{n'}$, but $I_{n'} \subseteq I = I_n = <i_1, i_2, \ldots, i_m>$. Therefore, $I_n = I_{n'}$, so $R$ is Noetherian.
    \item If $R$ is a PID, then every ideal of $R$ can be generated by one element. By part 3, $R$ is Noetherian.
\end{enumerate}


\end{proof}
Notation: throughout section $4$ and $5$, $<r_1, r_2, \ldots, r_n>$ or $\langle r_1, r_2, \ldots, r_n \rangle$ is used to denote the ideal or the submodule generated by $r_1, r_2, \ldots, r_n$. If it is clear whether the algebraic structure denoted is an ideal or submodule, we will not specify the algebraic structure.
\newpage 

\section{Problem 4.1.2}

\begin{proof}
\begin{enumerate}
    \item Let $M$ be a Noetherian module. Suppose for the sake of contradiction that there is a submodule $M'$ of $M$ that was not generated by finitely many elements. Let $m_1 \in M'$, and $I_1 = <m_1>$. Suppose $M_n = <m_1, m_2, \ldots, m_n>$. Since $M'$ is not finitely generated, $M_n \subset M'$ but $M_n \neq M'$. Hence, there is an element $m_{n+1}$ such that $m_{n+1} \notin M_n$. Let $M_{n+1} = <m_1, m_2, \ldots, m_n, m_{n+1}>$. In this way, we have created an ascending chain $M_1 \subset M_2 \subset M_3 \subset \cdots $ such that for each $n \in \N$, $M_{n+1} \neq M_n$. Hence, $M$ is not a Noetherian module, contradiction. Therefore, every submodule can be generated by a finite set of elements in $M$.
    \item We first prove that for any chain $M_1 \subseteq M_2 \subseteq M_3 \subseteq \cdots$, $\bigcup_{i=1}^\infty M_i$ is a submodule.\\
    
    Let $M' = \bigcup_{i=1}^\infty M_i$. Since addition and multiplication in $M'$ are the same as those in $M$, they follow commutative, associative, and distributive law. Since $M' \subseteq M$, for all $m \in M'$, $0+m=m$ and $1\cdot m = m$. If $m \in M'$, then there exists $n \in \N$ such that $m \in M_n$. Then, $m$ has an additive inverse in $M_n$, which implies that $m$ has an additive inverse in $M'$.\\
    
    If $m_1, m_2 \in M'$, then $m_1 \in M_{n_1}$, $m_2 \in M_{n_2}$ for some $n_1, n_2 \in \N$. Without loss of generality, assume $n_1 \leq n_2$. Then $I_{n_1} \subseteq I_{n_2}$, so $m_1 \in M_{n_2}$. By the definition of submodule, $m_1 + m_2 \in M_{n_2} \subset M'$. Hence, $M'$ is closed under addition. If $m \in M$, then there exists $n \in \N$ such that $m \in M_n$. If $r \in R$, then by definition, $r \cdot m \in M_n \subset M'$. Hence, $M'$ is a submodule of $M$. \\
    
    Suppose $M_1 \subseteq M_2 \subseteq M_3 \subseteq \ldots$ be a chain of ideals. Let $M' = \bigcup_{i=1}^\infty M_i$. $M'$ is a submodule of $M$. From the condition we know that $M'$ can be generated by a finite set of elements $\{m_1, m_2, \ldots, m_k\}$. Since $m_1, m_2, \ldots, m_k \in M'$, there is $n_1, n_2, \ldots, n_k \in \N$ such that $m_1 \in M_{n_1}, m_2 \in M_{n_2}, \ldots, m_{n_k} \in M_{n_k}$. Choose $n = \max \{n_1, n_2, \ldots, n_k\}$. Then, $m_1, m_2, \ldots, m_k \in M_n$, so $<m_1, m_2, \ldots, m_k> \subseteq M_n$. But $M_n \subseteq M' = <m_1, m_2, \ldots, m_k>$. Hence, $M_n = <m_1, m_2, \ldots, m_k>$. For all $n' > n $, $M_n = <m_1, m_2, \ldots, m_k> \subseteq M_{n'}$, but $M_{n'} \subseteq M' = M_n = <m_1, m_2, \ldots, m_k>$. Therefore, $M_n = M_{n'}$, so $M$ is Noetherian.
\end{enumerate}

