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Primary ideals in Z

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
In Dummit and Foote on page 682 Example 1 reads as follows:
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The primary ideals in [TEX] \mathbb{Z} [/TEX] are 0 and the ideals [TEX] (p^m) [/TEX] for p a prime and [TEX] m \ge 1 [/TEX].


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So given what D&F say, (4) is obviously not primary.

I began trying to show from definition that (4) was not a primary from the definition, but failed to do this

Can anyone help in this ... and come up with an easy way to show that (4) is not primary?

Further, can anyone please help me prove that the primary ideals in [TEX] \mathbb{Z} [/TEX] are 0 and the ideals [TEX] (p^m) [/TEX] for p a prime and [TEX] m \ge 1 [/TEX].


Peter


Note: the definition of a primary idea is given in D&F as follows:

Definition. A proper ideal Q in the commutative ring R is called primary if whenever [TEX] ab \in Q [/TEX] and [TEX] a \notin Q [/TEX] then [TEX] b^n \in Q [/TEX] for some positive integer n.

Equivalently, if [TEX] ab \in Q [/TEX] and [TEX] a \notin Q [/TEX] then [TEX] b \in rad \ Q [/TEX]
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
In Dummit and Foote on page 682 Example 1 reads as follows:
----------------------------------------------------------------------------------------------------------------------------


The primary ideals in [TEX] \mathbb{Z} [/TEX] are 0 and the ideals [TEX] (p^m) [/TEX] for p a prime and [TEX] m \ge 1 [/TEX].


-----------------------------------------------------------------------------------------------------------------------------

So given what D&F say, (4) is obviously not primary.
But (4) is primary, because $4 = 2^2$, which is a power of a prime.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
But (4) is primary, because $4 = 2^2$, which is a power of a prime.
oh! You are right of course!

Thanks so much for that! :-(

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
oh! You are right of course!

Thanks so much for that! :-(

Peter
Are you able to indicate how a proof regarding the nature of the ideals in Z in general might go?


Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Are you able to indicate how a proof regarding the nature of the ideals in Z in general might go?
To show that $(4)$ is primary, we have to show that if [TEX] ab \in (4) [/TEX] and [TEX] a \notin (4) [/TEX] then [TEX] b^n \in (4) [/TEX] for some positive integer $n.$

So suppose that $ab\in(4)$, in other words $ab$ is a multiple of $4$, and that $a\notin(4)$. Thus $a$ could be an odd number, or an odd multiple of $2$, but not a multiple of $4$. It follows that $b$ cannot be an odd number, so it must be a multiple of $2$, say $b=2c.$ Then $b^2 = 4c^2\in(4)$.

The same argument applies for any prime power $p^m$. If $ab\in(p^m)$ but $a\notin(p^m)$ then $b$ must be a multiple of $p$ and so $b^m\in(p^m).$

To show that $(6)$, for example, is not primary, use the fact that $6=2\times3$. If you take $a=2$ and $b=3$ in the definition of a primary ideal, you see that $2$ is not a multiple of $6$, so the definition asks that some power of $3$ should be a multiple of $6$. But that cannot happen, because every power of $3$ is an odd number. Therefore the definition of primary is not satisfied and so $(6)$ is not a primary ideal of $\mathbb Z.$
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
To show that $(4)$ is primary, we have to show that if [TEX] ab \in (4) [/TEX] and [TEX] a \notin (4) [/TEX] then [TEX] b^n \in (4) [/TEX] for some positive integer $n.$

So suppose that $ab\in(4)$, in other words $ab$ is a multiple of $4$, and that $a\notin(4)$. Thus $a$ could be an odd number, or an odd multiple of $2$, but not a multiple of $4$. It follows that $b$ cannot be an odd number, so it must be a multiple of $2$, say $b=2c.$ Then $b^2 = 4c^2\in(4)$

The same argument applies for any prime power $p^m$. If $ab\in(p^m)$ but $a\notin(p^m)$ then $b$ must be a multiple of $p$ and so $b^m\in(p^m).$

To show that $(6)$, for example, is not primary, use the fact that $6=2\times3$. If you take $a=2$ and $b=3$ in the definition of a primary ideal, you see that $2$ is not a multiple of $6$, so the definition asks that some power of $3$ should be a multiple of $6$. But that cannot happen, because every power of $3$ is an odd number. Therefore the definition of primary is not satisfied and so $(6)$ is not a primary ideal of $\mathbb Z.$
Thanks for the helpful post, Opalg

Appreciate your help,


Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I have been reflecting on Opalg's helpful post above ... and I think I need a bit more help ...

I can follow the particular numerical cases but I am having trouble in showing, formally and explicitly, that ...

\(\displaystyle ab \in (p^m) \) and \(\displaystyle a \notin (p^m) \)

\(\displaystyle \Longrightarrow b \) is a multiple of \(\displaystyle p \)

Would appreciate help in this matter.

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
I am having trouble in showing, formally and explicitly, that ...

\(\displaystyle ab \in (p^m) \) and \(\displaystyle a \notin (p^m) \)

\(\displaystyle \Longrightarrow b \) is a multiple of \(\displaystyle p \)

Would appreciate help in this matter.r
Think about the prime factorisation of $ab$. It has to contain the prime $p$ at least $m$ times. If $a\notin(p^m)$ then the prime factorisation of $a$ contains $p$ fewer than $m$ times. Therefore at least one $p$ must occur in the prime factorisation of $b$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Think about the prime factorisation of $ab$. It has to contain the prime $p$ at least $m$ times. If $a\notin(p^m)$ then the prime factorisation of $a$ contains $p$ fewer than $m$ times. Therefore at least one $p$ must occur in the prime factorisation of $b$.
Thanks Oplalg, that post was REALLY helpful!!!

Peter