Primary ideals of $mathbb{Z}$
$begingroup$
Let $R=mathbb{Z}$ I must prove that
In $mathbb{Z}$ the primary ideals s are precisely the ideals $(p^k)$ where $p$ is a prime number and $nge 1$.
$$$$
The definition of primary ideal.
Definition. An ideal $I$ of the ring $R$ is called primary if for all $a,bin R$ $$(abin Iquadtext{and}quad anotin I)Rightarrow b^nin I, text{for some positive integer $n$}.$$
Therefore I must prove that
$$(n);text{is primary in};mathbb{Z}iff n=p^k, kge1,pge2,;text{where};p;text{is a prime}.$$
Proof ($Leftarrow$) We suppose that $n=p^k$ for same prime $p$ and $kge 1$. Let $a,binmathbb{Z}$ such that $abin(p^k)$ and $anotin (p^k)$. Then $p^knmid a$, but $p^k mid ab$ I no longer know how to proceed.
$(Rightarrow)$ Let $(n)$ a primary ideal and we suppose for absurd that $n$ is not a power of a prime, then? how could I proceed?
Now, let $abin (p^k)$ with $anotin(p^k)$, then $p^kmid ab.$ Can I conclude that $p^kmid b$?
How can I set up the other implication? Thanks
proof-verification
asked Dec 5 '18 at 20:50
Jack J.Jack J.
4392419
$endgroup$
add a comment |
$begingroup$
Let $R=mathbb{Z}$ I must prove that
In $mathbb{Z}$ the primary ideals s are precisely the ideals $(p^k)$ where $p$ is a prime number and $nge 1$.
$$$$
The definition of primary ideal.
Definition. An ideal $I$ of the ring $R$ is called primary if for all $a,bin R$ $$(abin Iquadtext{and}quad anotin I)Rightarrow b^nin I, text{for some positive integer $n$}.$$
Therefore I must prove that
$$(n);text{is primary in};mathbb{Z}iff n=p^k, kge1,pge2,;text{where};p;text{is a prime}.$$
Proof ($Leftarrow$) We suppose that $n=p^k$ for same prime $p$ and $kge 1$. Let $a,binmathbb{Z}$ such that $abin(p^k)$ and $anotin (p^k)$. Then $p^knmid a$, but $p^k mid ab$ I no longer know how to proceed.
$(Rightarrow)$ Let $(n)$ a primary ideal and we suppose for absurd that $n$ is not a power of a prime, then? how could I proceed?
Now, let $abin (p^k)$ with $anotin(p^k)$, then $p^kmid ab.$ Can I conclude that $p^kmid b$?
How can I set up the other implication? Thanks
proof-verification
asked Dec 5 '18 at 20:50
Jack J.Jack J.
4392419
$endgroup$
$begingroup$
You take each pie a day. Can I say that you take two pie at one time?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 5 '18 at 20:54
add a comment |
$begingroup$
Let $R=mathbb{Z}$ I must prove that
In $mathbb{Z}$ the primary ideals s are precisely the ideals $(p^k)$ where $p$ is a prime number and $nge 1$.
$$$$
The definition of primary ideal.
Definition. An ideal $I$ of the ring $R$ is called primary if for all $a,bin R$ $$(abin Iquadtext{and}quad anotin I)Rightarrow b^nin I, text{for some positive integer $n$}.$$
Therefore I must prove that
$$(n);text{is primary in};mathbb{Z}iff n=p^k, kge1,pge2,;text{where};p;text{is a prime}.$$
Proof ($Leftarrow$) We suppose that $n=p^k$ for same prime $p$ and $kge 1$. Let $a,binmathbb{Z}$ such that $abin(p^k)$ and $anotin (p^k)$. Then $p^knmid a$, but $p^k mid ab$ I no longer know how to proceed.
$(Rightarrow)$ Let $(n)$ a primary ideal and we suppose for absurd that $n$ is not a power of a prime, then? how could I proceed?
Now, let $abin (p^k)$ with $anotin(p^k)$, then $p^kmid ab.$ Can I conclude that $p^kmid b$?
How can I set up the other implication? Thanks
proof-verification
asked Dec 5 '18 at 20:50
Jack J.Jack J.
4392419
$endgroup$
Let $R=mathbb{Z}$ I must prove that
In $mathbb{Z}$ the primary ideals s are precisely the ideals $(p^k)$ where $p$ is a prime number and $nge 1$.
$$$$
The definition of primary ideal.
Definition. An ideal $I$ of the ring $R$ is called primary if for all $a,bin R$ $$(abin Iquadtext{and}quad anotin I)Rightarrow b^nin I, text{for some positive integer $n$}.$$
Therefore I must prove that
$$(n);text{is primary in};mathbb{Z}iff n=p^k, kge1,pge2,;text{where};p;text{is a prime}.$$
Proof ($Leftarrow$) We suppose that $n=p^k$ for same prime $p$ and $kge 1$. Let $a,binmathbb{Z}$ such that $abin(p^k)$ and $anotin (p^k)$. Then $p^knmid a$, but $p^k mid ab$ I no longer know how to proceed.