\end{proof}
Notation: throughout section $4$ and $5$, $<r_1, r_2, \ldots, r_n>$ or $\langle r_1, r_2, \ldots, r_n \rangle$ is used to denote the ideal or the submodule generated by $r_1, r_2, \ldots, r_n$. If it is clear whether the algebraic structure denoted is an ideal or submodule, we will not specify the algebraic structure.
\newpage 

\section{Problem 4.1.3}

\begin{proof}
	Let $M' \subseteq N$ be a submodule of $N$. Since $N \subseteq M$ is a submodule of $M$, $M'$ is also a submodule of $M$. Since $M$ is Noetherian, $M'$ finitely generated. Therefore, every submodule of $N$ is finitely generated, implying that $N$ is Noetherian. \\
 
    I claim that if $M' \in M/N$ is a submodule of $M/N$, then $M_0 \in M$, the preimage of $M'$ under $\kappa_{M,N}$, is a submodule of $M$. Notice that if $m, m_1, m_2 \in M_0$, then $m = m' + n, m_1 = m'_1 + n_1, m_2 = m'_2 + n_2$, where $m', m'_1, m'_2 \in M', n, n_1, n_2 \in N$. Then $m_1 + m_2 = (m'_1+m'_2) + (n_1+n_2)$, and if $r \in R$, then $rm = rm' + rn$. So $\kappa_{M,N}(m_1+m_2) = (m'_1+m'_2) + N = (m'_1 + N) + (m'_2 + N) \in M'$ and $\kappa_{M,N}(rm) = (rm) + N = r(m+N) \in M'$, since $M'$ is a submodule. Hence, $m_1+m_2, rm \in M_0$. Moreover, the commutative, associative, and distributive law, the existence of $0$, $1$, and the additive inverse carry over from the corresponding properties of $M$. Therefore, $M_0$ is a submodule of $M$. \\

    Let $M'_1 \subseteq M'_2 \subseteq M'_3 \subseteq \cdots$ be a chain of submodule of $M/N$. Let $M_1, M_2, M_3, \ldots$ be the preimage of $M'_1, M'_2, M'_3, \ldots$ under $\kappa_{M,N}$, respectively. Then, $M_1 \subseteq M_2 \subseteq M3 \subseteq \cdots$. By definition, there exists $n \in \N$ such that for all $m \geq n$, $M_m = M_n$. Therefore, for all $m \geq n$, $M'_m = M'_n$. This shows that $M/N$ is Noetherian. 

\end{proof}

Notation: throughout section $4$ and $5$, $<r_1, r_2, \ldots, r_n>$ or $\langle r_1, r_2, \ldots, r_n \rangle$ is used to denote the ideal or the submodule generated by $r_1, r_2, \ldots, r_n$. If it is clear whether the algebraic structure denoted is an ideal or submodule, we will not specify the algebraic structure.
\newpage 

