Proof of Uniqueness of Ideal Let $R$ be the subring of the ring $M_3$$(\mathbb{R})$ of real $3\times3$ matrices consisting of matrices of the form
$R=\begin{pmatrix} a & b & c \\ 0 & a & b \\ 0 & 0 & a \end{pmatrix}$ where $a,b,c \in \mathbb{R}$.
Let $X$ be the set of non-invertible elements of R.
This was a multi-step question. First, it was asked to show that $X$ is an ideal of $R$ which was done by first defining X as the set of matrices with $detX=0$ showing that the elements that belong to $X$ are of the form $X= \begin{pmatrix} 0 & b & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}$. Then I showed that $X$ is closed under subtraction and obviously $X \neq$ $\emptyset$. Then I showed that $X$ is a left ideal and a right ideal, and thus an ideal.
Then, I was asked to exhibit an ideal of $R$ which is different from zero, $X$, and $R$. In order to do show, I followed a "trial-and-error" approach, namely, trying left-multiplication and right multiplication of matrices having the product belong to the ideal we are testing: $y*r=yr\in Y$ and $r*y=ry \in Y$ and $yr=ry$. After some work I showed that
$Y=\begin{pmatrix} 0 & 0 & c \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ is also an ideal of $R$.
The question about which I am confused is to show that $Y$ is the only ideal of $R$ different from zero, $X$, and $R$. Is my work of trial and error enough or should I try a different approach?
 A: Trial-and-error is a way to come up with solutions, or to construct counterexamples (for disproofs) but it is rarely (never?) adequate for proof.  But it may point to a proof.
Let $I$ be an ideal. If it has a nonzero value for $a$ (which means it contains a unit), then it's all of $R$.
If not, then it must be contained in $X$. Show that if it contains an element with nonzero value for $b$, then it is all of $X$.
If it does not have a nonzero value for $b$, then it is contained in $Y$. It is not hard to show that if it has a nonzero value for $c$ then it is all of $Y$.
Otherwise it has zero values for $a$, $b$, and $c$ in which case it is finally $\{0\}$.
Really the best description of this is process of elimination as opposed to trial-and-error.

It may also be interesting for you to note that this ring is isomorphic to $\mathbb R[x]/(x^3)$ via the map $a+bx+cx^2+(x^3)\mapsto \begin{bmatrix}a&b&c\\0&a&b\\0&0&a\end{bmatrix}$.
If you are familiar with the fact that because $\mathbb R[x]$ is a principal ideal domain, the ideals of $\mathbb R[x]/I$ correspond to distinct divisors of the element generating $I$, then you can also note that the ideals of your ring would correspond to divisors of $x^3$: $1, x,x^2,x^3$ corresponding to $R, X ,Y$ and $\{0\}$ respectively.
