This page covers section 3.5 (“More Ideals and Quotient Rings”). Throughout, $R$ is a ring.
Lemma 3.5.1: Let $R$ be commutative and unital and only have the ideals $\left(0\right)$ and $R$. Then $R$ is a field.
Definition: An ideal $I\subset R$ is a maximal ideal if any other ideal $J$ such that $I\subset J\subset R$ is either $J=I$ or $J=R$.
The prime ideals of $\mathbb{Z}$ are exactly the ideals $\left(p\right)=\{kp\mid k\in \mathbb{Z}\}$ for $p$ a prime integer. See also: problem 3.4.8.
Theorem 3.5.1: Let $R$ be commutative and unital and let $I$ be an ideal of $R$. Then $R\mathrm{/}M$ is maximal if and only if $R\mathrm{/}M$ is a field.
The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book.
Let $a\mathrm{\ne}0$ be an element of $R$ and consider $aR=\{ar\mid r\in R\}$ which was shown to be a right ideal in problem 3.4.14. Because $1\in R$, we have that $a1\mathrm{\ne}0$ is a non-zero element of $aR$. By the conditions of the problem, we conclude that $aR=R$. In particular, there exists $b\in R$ such that $ab=1$. We would like to show also that $ba=1$. If that is the case, then we have exhibited a multiplicative inverse in $R$ for an arbitrary non-zero ring element.
Considering $bR$ and arguing as above, we find an element $c\in R$ such that $bc=1$. However, we see that $abc=a\left(bc\right)=a1=a$ while $abc=\left(ab\right)c=1c=c$ so that $a=c$. Therefore, $ab=ba=1$ and we have proved that every non-zero element of $R$ has a multiplicative inverse, so that $R$ is a division ring.
This is related to problem 3.5.1 in that we only relax the condition on $R$ being unital. My solution was heavily influenced by the nice solution by AllTheCheese in this forum thread.
Case 1: there exists some $a\in R$ such that $aR=R$. In that case, there exists an element $e\in R$ such that $ae=a$. We would like to show that this $e$ is the multiplicative unit for the ring, which will put us in the realm of problem 3.5.1. First recall the annihilator $\mathrm{A}\mathrm{n}\mathrm{n}\left(a\right)=\{x\in R\mid ax=0\}$ shown in problem 3.4.17 to be a right ideal of $R$. As $e\ue020\in \mathrm{A}\mathrm{n}\mathrm{n}\left(a\right)$, we must have that $\mathrm{A}\mathrm{n}\mathrm{n}\left(a\right)=\left(0\right)$, so $a$ annihilates nothing but the zero element. However, right-multiplying the equation $ae=a$ by $a$, we find $0=aea-aa=a(ea-a)\mathrm{.}$
In light of the comment about $\mathrm{A}\mathrm{n}\mathrm{n}\left(a\right)$, we have that $ea=a$.
Now take any other element $b\in R$; we would like to show that $eb=be=e$ so that $e$ is truly the unit element. Recalling that $aR=R$, there exists $x\in R$ such that $ax=b$, so $eb=eax=ax=b$. It is more difficult to show that $be=b$.
Let $be=c$ and multiply on the right by $e$ to see that $be=ce$ (${e}^{2}=e$, see below for the proof*). Then $(b-c)e=0$. Recall problem 3.4.18 which tells us that $\lambda \left(Re\right)=\{x\in R\mid xre=0$ for all $r\in R\}$ is a right ideal. Putting $x=a$ and $r=e$, we see that $aee=ae=a\mathrm{\ne}0$, so there exists at least one element of $R$ which doesn’t belong to $\lambda \left(Re\right)$. With our strict condition on ideals, this means that $\lambda \left(Re\right)=\left(0\right)$. In other words, there does not exist a non-zero element $x\in R$ for which $xre=0$ for all $r\in R$. Letting $x=b-c$, and assuming $x\mathrm{\ne}0$, we see that there is at least some $r\in R$ such that $(b-c)re\mathrm{\ne}0$, which forces us to conclude that $\mathrm{A}\mathrm{n}\mathrm{n}(b-c)=\left(0\right)$. However, we have already seen that $(b-c)e=0$, a contradiction. The only way out is to have that $c=b$, which shows that $be=b$, after all. Now we have shown that case 1 leads to the ring having a unit, and therefore being a division ring by problem 3.5.1.
