2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – rschwieb Jan 9 '17 at 14:42
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @rschwieb, I have corrected, thanks. $\endgroup$ – Andreas Caranti Jan 10 '17 at 8:43
  • $\begingroup$ looks good now! $\endgroup$ – rschwieb Jan 10 '17 at 12:46
  • $\begingroup$ @rschwieb, thank you once more. $\endgroup$ – Andreas Caranti Jan 10 '17 at 12:47
  • $\begingroup$ ok so what is the problem with the usual definition of the prime ideal? $\endgroup$ – Kushal Bhuyan Jan 10 '17 at 14:16
  • $\begingroup$ You could see this discussion. $\endgroup$ – Andreas Caranti Jan 10 '17 at 14:17
2
$\begingroup$

$P$ 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).

$\endgroup$
  • 2
    $\begingroup$ 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.) $\endgroup$ – rschwieb Jan 9 '17 at 14:56
  • $\begingroup$ Thanks a lot for setting me straight! I was not aware that the definition of being prime is different in the non-commutative case. $\endgroup$ – Claudius Jan 9 '17 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.