Let $R$ be a ring and $I$ be the ideal of $R$ as follows. Edit: The matrices is $2\times 2$.
Let $R$ be a ring of upper triangular matrix with entries $\mathbb{Z}_6$ and $I$ be an ideal in $R$ with entries $3\mathbb{Z}_6$. Find the quotient ring $R/I$. Is $R$ field?
To start, first I note that $R$ has 125 elements, but $I$ has 8 elements. So, I can't find the $R/I$ since 8 is not divides 125. Any ideas?
 A: If $R/I$ is a field then it is at least commutative, hence
$$(A+I)(B+I)=(B+I)(A+I)\iff AB+I=BA+I\\\iff AB-BA=I\iff AB-BA\in I,$$
for any $A,B\in R$. So let
$$A=\pmatrix{a_1 & a_2 \\ 0 & a_3}, B=\pmatrix{b_1 & b_2 \\ 0 & b_3}.$$
After multiplying and subtracting we get
$$AB-BA=\pmatrix{0 & a_1 b_2+a_2b_3 - b_1a_2-b_2a_3 \\ 0& 0}.$$
We can rewrite the top-right entry as $b_2(a_1-a_3)+a_2(b_3-b_1)$, so $AB-BA$ will be in $I$ if and only if $b_2(a_1-a_3)+a_2(b_3-b_1)\equiv 0,3 \mod 6$.
But we can take $b_2=0$, $a_2=2$, $b_3=2$, $b_1=1$ to get $0+2(2-1)=2$ which gives
$$AB-BA=\pmatrix{0 & 2 \\ 0& 0}$$
which is not in $I$.
So the elements of $R/I$ represented by $\pmatrix{p & 2 \\ 0 & q}$ and $\pmatrix{1 & 0 \\ 0 & 2}$ do not commute, therefore $R/I$ is not a field.
A: 
Let $R$ be a ring of upper triangular matrix with entries $\mathbb{Z}_6$ and $I$ be an ideal in $R$ with entries $3\mathbb{Z}_6$. Find the quotient ring $R/I$. Is $R$ field?

Hm, I guess you meant "is $R/I$ a field?" Either way, there is a simple way to see why it is not.
In both $R$ and $R/I$ you have an element that looks like $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$. Really in the case of $R$ it should be the element of $\mathbb Z_6$, and in the case of $R/I$ it would be an element of $\mathbb Z_6/3\mathbb Z_6$, but either way $A$ isn't zero.
Now notice $A^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$.  But of course a field can't have nonzero elements that square to zero. Why? Multiplying $A^2=0$ on the left with the inverse of $A$ would create a contradiction.