Case 2: there is no $a\in R$ such that $aR=R$; this is the complement of case 1. Then $aR=\left(0\right)$ always so that $ab=0$ for any $a,b\in R$. Now, let $a\in R$ be a non-zero element. If we consider the set $I=\{0,a,2a,3a,\dots \}$, we see that it is a right ideal because multiplication from the outside always yields $0$ which belongs to $I$, while closure under addition is clear. Also, $I$ contains the non-zero element $a$, so $I=R$. Now, if $\mathrm{\mid}R\mathrm{\mid}$ is finite, the sequence $0,\text{\hspace{0.17em}}a,\text{\hspace{0.17em}}2a\text{\hspace{0.17em}}\dots $ will yield $\mathrm{\mid}R\mathrm{\mid}$ distinct elements and then loop around to elements it has already visited. In particular, if $\mathrm{\mid}R\mathrm{\mid}$ is composite, say $\mathrm{\mid}R\mathrm{\mid}=mn$ with $m,n>1$ integers, then $ma$ is non-zero, and $J=\{0,ma,2ma,\dots \}$ is a right ideal with a non-zero element, but it is a proper subset of $I$. By the condition of the problem, such a $J$ may not exist. Therefore if $R$ is in case 2 and is finite, then $\mathrm{\mid}R\mathrm{\mid}$ must be prime.
Finally, we must also consider case 2 with $\mathrm{\mid}R\mathrm{\mid}$ infinite. In fact, this is not possible. Simply consider the ideal $J=\{0,2a,4a,\dots \}$ which is a proper subset of $I$ with a non-zero element. As before, this is a contradiction of the terms of the problem.
Here is a brief digression from the main course of the proof to show that ${e}^{2}=e$ for the alleged identity element $e$ of $R$. Recall that the non-zero element $a$ satisfied $ae=ea=a$ and $ax=0$ implied $x=0$. Now, $ea\mathrm{\ne}0$ implies that $eR\mathrm{\ne}\left(0\right)$, so $eR=R$. In particular, there exists $d\in R$ with $ed=e$. Multiplying on the left by $a$ gives $0=aed-ae=ad-ae=a(d-e)$. As $a$ only annihilates the zero element, we must have that $d=e$, so ${e}^{2}=e$ as claimed.
(a) There is a natural homomorphism to consider, namely $\varphi :\mathbb{Z}\mathrm{/}\left(p\right)\to {\mathbb{Z}}_{p}$
given by $\varphi (a+(p\left)\right)=a\phantom{\rule{0ex}{0ex}}\phantom{\rule{0.6666666666666666em}{0ex}}\mathrm{m}\mathrm{o}\mathrm{d}\text{\hspace{0.17em}}\text{\hspace{0.17em}}p$, where on the right hand side $a$ is reduced modulo $p$ so as to live in ${\mathbb{Z}}_{p}$. To show that this is well-defined, let $a,{a}^{\mathrm{\prime}}\in R$ be such that $a+\left(p\right)={a}^{\mathrm{\prime}}+\left(p\right)$. Then $a-{a}^{\mathrm{\prime}}\in \left(p\right)$ so they differ by a multiple of $p$ and the map is well-defined. It is also easily seen to be a homomorphism. Because the kernel is trivial, $\varphi $ is an isomorphism.
(b) By problem 3.4.8, we know that $\left(p\right)$ is a maximal ideal of $\mathbb{Z}$, and by theorem 3.5.1, we therefore have that $\mathbb{Z}\mathrm{/}\left(p\right)$ is a field, because it is a commutative unital ring ($\mathbb{Z}$) modded out by a maximal ideal.
First we note that an invertible element in this ring is one which has no zeroes. The multiplicative identity element is the constant function $x\mapsto 1$, and if $f$ has no roots then it has a multiplicative inverse given by $x\mapsto 1\mathrm{/}f\left(x\right)$. If an ideal contains an invertible element, then it is all of $R$. Therefore every element of a proper ideal must have a root.