\section{Problem 4.1.4}

\begin{proof}
\begin{enumerate}
    \item If $m \in M_2$, then $\kappa_{M,N}(m) = m + N$. Since $\kappa_{M,N}(M_1) = \kappa_{M,N}(M_2)$, we have $m + N \in \kappa_{M,N}(M_2) = \kappa_{M,N}(M_1)$. Therefore, $m + N = \kappa_{M,N}(m') = m' + N$, where $m' \in M_1 \subseteq M_2$. Then, $m'-m \in N$, and since $m', m \in M_2$, we have $m'-m \in M_2 \cap N$. But $M_2 \cap N = M_1 \cap N$, so $m'-m \in M_1 \cap N \subseteq M_1$. Hence, $m\in M_1$. Therefore, $M_1 = M_2$.
    \item Let $M'_1 \subseteq M'_2 \subseteq M'_3 \subseteq \cdots$ be a chain of submodule of $M/N$. Let $M_1, M_2, M_3, \ldots$ be the preimage of $M'_1, M'_2, M'_3, \ldots$ under $\kappa_{M,N}$, respectively. By definition, there exists $n_1 \in \N$ such that for all $m \geq n_1$, $M'_m = M'_n$, so $\kappa_{M,N}(M_m) = \kappa_{M,N}(M_{n_1})$. Consider the chain of submodule $M_1 \cap N \subseteq M_2 \cap N \subseteq M_3 \cap N \subseteq \cdots$. Since each element in the chain of submodule is a submodule of $N$, there exists $n_2 \in \N$ such that for all $m \geq n_2$, $M_m \cap N = M_{n_2} \cap N$. Choose $n$ to be the greater of $n_1$ and $n_2$. Then, for all $m \geq n$, we have $M_m \cap N = M_n \cap N$ and $\kappa_{M,N}(M_m) = \kappa_{M,N}(M_n)$. By the previous part, $M_m=M_n$. This implies that $M$ is Noetherian.
\end{enumerate}
\end{proof}
Notation: throughout section $4$ and $5$, $<r_1, r_2, \ldots, r_n>$ or $\langle r_1, r_2, \ldots, r_n \rangle$ is used to denote the ideal or the submodule generated by $r_1, r_2, \ldots, r_n$. If it is clear whether the algebraic structure denoted is an ideal or submodule, we will not specify the algebraic structure.
\newpage 

\section{Problem 4.1.5}

\begin{proof}
\begin{enumerate}
    \item By Problem 4.1.2 part 1, if $M$ is a Noetherian module, then every submodule of $M$ is finitely generated. Since $M$ is a submodule of itself, $M$ is finitely generated.
    \item  If $M$ is a submodule of $R$, then $M$ is an ideal of $R$. This is because if $m, m_1, m_2 \in M$, $r \in R$, then $m_1+m_2 \in M$, $rm \in M$, and by definition, $M$ is an ideal of $R$. If $M_1 \subseteq M_2 \subseteq M_3 \subseteq \cdots$ is a chain of submodule of $R$, then it is also a chain of ideals of $R$. Since $R$ is a Noetherian ring, there exists $n \in \N$ such that for all $m \geq n$, we have $M_m = M_n$. Therefore, $R$ is also a Northerian $R$-module.
    \item I claim that $R^{n-m}$ is a free module generated by a free basis of $n-m$ elements. Let $r_1 = (1, 0, 0, \ldots, 0), r_2 = (0, 1, 0, \ldots 0), r_3 = (0, 0, 1, \ldots, 0), \ldots, r_n = (0, 0, 0, \ldots, 1)$. It is clear that $\{r_1, r_2, \ldots, r_n\}$ is a free basis for $R^n$, and $\{r_1, r_2, \ldots, r_m\}$ is a free basis for $R^m$, treated as a submodule of $R^n$. Consider $\{r_{m+1} + R^m, r_{m+2} + R^m, \ldots, r_n + R^m\}$. Suppose $r' + R^m \in R^n/R^m$. Since $r' \in R^n$, $r' = c_1r_1 + c_2r_2 + \cdots + c_nr_n$ for some $c_1, c_2, \ldots, c_n \in R$. Since $c_1r_1 + c_2r_2 + \cdots + c_mr_m \in R^m$, $r'+R^m = (c_{m+1}r_{m+1}+ c_{m+1}r_{m+1}+\cdots+r_nr_n) + R^m = c_{m+1}(r_{m+1}+R^m)+c_{m+2}(r_{m+2}+R^m) + \ldots + c_n(r_n+R^m)$. Hence, $\{r_{m+1} + R^m, r_{m+2} + R^m, \ldots, r_n + R^m\}$ is a generating set. If $c_{m+1}(r_{m+1}+R^m)+c_{m+2}(r_{m+2}+R^m) + \ldots + c_n(r_n+R^m) = 0$, then $c_{m+1}r_{m+1} + c_{m+2}r_{m+2} + \cdots + c_nr_n \in R^m$. But this could only happen when $c_{m+1} = c_{m+2} = \cdots = c_n = 0$. Hence, $\{r_{m+1} + R^m, r_{m+2} + R^m, \ldots, r_n + R^m\}$ is linearly independent, and so a free basis of $n-m$ elements. \\

