an elementary ring problem Problem:
Let $a, b \in \mathbb{Z}$. the goal of this problem is to define the set $E$ of elements of  $\mathbb{Z}$ congrus to $a$ modulo 2 and to $b$ modulo 3 .

*

*Show that
$$
\begin{aligned}
\varphi: \mathbb{Z} & \longrightarrow \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z} \\
& x \quad \mapsto \quad(x+2 \mathbb{Z}, x+3 \mathbb{Z})
\end{aligned}
$$
is a ring homomorphism


*Determine the kernel of $\varphi$.


*Show that the couple $(a+2 \mathbb{Z}, b+3 \mathbb{Z}) \in \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}$ admits an antecedent by $\varphi$. (we can use Bézout's identity)
What I could answer:
1.
we know that $(\mathbb{Z} / n \mathbb{Z},+, \times) $ is a commutative ring, in particular $(\mathbb{Z} / 2 \mathbb{Z},+, \times)$ and $(\mathbb{Z} / 3 \mathbb{Z},+, \times)$ are commutative rings,
we need to prove that
\begin{array}{l}
\text { - } \forall x, y \in Z, \varphi(x+y)=\varphi(x)+\varphi(y) \\
\text { - } \forall x, y \in Z, \varphi(x \times y)=\varphi(x) \times \varphi(y) \\
\text { - } \varphi(1_Z)=1_{ \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}}
\end{array}
$\varphi(x+y) =  (x+y + 2 \mathbb{Z}, x+y + 3 \mathbb{Z}) \\ \iff \varphi(x+y) =  (x+2\mathbb{Z} + y+2\mathbb{Z}, x+3 \mathbb{Z} + y+3 \mathbb{Z}) \\ \iff \varphi(x+y) =  \varphi(x) + \varphi(y)$
since in $\mathbb{Z} / 2 \mathbb{Z}$ and in $\mathbb{Z} / 3 \mathbb{Z}$ we have:
$\forall (\overline{a},\overline{b} ) \in (\mathbb{Z} / 2 \mathbb{Z})²$ or in $(\mathbb{Z} / 3 \mathbb{Z})² \\ \overline{a} + \overline{b}= \overline{a + b}$
since $\exists k, l \in \mathbb{Z}, x=a+k n \text { et } y=b+\ln $
so $x+y = a+b + (k+l) n $
Also, in the same way, we also have $\overline{a} \times \overline{b} = \overline{a \times b} $ since
$\exists k, l \in \mathbb{Z}, x=a+k n \text { et } y=b+\ln \\ x \times y=a \times b+(k b+a l+k l) n$
For the neutral element with respect to the multiplicative operation,
We have
$\varphi(1_Z) = (1,1) = 1_{ \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}}$
2.(not sure if this  is right or if I can write it better)
The kernel of $\varphi$ is the set of elements that are multiples of $3$ and $2$ at the same time so it's the set { $x \quad  | \quad x = 6 k, \forall k \in Z$ }
3. (I don't know what to prove here with Bézout's identity that states : $\exists u,v \in Z² \quad au+bv=\mathbf{1} \Leftrightarrow pgcd(a ; b)=1$)
we have that $card(\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}) \leq card(Z)$ and by definition of $\varphi$ we can see that each element of $\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}$ has at least one antecedent from $Z$.
Any mistake pointing , indications or suggestions are welcome, thank you so much.
 A: I think you have all the right ideas, but your phrasing is a bit awkward and hard to read in a few places. For example in the first section, you don't need to introduct $k,l,n$ and all these extra letters. E.g. for additivity, you can just say something like:
\begin{align*}
\varphi(x+y) &= ((x+y) + 2\mathbb{Z},(x+y) + 3\mathbb{Z})\\
&=((x + 2\mathbb{z}) + (y + 2\mathbb{Z}), (x + 3\mathbb{z}) + (y + 3\mathbb{Z}))\\
&=(x + 2\mathbb{z}, x + 3\mathbb{z}) + (y + 2\mathbb{z}, y + 3\mathbb{z})\\
&=\varphi(x) + \varphi(y)
\end{align*}
Which is basically what you wrote in the first few lines of your explanation. Presumably you're only working on this problem after having spent some time at some point working with rings of the form $n\mathbb{Z}$, and you don't need to go back to the essentials every time.
For 2), I think you can make the idea more precise in the following way:
Fix $x \in \ker\varphi$. Then
$$
(\bar{x},\bar{x}) = (\bar{0},\bar{0})
$$
So we have that $x \in 2\mathbb{Z}$ and $3\mathbb{Z}$, which implies $x \in 6\mathbb{Z}$.
For the third problem, fix some $(\bar{a},\bar{b}) \in \mathbb{Z}/2\mathbb{Z}, \times \mathbb{Z}/3\mathbb{Z}$. We want to find prove that there exists some $x \in \mathbb{Z}$ such that $x \equiv a \mod 2$ and $x \equiv b \mod 3$. A really simple way to do this is to let $x=b$ if $a$ and $b$ have the same parity and let $x = b+3$ if they have different parities. In either case $\varphi(x) = (a+2\mathbb{Z},b+\mathbb{Z})$ as desired.
Your argument about cardinality does not make sense to me.
