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\section{Problem 7}
\paragraph{(a)}
\subparagraph{$\implies$} Let $p\in R$ be prime. Then, consider $R/(p)$. We want to show that $R/(p)$ is an integral domain, so let us consider $(p)+x,(p)+y\in R/(p)$ such that $((p)+x)((p)+y)=(p)+0$. Our definition of coset multiplication gives us that this holds only if $(p)+xy=(p)+0\implies xy\in (p)$. Then, we have that $p|xy\implies p|x$ or $p|y$, which means that $x\in (p)$ or $y\in (p)$, implying that $(p)+x=(p)+0$ or that $(p)+y=(p)+0$ respectively. In either case, we have that one of our operands is the zero coset in $R/(p)$, meaning that $R/(p)$ is an integral domain and therefore that $(p)$ is a prime ideal of $R$.
\subparagraph{$\impliedby$} Let $(p)$ be a prime ideal of $R$. Then, note that $R/(p)$ is an integral domain and suppose that $\exists  a,b\in R$ such that $p|ab$. Then, consider the coset product $(p)+0=(p)+ab=((p)+a)((p)+b)$. Since $R/(p)$ is an integral domain, we have that $a\in (p)$ or $b\in (p)$, which is equivalent to our definition of prime that $p|a$ or $p|b$.
\paragraph{(b)}
\subparagraph{$\implies$} Let $p\in R$ be irreducible and consider the set $S$ of all proper principal ideals of $R$. Suppose that there existed some non-unit $x\neq p$ with $(x)\in S$ such that $(p)\subseteq (x)$. Then, we have that $p=ax$ for some $a\in R$, and since $p$ is irreducible, we have that $a$ is a unit. Thus, $p$ and $x$ must be associates, meaning that $(x)\subseteq (p)$ also, meaning that $(p)=(x)$. Thus, no element of $S$ can properly contain $(p)$ since any element that contains $(p)$ must itself be $(p)$.
\subparagraph{$\impliedby$} We can prove this by the contrapositive. Let $p\in R$ be reducible. Then, consider non-units $x,a\in R$ such that $p=ax$. Since $p=ax$, we have clearly that $(p)\subseteq(x)$. However, since $x$ is a proper factor of $p$, we have that $x\not\in (p)$ (since they would otherwise be associates, which is impossible since they are multiplicatively related by a non-unit). Thus, we have that $(p)\subset (x)$, which is sufficient to give us the contrapositive of our desired implication.
\paragraph{(c)} Let $p\in R$ be prime and that $p=xy$ for non-units $x,y\in R$. Then, since $x$ and $y$ are non-units, we have that $p\neq x, p\neq y$, and we have too that $x|p$ and $y|p$. Also, by the definition of prime, we have that $p|x$ or $p|y$. Without loss of generality, suppose that $p|x$. Since $p|x$ and $x|p$, we have that $p$ and $x$ are associates, which means that $y$ is a unit. This is a contradiction, so no two non-units $x$ and $y$ can exist, which is equivalent to our definition of irreducible, and so we are done.
\paragraph{(d)} Part (c) gives us the forward direction; then, for any irreducible element $p\in R$ a PID, we have that $(p)$ is maximal in the set of all proper principal ideals of $R$ (which in our case means that it is maximal in $R$ since every proper ideal of $R$ is principal). Then, since all maximal ideals are prime, we have that $(p)$ is prime in $R$, which by part (a) implies that $p\in R$ is prime.
\paragraph{(e)}
\subparagraph{Irreducible} Consider an irreducible element $p\in R$ and an associate $q\in R$ such that $p=aq$ and $q=bp$ for units $a,b\in R$. Then, if we have that $q=cx$ for $c,x\in R$, we find that $q=bp=cx\implies p=b^{-1}cx$. Since $p$ is irreducible, this gives us that either $x$ is a unit or $b^{-1}c$ is a unit. If $x$ is a unit, we are done, so suppose that $b^{-1}c$ is a unit. Since $b^{-1}$ is a unit, this necessarily implies that $c$ is a unit, giving us the result that $q=cx\implies c$ a unit or $x$ a unit, meaning that $q$ is irreducible, as required.
\subparagraph{Prime} Consider a prime element $p\in R$ an an associate $q\in R$ such that $p=aq$ and $q=bp$ for units $a,b\in R$. Then, consider $cd\in R$ such that $q|cd$. Next, since $q=bp$, we have that $bp|cd\implies p|cd\implies p|c$ or $p|d$ since $p$ is prime. Similarly, we then find that since $p=aq$, we have that $aq|c$ or $aq|d\implies q|c$ or $q|d$, meaning that $q$ is prime as required.
\paragraph{(f)} Consider an irreducible element $p\in R$ and $u,x\in R$ such that $p=ux$. Then, since $p$ is irreducible, let $u$ be a unit without loss of generality. Since $p=ux$, we have that $x=u^{-1}p$, which is the definition of associate. Thus, for any two elements that multiply to $p$, we have that one of them is a unit and the other an associate, which is equivalent to the claim we sought to prove.
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