    Proceed as it is in Problem 3.0.2, we can show that $R^n/R^m$ is isomorphic to $^{n-m}$.
    \item We use induction. By part 2, we know that $R$ is an Noetherian $R$-module. Assume by induction that $R^n$ is a Noetherian $R$-module. Then, since $R^{n+1}/R$ is isomorphic to $R^n$, and both $R$ and $R^n$ are Noetherian, by Problem 4.1.4 part 2, $R^{n+1}$ is also Noetherian. This completes the induction and shows that $R^n$ is Noetherian for every positive integer $n$.
    \item We use induction. For the base case, let $M$ is a $R$-module generated by a single element, $m$. Then, $M$ is isomorphic to $R$, a Noetherian $R$-module. Hence $M$ is a Noetherian $R$-module. Assume by strong induction that if $M$ is a $R$-module generated by $n$ or less elements, then $M$ is Noetherian. Let $M$ be a $R$-module generated by $n+1$ elements, $m_1, m_2, \ldots, m_{n+1}$. Consider $M_1$ be the submodule of $M$ generated by $m_1$. Then, if $m' + M_1 \in M/M_1$, and $m' = r_1m_1 + r_2m_2 + \cdots + r_{n+1}m_{m+1}$, where $r_1, r_2, \ldots, r_{n+1} \in R$, then $m'+M_1 = (r_1m_1 + r_2m_2 + \cdots + r_{n+1}m_{m+1}) + M_1 = r_1(m_1+M_1) + r_2(m_2+M_1) + \cdots + r_{n+1}(m_{n+1}+M_1) = r_2(m_2+M_1) + r_3(m_3+M_1) + \cdots + r_{n+1}(m_{n+1}+M_1)$. Hence, $M/M_1$ is a submodule generated by $\{m_2 + M_1, m_3 + M_1, m_{n+1} + M_1\}$, a set of $n$ elements. By our induction hypothesis, $M/M_1$ is Noetherian. Since $M_1$ is also Noetherian by our base case, $M$ is Noetherian. This completes the induction and shows that if $M$ is finitely generated, then $M$ is a Noetherian $R$-module.
\end{enumerate}
\end{proof}
Notation: throughout section $4$ and $5$, $<r_1, r_2, \ldots, r_n>$ or $\langle r_1, r_2, \ldots, r_n \rangle$ is used to denote the ideal or the submodule generated by $r_1, r_2, \ldots, r_n$. If it is clear whether the algebraic structure denoted is an ideal or submodule, we will not specify the algebraic structure.
\newpage 

\section{Problem 4.2.1}

\begin{proof}
Denote $\sum_{i=1}^n a_{i,j}e_i$ to be $A_j$. This corresponds to the $j$th column of $A$. If $S_i = (s_{1,i}, s_{2, i}, \ldots, s_{m, i}$, the submodule generated by $\{ s_{1,i}, s_{2, i}, \ldots, s_{m, i} \}$, then $AS_i = s_{1,i}A_1 + s_{2,i}A_2 + \cdots + s_{m,i}A_m$. Since $A_1, A_2, \ldots, A_m \in N(A)$, $AS_i \in N(A)$. Notice that the column of $A'=AS$ is $AS_1, AS_2, \ldots, AS_m$. If $a' \in N(A')$, then $a' = r_1 AS_1 + r_2 AS_2 + \cdots + r_m AS_m$ for some $r_1, r_2, \ldots, r_m \in R$. This implies that $a' \in N(A)$. So $N(A') \subseteq N(A)$. \\

Notice that $A = (AS)S'$, where $S'$ is the inverse of $S$. Denote $S'_i$ to be the $i$th column of $S'$. If $S'_i = (s'_1, s'_2, \ldots, s'_m)$, then $A_i = s'_1(AS_1)+s'_2(AS_2)+\cdots+s'_m(AS_m)$. Since $AS_1, AS_2, \ldots, AS_m \in N(A')$, $A_i \in N(A')$. If $a \in N(A)$, then $a' = r_1 A1 + r_2 A_2 + \cdots + r_m A_m$ for some $r_1, r_2, \ldots, r_m \in R$. This implies that $a \in N(A')$. So $N(A) \subseteq N(A')$. Hence, $N(A) = N(A')$.
\end{proof}
Notation: throughout section $4$ and $5$, $<r_1, r_2, \ldots, r_n>$ or $\langle r_1, r_2, \ldots, r_n \rangle$ is used to denote the ideal or the submodule generated by $r_1, r_2, \ldots, r_n$. If it is clear whether the algebraic structure denoted is an ideal or submodule, we will not specify the algebraic structure.
\newpage 