The sketch of the proof is as follows:
(1) Let $f,g\in I$. We have that ${f}^{2}$, ${g}^{2}$ and therefore ${f}^{2}+{g}^{2}$ are also in $I$ because it is an ideal. Suppose that $f$ and $g$ share no root. Then at any root $x$ of $f$, we have $f(x{)}^{2}+g(x{)}^{2}=g(x{)}^{2}>0$. Similarly $f(x{)}^{2}+g(x{)}^{2}=f(x{)}^{2}>0$ at any root $x$ of $g$. Of course, ${f}^{2}+{g}^{2}>0$ elsewhere. Therefore if $f$ and $g$ share no root, then $I$ contains the invertible element ${f}^{2}+{g}^{2}$, and hence $I=R$. We are interested in proper ideals of $R$, in which, as we have just seen, any pair of elements has a root in common.
We can say something more general than the pairwise statement, though. Given any finite subset of a proper ideal $I$, say $\{{f}_{1},\dots ,{f}_{n}\}\subset I$, the same argument shows that all members of that subset share at least one zero. Otherwise ${\sum}_{i}{f}_{i}^{2}$ would be an invertible element which forces $I=R$.
With $f\in I$ and writing ${Z}_{f}$ for the (non-empty) set of zeroes of $f$, we have just shown that the collection $\{{Z}_{f}\mid f\in I\}$ has the finite intersection property. Any finite subcollection of that collection has non-empty intersection.
(2) We should now like to extend the statement above to cover the existence of a root shared by everything in the ideal. That is, we would like to prove that ${Z}_{I}=\bigcap _{f\in I}{Z}_{f}\mathrm{\ne}\mathrm{\varnothing}\mathrm{.}$
It is not true in general that the finite intersection property extends to a non-empty intersection on the whole collection. However, we have additional conditions at our disposal.
Let $f\in R$. Because $f$ is continuous, we know that the inverse image under $f$ of an open set is open. In particular, ${f}^{-1}(\mathbb{R}\setminus 0)$ is open. Its complement, ${f}^{-1}\left(0\right)={Z}_{f}$, is therefore closed in $[0,1]$. That is, every zero set ${Z}_{f}$ that we consider is topologically closed.
Additionally, $[0,1]$ is a compact set. In a compact metric space such as $[0,1]$, any collection of closed sets with the finite intersection property has a non-empty intersection (proof). This follows simply by taking the definition that “every open cover of a compact set has a finite subcover” and turning it into a statement about the closed sets given by the complements of the sets in the cover. Therefore we have that, for any proper ideal $I$ of $R$, there is a nonempty set ${Z}_{I}$ on which every member of $I$ takes the value zero.
(3) Now we want to determine what a maximal ideal $M$ looks like. If $\mathrm{\mid}{Z}_{M}\mathrm{\mid}>1$, then $M$ may not be maximal. If $\mathrm{\mid}{Z}_{M}\mathrm{\mid}>1$, then we could take a proper, non-empty subset ${Z}_{J}\subset {Z}_{M}$ and construct the proper ideal $J=\{f\in R\mid f({Z}_{J})=0\}$. If $f\in M$, then $f\left({Z}_{J}\right)\subset f\left({Z}_{M}\right)=0$, so we also have $f\in J$ and therefore $M\subset J$. In other words, reducing the size of the zero set makes the ideal bigger — the bigger ideal contains all the functions that were zero on the expanded zero set, but it also contains new functions that take other values on the roots that were excluded. On a side note, it may be true that zero sets are in one-to-one correspondences with ideals, but I do not know at this time. For the purposes of this exercise, it suffices to construct one ideal from a given zero set, as we constructed $J$ earlier in this paragraph.
Thus it must be the case that a maximal ideal $M$ has exactly one shared root among its elements, $\mathrm{\mid}{Z}_{M}\mathrm{\mid}=1$. Calling that root $\gamma $, we must have $M$ as a subset of ${M}^{\mathrm{\prime}}=\{f\in R\mid f(\gamma )=0\}$. However, we easily verify that ${M}^{\mathrm{\prime}}$ is itself an ideal, so any proper subset of it would clearly not be maximal. Therefore, our maximal ideal is ${M}^{\mathrm{\prime}}$ and the claim is proven.