Proof .

（a）
Solution: It is clear that the set is closed under addition, and the following computation checks closure under multiplication.
$$
\left[\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
b & c & a
\end{array}\right]\left[\begin{array}{lll}
x & 0 & 0 \\
0 & x & 0 \\
y & z & x
\end{array}\right]=\left[\begin{array}{ccc}
a x & 0 & 0 \\
0 & a x & 0 \\
b x+a y & c x+a z & a x
\end{array}\right]
$$
Because of the symmetry $a \leftrightarrow x, b \leftrightarrow y, c \leftrightarrow z$, the above computation also checks commutativity.

（b）

Solution: Four of the matrices in $R$ have 1’s on the diagonal, and these are invertible since their determinant is nonzero. Squaring each of the other four matrices gives the zero matrix, and so they are nilpotent.

Proof . Solution: In $R$ we have $a+a=0$, for all $a \in R$, so $R$ is not isomorphic to $\mathbf{Z}_{4}$. On the other hand, in $R$ we have $[x+1] \neq[0]$ but $[x+1]^{2}=\left[x^{2}+1\right]=[0]$. Thus $R$ cannot be isomorphic to $\mathbf{Z}_{2} \oplus \mathbf{Z}_{2}$, since in that ring $(a, b)^{2}=(0,0)$ implies $a^{2}=0$ and $b^{2}=0$, and this implies $a=0$ and $b=0$ since $\mathbf{Z}_{2}$ is a field.

Proof .

Solution: Let $\phi: \mathbf{Z}_{120} \rightarrow \mathbf{Z}_{42}$ be a ring homomorphism. The additive order of $\phi(1)$ must be a divisor of $\operatorname{gcd}(120,42)=6$, so it must belong to the subgroup $7 \mathbf{Z}_{42}=\{0,7,14,21,28,35\} .$ Furthermore, $\phi(1)$ must be idempotent, and it can be checked that in $7 \mathbf{Z}_{42}$, only $0,7,21,28$ are idempotent.

If $\phi(1)=7$, then the image is $7 \mathbf{Z}_{42}$ and the kernel is $6 \mathbf{Z}_{120}$. If $\phi(1)=21$, then the image is $21 \mathbf{Z}_{42}$ and the kernel is $2 \mathbf{Z}_{120}$. If $\phi(1)=28$, then the image is $14 \mathbf{Z}_{42}$ and the kernel is $3 \mathbf{Z}_{120}$.

Proof .

Solution: The answer is no. The argument can be given using either addition or multiplication. Addition in the two rings is different, since the additive group of $\mathbf{Z}_{9}$ is cyclic, while that of $\mathbf{Z}_{3} \oplus \mathbf{Z}_{3}$ is not. Multiplication is also different, since in $\mathbf{Z}_{9}$ there is a nonzero solution to the equation $x^{2}=0$, while in $\mathbf{Z}_{3} \oplus \mathbf{Z}_{3}$ there is not. $\left(\operatorname{In} \mathbf{Z}_{9}\right.$ let $x=3$, while in $\mathbf{Z}_{3} \oplus \mathbf{Z}_{3}$ the equation
$(a, b)^{2}=(0,0)$ implies $a^{2}=0$ and $b^{2}=0$, and then $a=0$ and $\left.b=0 .\right)$

Proof .

Solution: If $P$ is a prime ideal of $\mathbf{Z}_{36}$, then $\mathbf{Z}_{36} / P$ is a finite integral domain, so it is a field, and hence $P$ is maximal. Thus we only need to find the maximal ideals of $\mathbf{Z}_{36}$. The lattice of ideals of $\mathbf{Z}_{36}$ is exactly the same as the lattice of subgroups, so the maximal ideals of $\mathbf{Z} 36$ correspond to the prime divisors of
36. The maximal ideals of $\mathbf{Z}_{36}$ are thus $2 \mathbf{Z}_{36}$ and $3 \mathbf{Z}_{36}$.
An alternate approach we can use Proposition 5.3.7, which shows that there is a one-to-one correspondence between the ideals of $\mathbf{Z} / 36 \mathbf{Z}$ and the ideals of $\mathrm{Z}$ that contain $36 \mathbf{Z}$. In $\mathbf{Z}$ every ideal is principal, so the relevant ideals correspond to the divisors of 36 . Again, the maximal ideals that contain $36 \mathbf{Z}$ are $2 \mathbf{Z}$ and $3 \mathbf{Z}$, and these correspond to $2 \mathbf{Z}_{36}$ and $3 \mathbf{Z}_{36}$.

Proof .

(a) Show that if $e$ is an idempotent element of $R$, then $1-e$ is also idempotent.
Solution: We have $(1-e)^{2}=(1-e)(1-e)=1-e-e+e^{2}=1-e-e+e=1-e$.
(b) Show that if $e$ is idempotent, then $R \cong R e \oplus R(1-e)$.
Solution: Note that $e(1-e)=e-e^{2}=e-e=0 .$ Define $\phi: R \rightarrow R e \oplus R(1-e)$ by $\phi(r)=($ re,$r(1-e))$, for all $r \in R$. Then $\phi$ is one-to-one since if $\phi(r)=\phi(s)$, then $r e=s e$ and $r(1-e)=s(1-e)$, and adding the two equations gives $r=s .$ Furthermore, $\phi$ is onto, since for any element $(a e, b(1-e))$ we have $(a e, b(1-e))=\phi(r)$ for $r=a e+b(1-e) .$ Finally, it is easy to check that $\phi$ preserves addition, and for any $r, s \in R$ we have $\phi(r s)=(r s e, r s(1-e))$ and $\phi(r) \phi(s)=(r e, r(1-e))(s e, s(1-e))=\left(r s e^{2}, r s(1-e)^{2}\right)=(r s e, r s(1-e))$