\section{Problem 4.2.2}

\begin{proof}
Let $\phi: R^n/N(A) \longrightarrow R^n/N(A')$, $r+N(A) \rightsquigarrow Sr+N(A')$. If $r_1 + N(A) = r_2 + N(A) \in R^n/N(A)$, then $r_2 = r_1 + Ar_0$, since any element of $N(A)$ can be expressed as $Ar_0$ for some $r_0 \in R$. Then, $\phi(r_2 + N(A)) = Sr_2 + N(A') = S(r_1 + Ar_0) + N(A')= Sr_1 + A'r_0 + N(A') = Sr_1 + N(A') = \phi(r_1 + N(A))$. Therefore, this map is well-defined. Furthermore, $\phi((r_1+N(A))+(r_2+N(A))) = \phi((r_1+r_2)+N(A)) = S(r_1+r_2)+N(A') = (Sr_1+N(A')) + (Sr_2+N(A')) = \phi(r_1+N(A))+\phi(r_2+N(A))$. And if $r\in R$, then $\phi(r(r_1+N(A))) = S(r(r_1)) + N(A')= r(Sr_1 + N(A')) = r\phi(r_1 + N(A))$. Hence, $\phi$ is a homomorphism. The fact $\phi$ is an isomorphism is implied from the fact that $\phi': R^n/N(A') \rightsquigarrow R^n/N(A), r+N(A') \rightsquigarrow S^{-1}r + N(A)$ is the inverse of $\phi$.
\end{proof}
Notation: throughout section $4$ and $5$, $<r_1, r_2, \ldots, r_n>$ or $\langle r_1, r_2, \ldots, r_n \rangle$ is used to denote the ideal or the submodule generated by $r_1, r_2, \ldots, r_n$. If it is clear whether the algebraic structure denoted is an ideal or submodule, we will not specify the algebraic structure.
\newpage 

\section{Problem 4.2.3}

\begin{proof}
\begin{enumerate}
    \item Since $R$ is a PID and $r_1, r_2$ are not all $0$, $\langle r_1, r_2 \rangle = \langle r \rangle$ for some $r \neq 0 \in R$. Hence, $r = ar_1 + br_2$, and $r_1 = c r, r_2 = d r$ for some $a, b, c, d \in R$. Let $S$ be the matrix $\begin{bmatrix} a & b\\ -d & c \end{bmatrix}$. Then, $S\begin{bmatrix} r_1\\ r_2 \end{bmatrix} = \begin{bmatrix} ar_1+br_2\\ -d(cr)+c(dr) \end{bmatrix} = \begin{bmatrix} r \\ 0 \end{bmatrix}$. Notice that $r = ar_1 +br_2 = acr+bdr = (ac+bd)r$. Since $R$ is an integral domain and $r \neq 0, ac+bd=1$, so $\begin{bmatrix} c & -b\\ d & a \end{bmatrix}$ is the inverse of $S$. Hence, $S$ is the desired invertible matrix.
    \item Let $i \in \{2, 3, \ldots, n\}$. By the previous part, there is a $2$ by $2$ invertible matrix $\begin{bmatrix} a & b\\ -d & c \end{bmatrix}$ such that $\begin{bmatrix} a & b\\ -d & c \end{bmatrix} \begin{bmatrix} r_1\\ r_i \end{bmatrix} = \begin{bmatrix} r \\ 0 \end{bmatrix}$, where $\langle r_1, r_i \rangle = \langle r \rangle$. Consider a matrix $S^i$ defined such that for all $j \neq 1, i$, $S^i_{j, j} = 1$, and $S^i_{1,1} = a, S^i_{1,i} = b, S^i_{i,1} = -d, S^i_{i,i} = c$, and all other entries are $0$. Let $x =  \begin{bmatrix} x_1\\ x_2 \\ \vdots \\ x_n \end{bmatrix}$ such that $x_1=r_1$ and $x_i=r_i$. By matrix multiplication, 
    $$S^ix = \begin{bmatrix} r\\ x_2 \\ \vdots \\ x_{i-1} \\ 0 \\ x_{i+1} \\ \vdots \\ x_n \end{bmatrix}$$ 
    Moreover, Let $S'^i$ with $S^i_{j, j} = 1$ for all $j \neq 1, i$, $S^i_{j, j} = 1$, and $S^i_{1,1} = c, S^i_{1,i} = -b, S^i_{i,1} = d, S^i_{i,i} = a$, and all other entries $0$. By matrix multiplication, $S'^i$ is the inverse of $S^i$. Hence, $S^i$ is invertible. \\

