Ideal of the ring of upper triangular matrices 
Let $S=\left\{\begin{bmatrix}a&b\\0&c\end{bmatrix}: a,b,c\in \mathbb{R} \right\}$ be a ring under matrix addition and multiplication. Then the subset $P=\left\{\begin{bmatrix}0& p\\ 0&0\end{bmatrix}:p\in \mathbb{R}\right\}$ is 
  
  
*
  
*not an ideal of $S$
  
*an ideal but not a prime ideal of $S$
  
*is a prime ideal but not a maximal ideal of $S$
  
*is a maximal ideal of $S$.
  

It is obvious that $P$ is an ideal of $S$. So all we need to determine is whether it is prime ideal or maximal ideal of $S$. Now we see that-
$$\begin{bmatrix}0&a\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&a\end{bmatrix}\in P$$ but $\begin{bmatrix}0&0\\0&a\end{bmatrix}\notin P$. Here I have a minor confusion. For and ideal to be a prime ideal, if $ab\in P$ then either $a$ or $b$ has to be in $P$. In this case, the above mentioned matrix is not in $P$, does it implies $P$ is not a prime ideal? Also I don't know how to show $P$ is maximal or not. So can anyone help me on this? Thanks.
 A: This is a fix of a horribly wrong first attempt - thanks to @rschwieb for notifying me.
We are taking the following definition of prime ideal in a possibly non-commutative ring.

The ideal $P$ of the ring $S$ is prime iff for each $x, y \in S$, if $x S y \subseteq P$, then either $x \in P$ or $y \in P$.

We have
$$
x = \begin{bmatrix}1&0\\0&0\end{bmatrix} \notin P \qquad y = \begin{bmatrix}0&0\\0&1\end{bmatrix} \notin P,
$$
but 
$$
x \begin{bmatrix}a&c\\0&b\end{bmatrix} y
=
\begin{bmatrix}0&c\\0&0\end{bmatrix} \in P
$$
for all $\begin{bmatrix}a&c\\0&b\end{bmatrix} \in S$.
A: $P$ is not a prime ideal. Consider for example the homomorphism of rings
$$
S\longrightarrow \mathbb R\times \mathbb R,\quad \begin{bmatrix}a & b\\ 0 & d\end{bmatrix} \longmapsto (a,d),
$$
where the operations on $\mathbb R\times \mathbb R$ are componentwise. This homomorphism is clearly surjective with kernel $P$. Hence $S/P\cong \mathbb R\times \mathbb R$, which is not an integral domain. Therefore, $P$ is not a prime ideal (and in particular not maximal).
