# prime ideals and maximals

• Jul 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
• Jul 31st 2008, 07:02 PM
JaneBennet
Quote:

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

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

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

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

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

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

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

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

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 $\displaystyle R$ here is not a PID.)