# prime ideals and maximals

• July 31st 2008, 10:52 AM
Ryan0710
prime ideals and maximals
Let
R ={
a b
0 a| a, b Z}

��
Then
R is a ring with unity. Show that

P
={

0
b

0 0
| b Z}

is a prime ideal which is not maximal. Carefully justify all your claims.
[Hint:
Use the mapping ψ : R Z given by

ψ{

a b
0
a}

= a.]
a b
0 a are all matrices
• July 31st 2008, 07:02 PM
JaneBennet
Quote:

Originally Posted by Ryan0710
Let
$R =\left\{\begin{pmatrix}a & b\\0&a\end{pmatrix}:a,b\in\mathbb{Z}\right\}$

��
Then
R is a ring with unity. Show that

$P =\left\{\begin{pmatrix}0 & b\\0&0\end{pmatrix}:b\in\mathbb{Z}\right\}$

is a prime ideal which is not maximal. Carefully justify all your claims.

Hint:
Use the mapping $\psi:R\to\mathbb{Z}$ given by

$\psi\begin{pmatrix}a & b\\0 & a\end{pmatrix}=a$

$P$ is the kernel of the homomorphism $\psi$; hence the quotient ring $R/P$ is isomorphic to $\psi(R)=\mathbb{Z}$. Since $\mathbb{Z}$ is an integral domain, so is $R/P$, i.e. $P$ is a prime ideal.

Note that $Q=\left\{\begin{pmatrix}2a & b\\0&2a\end{pmatrix}:a,b\in\mathbb{Z}\right\}$ is an ideal of $R$ such that $P\subsetneq Q\subsetneq R$. Hence $P$ is not maximal in $R$.

Thus while every maximal ideal is prime, this example shows that the converse is not true in general. (It’s true in a principal-ideal domain, but $R$ here is not a PID.)