# 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

1. not an ideal of $$S$$
2. an ideal but not a prime ideal of $$S$$
3. is a prime ideal but not a maximal ideal of $$S$$
4. 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.

• You are using the wrong definition of "prime." In a noncommutative ring like this one, it means that for any two ideals $I,J$ with $IJ\subseteq P$, either $I\subseteq P$ or $J\subseteq P$. That these two matrices which aren't in $P$ multiply to something in $P$ is inconclusive. In the full ring of $2\times 2$ matrices, the zero ideal is prime, but there are nonzero things that multiply to zero, nevertheless. – rschwieb Jan 9 '17 at 14:42

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 $\begin{bmatrix}a&c\\0&b\end{bmatrix} \in S$.

• @rschwieb, I have corrected, thanks. – Andreas Caranti Jan 10 '17 at 8:43
• looks good now! – rschwieb Jan 10 '17 at 12:46
• @rschwieb, thank you once more. – Andreas Caranti Jan 10 '17 at 12:47
• ok so what is the problem with the usual definition of the prime ideal? – Kushal Bhuyan Jan 10 '17 at 14:16
• You could see this discussion. – Andreas Caranti Jan 10 '17 at 14:17

$P$ 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).

• This is correct, and avoids the problem existing in the other answer, but it leaves a window for misinterpretation. $P$ can be prime without $S/P$ being a domain (although in this case since $S/P$ is commutative, the two are equivalent.) You can close this window by rewording to "which is not a prime ring (a commutative prime ring is an integral domain, and it is not an integral domain.) – rschwieb Jan 9 '17 at 14:56
• Thanks a lot for setting me straight! I was not aware that the definition of being prime is different in the non-commutative case. – Claudius Jan 9 '17 at 15:44