    Let $S = S^{n}S^{n-1}\cdots S^2$. Since the product of invertible matrices is invertible, $S$ is invertible. Moreover, multiplication of a vector by $S^i$ changes only the first entry and reduces the $i$th entry to $0$. Hence, $S\begin{bmatrix} r_1\\ r_2 \\ \vdots \\ r_n \end{bmatrix} = \begin{bmatrix} r\\ 0 \\ \vdots \\ 0 \end{bmatrix}$. So $S$ is the desired invertible matrix. \\

    Let $\textbf{r} = \begin{bmatrix} r_1\\ r_2 \\ \vdots \\ r_n \end{bmatrix}$. Multiplication of $\textbf{r}$ by $S^2$ changes $r_1$ to $r'_2$, where $\langle r'_2 \rangle = \langle r_1, r_2 \rangle$. Assume by induction that multiplication of $\textbf{r}$ by $S^kS^{k-1}\cdots S^2$, where $2\leq k \leq n-1$, changes $r_1$ to $r'_k$ such that $\langle r'_k \rangle = \langle r_1, r_2, \ldots, r_k \rangle$. Then, multiplication of $\textbf{r}$ by $S^{k+1}S^kS^{k-1}\cdots S^2$ changes $r_1$ to $r'_{k+1}$, where $\langle r'_{k+1} \rangle = \langle r'_k, r_{k+1} \rangle$. But since $\langle r'_k \rangle = \langle r_1, r_2, \ldots, r_k \rangle$, we have $\langle r'_{k+1} \rangle = \langle r_1, r_2, \ldots, r_{k+1} \rangle$. This completes the induction and shows multiplication of $\textbf{r}$ by $S$ changes $r_1$ to $r$ such that $\langle r \rangle = \langle r_1, r_2, \ldots, r_n \rangle$.

    \item Let $A_i$ be an $n$ by $m$ matrix, $A_1=A$. Let the sequence $A_1, A_2, \ldots$ be constructed as following. If $A_{i,1} = \begin{bmatrix} r_i \\ a_{i,2} \\ \vdots \\ a_{i, n}\end{bmatrix}$ is the first column of $A_i$, then let $S_i$ be an $n$ by $n$ invertible matrix such that $S A_{i, 1} = \begin{bmatrix} r'_i \\ 0 \\ \vdots \\ 0\end{bmatrix}$, with $\langle r'_i \rangle = \langle r_i, a_{i,2}, \ldots, a_{i,n} \rangle$. Then consider the matrix $(SA)^t$, the transpose of $SA$. Let $A'_{i,1}=\begin{bmatrix} r'_i \\ a'_{i,2} \\ \vdots \\ a'_{i, m}\end{bmatrix}$ be the first column of $(SA)^t$.  Then there exists an $m$ by $m$ invertible matrix $T$ such that $T^t A'_{i,1} = \begin{bmatrix} r_{i+1} \\ 0 \\ \vdots \\ 0\end{bmatrix}$, where $\langle r_{i+1} \rangle = \langle r'_i, a'_{i,2}, \ldots, a'_{i,n} \rangle$. Let $A_{i+1} = S_iA_iT_i$. Notice that $\langle r_i \rangle \subseteq \langle r_i, a_{i,2}, \ldots, a_{i,n} \rangle = \langle r'_i \rangle \subseteq \langle r'_i, a'_{i,2}, \ldots, a'_{i,n} \rangle = \langle r_{i+1} \rangle$. Notice also that by this method of construction, the only nonzero entry in first row of the matrix $A_i$ is $r_i$. \\

