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which of the following rings is a PID?

$1$. $\mathbb{Q}[x,y]/(x)$.
$2$. $\mathbb{Z} \times \mathbb{Z} $
$3$.$\mathbb{Z}[x]$
$4$. $M_2(\mathbb{Z})$,the ring of $2 \times 2$ matrices with entries in $\mathbb{Z}$


My thoughts:-

1.$\mathbb{Q}[x,y]/(x)$ is isomorphic to $\mathbb{Q}[x]$.since $\mathbb{Q}$ is field it is a PID. 2.not sure.
3.since $\mathbb{Z}$ is not a field it is a not a PID.
4.not sure.

can someone help me please

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3  
2. and 4. aren't domains. –  martini Jun 27 '13 at 11:41
    
But only one answer need to be true.so $1$ is true. am I right? –  sumo Jun 27 '13 at 11:43
1  
1 is true, correct, as $\mathbb Q[x,y]/(x) \cong \mathbb Q[y]$ is a pid, since $\mathbb Q$ is a field, as you wrote above. –  martini Jun 27 '13 at 11:44
    
A good exercise is finding an explicit non-principal ideal in $\Bbb Z[x]$ –  Andrea Mori Jun 27 '13 at 12:00
1  
@rschwieb done. –  martini Jun 27 '13 at 12:40

1 Answer 1

To be a PID two things have to be true: (1) The ring in question must be a domain, that is mustn't have zero divisors. (2) All ideals have to be principal ideals.

As you correctly write (1)+(2) are fulfilled for 1., as $\mathbb Q[x,y]/(x) \cong \mathbb Q[y]$, and (2) isn't fulfilled for 3. as $(2,X) \subseteq \mathbb Z[X]$ is no principal ideal. For the other two rings, (1) isn't true, as they have zero divisors.

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