Cardinality of the quotient of $M_2(\mathbb{Z})$ by the ideal of matrices with even entries 
Let $R=\left\{\begin{pmatrix}
a_1&a_2\\a_3&a_4
\end{pmatrix} \mid a_i\in \mathbb{Z} \right\}$ and let $I$ be the subset of $R$ with even entries. Show that $I$ is an ideal of $R$. What is the cardinality of $R/I$?

This is how I prove it:
Let $x=\{\left[\begin{array}{rr}
2b_1&2b_2\\2b_3&2b_4
\end{array}\right]|b_i\in Z\}\in I$ and let $y=\{\left[\begin{array}{rr}
a_1&a_2\\a_3&a_4
\end{array}\right]|a_i\in Z\}\in R$.
Note: To show that I is and ideal of R , $xy\subseteq I$ and $yx\subseteq I$ for all $y\in R$.
First, we try $xy\subseteq I$, that is
$xy=\left[\begin{array}{rr}
2b_1&2b_2\\2b_3&2b_4
\end{array}\right]\left[\begin{array}{rr}
a_1&a_2\\a_3&a_4
\end{array}\right]=\left[\begin{array}{rr}
2(b_1a_1+b_2a_3)&2(b_1a_2+b_2a_4)\\2(b_3a_1+b_4a_3)&2(b_3a_2+b_4a_4)
\end{array}\right]\in I$
Now for $yx\subseteq I$.
$yx=\left[\begin{array}{rr}
a_1&a_2\\a_3&a_4
\end{array}\right]\left[\begin{array}{rr}
2b_1&2b_2\\2b_3&2b_4
\end{array}\right]=\left[\begin{array}{rr}
2(a_1b_1+a_2b_3)&2(a_1b_2+a_2b_4)\\2(a_3b_1+a_4b_3)&2(a_3b_2+a_4b_4)
\end{array}\right]\in I$
Since its satisfy the condition, therefore I is an ideal of R.
Now for finding the cardinality of R/I, I'm having a hard time solving for it. Can someone please show me how to get it. Thank you so much!
 A: To compute the cardinality of the quotient, simply consider $R,I$ as abelian groups. $R$ is isomorphic to $\Bbb{Z}^4$, while $I$ is isomorphic to $2\Bbb{Z}^4$.
Every coset has a representative of the form
$$\left[ \begin{matrix} a_1 & a_2 \\ a_3 & a_4 \end{matrix} \right]$$
where $a_1 , a_2 , a_3 , a_4 \in \{ 0,1 \}$. This means that the cosets are exactly $2^4=16$.
Another way to see this is to find a group isomorphism
$$\Bbb{Z}^4/2\Bbb{Z}^4 \cong (\Bbb{Z}/2\Bbb{Z})^4$$
which can be found using the first isomorphism theorem on the projection $\Bbb{Z}^4 \to (\Bbb{Z}/2\Bbb{Z})^4$.
A: We can see that $I=2R$, and taking the quotient sends the matrices with all even entries to zero. The quotient ring consists of all the cosets of $I$, so all you need to do is identify the different ways a matrix can fail to have all even entries. For example, 
$\left[\begin{array}{rr}
2&3\\7&7
\end{array}\right]$ would be in the coset defined by $I+\left[\begin{array}{rr}
0&1\\1&1
\end{array}\right]$. Can you see how to find the other cosets?
A: It suffices to only consider $R/I$ as an abelian group. As an abelian group, we have $R=\mathbb Z^4$ and $I=2R$, i.e. $R/I=(\mathbb Z/(2))^4$ with $16$ elements.