    Consider the chain of ideals $\langle r_1 \rangle \subseteq \langle r_2 \rangle \subseteq \langle r_3 \rangle \subseteq \cdots$. Since $R$ is a Noetherian ring, there is a $k \in \N$ such that $\langle r_k \rangle = \langle r_{k+1} \rangle$. Since we have shown above that $\langle r_k \rangle \subseteq \langle r'_k \rangle \subseteq \langle r_{k+1} \rangle$, we also have $\langle r_k \rangle = \langle r'_k \rangle$. But $\langle r'_k \rangle = \langle r_i, a_{k,2}, \ldots, a_{k,n} \rangle$. Hence, for all $2 \leq j \leq n$, $a_{k, j} = x_j r_i$ for some $x_j \in R$. Consider a matrix $M$ such that its diagonal entries are all $1$, its first column is $\begin{bmatrix} 1 \\ -x_2 \\ \vdots \\ -x_n\end{bmatrix}$. This is an invertible matrix that takes the first column of $A_k$ to $\begin{bmatrix} r_k \\ 0 \\ \vdots \\ 0\end{bmatrix}$ without changing the entries in the first row. Let $S = MS_kS_{k-1}\cdots S_2S_1$, $T = T_1 T_2 \cdots T_{k-1} T_k$. Then, since $A_k = S_kS_{k-1}\cdots S_2S_1 A T_1 T_2 \cdots T_{k-1} T_k$, $SAT = MA_k$ has no nonzero entries in the first row or first column except for the $(1,1)$ entry.
\end{enumerate}
\end{proof}
Notation: throughout section $4$ and $5$, $<r_1, r_2, \ldots, r_n>$ or $\langle r_1, r_2, \ldots, r_n \rangle$ is used to denote the ideal or the submodule generated by $r_1, r_2, \ldots, r_n$. If it is clear whether the algebraic structure denoted is an ideal or submodule, we will not specify the algebraic structure.
\newpage 

\section{Problem 4.2.4}

\begin{proof}
Let $B$ be an $n$ by $n$ matrix, $C$ be an $n$ by $m$ matrix. Suppose $B = \begin{bmatrix} b & \textbf{0}_{n-1}^t \\ \textbf{0}_{n-1} & B' \end{bmatrix}$, where $b \in R$, $\textbf{0}_{n-1}$ be the zero-vector in $R^{n-1}$, and $B'$ be an $n-1$ by $n-1$ vector, and suppose $C = \begin{bmatrix} c & \textbf{0}_{m-1}^t \\ \textbf{0}_{n-1} & C' \end{bmatrix}$, where $c \in R$, $\textbf{0}_{m-1}$ be the zero-vector in $R^{m-1}$, and $C'$ be an $n-1$ by $m-1$ vector. By matrix multiplication, $BC = \begin{bmatrix} bc & \textbf{0}_{m-1}^t \\ \textbf{0}_{n-1} & B'C' \end{bmatrix}$. Notice that if $B = \begin{bmatrix} 1 & \textbf{0}_{n-1}^t \\ \textbf{0}_{n-1} & B' \end{bmatrix}$ and $B'$ is invertible, then $B$ is also invertible, since $\begin{bmatrix} 1 & \textbf{0}_{n-1}^t \\ \textbf{0}_{n-1} & (B')^{-1} \end{bmatrix}$ is the inverse of $B$.\\


\end{proof}

\newpage 

\section{Problem 4.2.5}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage

\section{Problem 4.3.1}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 4.3.2}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 4.3.3}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 4.3.4}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 5.1.1}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 5.2.1}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 5.2.2}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 5.2.3}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 5.2.4}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 5.2.5}

\begin{proof}
	(write your solution here)
\end{proof}

\newpage 

\section{Problem 5.3.1}

\begin{proof}
	(write your solution here)
\end{proof}
\newpage 

\section{Problem 5.3.2}

\begin{proof}
	(write your solution here)
\end{proof}
\newpage 

\section{Problem 5.3.3}

\begin{proof}
	(write your solution here)
\end{proof}
\newpage 

\section{Problem 5.3.4}

\begin{proof}
	(write your solution here)
\end{proof}

\end{document}