Show that $A = \{(3x,y)~|~ x,y \in Z\}$ is a maximal ideal of $Z \oplus Z$. My Attempt Shown Show that $A = \{(3x,y)~|~ x,y \in Z\}$ is a maximal ideal of $Z \oplus Z$. Here's my attempt, Please tell me where did I go wrong.
Attempt: When $R$ is a commutative ring with unity and $I$ is any ideal, then $R/I$ is a field if and only if $I$ is the maximal ideal
Now, $A = \{3Z \oplus Z\}$ and : 
$(Z \oplus Z) / (3Z \oplus Z) = \{(a,b) + (3Z \oplus Z) ~|~a,b \in Z\}$
$= \{(a+ 3Z) \oplus (b+Z)~|~a,b \in Z\}$
Now, since $Z + 3Z = Z/3Z \approx Z_3$ and $a \in Z$
$\implies (Z \oplus Z) / (3Z \oplus Z) \approx Z_3 \oplus (Z \oplus Z)$
Is my attempt correct? Thank you.
 A: The last line "$\implies (Z \oplus Z) / (3Z \oplus Z) \approx Z_3 \oplus (Z \oplus Z)$" does not make much sense. It certainly doesn't accomplish the goal you had of proving the quotient is a field, so it should definitely be reconsidered.
Given an element of $\{(a,b) + (3Z \oplus Z) ~|~a,b \in Z\}$, the element is equivalent to one of the form $(a,0)+(3Z+Z)$. This sugests a candidate for an homomorphism from $Z\oplus Z\to Z_3$ given by $(a,b)\mapsto a+3Z$. What's the kernel?

We could correct the line of thought you were pursuing this way:
$$(Z \oplus Z) / (3Z \oplus Z) = \{(a,b) + (3Z \oplus Z) \mid a,b \in Z\}$$
$$= \{(a+ 3Z) \oplus (b+Z)\mid a,b \in Z\}\\
= \{(a+ 3Z)\mid a \in Z\} \oplus \{(b+Z)\mid b \in Z\}\\
=Z_3\oplus \{0\}\cong Z_3$$
A: Try to find a field $\;\Bbb F\;$ s.t. you can define a homomorphism $\;\phi:\Bbb Z\oplus\Bbb Z\to \Bbb F\;$ with $\;\ker\phi= A\;$, and looking at the form of the elements in $\;A\;$ a good idea is to choose $\;\Bbb F=\Bbb F_3\;$:
$$\phi(n,m):=n\pmod 3$$
Do some checking:
$$\begin{align*}\bullet&\;\;\phi((n,m)+(n',m')):=\phi(n+n',m+m')=(n+n')\pmod 3=\\
&\;\;=n\pmod3+n'\pmod3=\phi(n,m)+\phi(n',m')\\{}\\
\bullet&\;\;\phi((n,m)(n',m'))=\phi(nn',mm'):=nn'\pmod 3=\\
&\;\;=n\pmod3\cdot n'\pmod3=\phi(n,m)\phi(n',m')\pmod3\\{}\\
\bullet&\;\;(n,m)\in\ker\phi\iff n=0\pmod 3\iff n=3x\,,\,\,x\in\Bbb Z\iff\\&\;\;(n,m)=(3x,m)\iff(n,m)\in A\end{align*}$$