$(Rightarrow)$ Let $(n)$ a primary ideal and we suppose for absurd that $n$ is not a power of a prime, then? how could I proceed?
Now, let $abin (p^k)$ with $anotin(p^k)$, then $p^kmid ab.$ Can I conclude that $p^kmid b$?
How can I set up the other implication? Thanks
proof-verification
proof-verification
asked Dec 5 '18 at 20:50
Jack J.Jack J.
4392419
asked Dec 5 '18 at 20:50
Jack J.Jack J.
4392419
edited Dec 5 '18 at 21:35
Jack J.
asked Dec 5 '18 at 20:50
Jack J.Jack J.
4392419
asked Dec 5 '18 at 20:50
Jack J.Jack J.
4392419
asked Dec 5 '18 at 20:50
Jack J.Jack J.
4392419
4392419
$begingroup$
You take each pie a day. Can I say that you take two pie at one time?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 5 '18 at 20:54
add a comment |
$begingroup$
You take each pie a day. Can I say that you take two pie at one time?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 5 '18 at 20:54
$begingroup$
You take each pie a day. Can I say that you take two pie at one time?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 5 '18 at 20:54
$begingroup$
You take each pie a day. Can I say that you take two pie at one time?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 5 '18 at 20:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No. $2 cdot 2^5 in (2^6)$ and $2 notin (2^6)$, but $2^6 not mid 2^5$. What you do know is there are $6$ powers of two among $2$ and $2^5$.
If $a b in (p^k)$, then $a b = m p^k$ for some $m in mathbb{Z}$. Since $a not in (p^k)$, $a = m' p^ell$ for $m,ell in mathbb{Z}$ where $m'|m$, $(m',p) = 1$, and $0 leq ell < p$. Then $b = frac{m}{m'}p^{k-ell}$. Since $k-ell geq 1$, $b^k in (p^k)$.
For the other direction, you should see how to use a composite, as indicated above, to show that a non-prime-power generator of the ideal can be split into parts preventing the ideal being primary.
answered Dec 5 '18 at 20:54
Eric TowersEric Towers
32.5k22369
$endgroup$
$begingroup$
Thanks for your ansewer. But, at this point, Can I formalize the fact that the primary ideals of $mathbb{Z}$ are $(p^k)$?
$endgroup$
– Jack J.
Dec 5 '18 at 20:56
$begingroup$
I don't see that you have shown any work to support that claim. Do you even have any reason to believe the primary ideals are principal ideals?
$endgroup$
– Eric Towers
Dec 5 '18 at 20:58
$begingroup$
So, I proved that $mathbb{Z}$ is a P.I.D and that the prime ideal of $mathbb{Z}$ are the ideals of the type $(p)$.
$endgroup$
– Jack J.
Dec 5 '18 at 21:01
$begingroup$
Every ideal of $mathbb{Z}$ is principal
$endgroup$
– Jack J.
Dec 5 '18 at 21:13
$begingroup$
I don't see that in any of the work you posted in your Question. We're not psychic -- you should tell us the progress you have already made.
$endgroup$
– Eric Towers
Dec 5 '18 at 21:23
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
No. $2 cdot 2^5 in (2^6)$ and $2 notin (2^6)$, but $2^6 not mid 2^5$. What you do know is there are $6$ powers of two among $2$ and $2^5$.
If $a b in (p^k)$, then $a b = m p^k$ for some $m in mathbb{Z}$. Since $a not in (p^k)$, $a = m' p^ell$ for $m,ell in mathbb{Z}$ where $m'|m$, $(m',p) = 1$, and $0 leq ell < p$. Then $b = frac{m}{m'}p^{k-ell}$. Since $k-ell geq 1$, $b^k in (p^k)$.
For the other direction, you should see how to use a composite, as indicated above, to show that a non-prime-power generator of the ideal can be split into parts preventing the ideal being primary.
answered Dec 5 '18 at 20:54
Eric TowersEric Towers
32.5k22369
$endgroup$
$begingroup$
Thanks for your ansewer. But, at this point, Can I formalize the fact that the primary ideals of $mathbb{Z}$ are $(p^k)$?
$endgroup$
– Jack J.
Dec 5 '18 at 20:56
$begingroup$
I don't see that you have shown any work to support that claim. Do you even have any reason to believe the primary ideals are principal ideals?
$endgroup$
– Eric Towers
Dec 5 '18 at 20:58
$begingroup$
So, I proved that $mathbb{Z}$ is a P.I.D and that the prime ideal of $mathbb{Z}$ are the ideals of the type $(p)$.
$endgroup$
– Jack J.
Dec 5 '18 at 21:01
$begingroup$
Every ideal of $mathbb{Z}$ is principal
$endgroup$
– Jack J.
Dec 5 '18 at 21:13
$begingroup$
I don't see that in any of the work you posted in your Question. We're not psychic -- you should tell us the progress you have already made.
$endgroup$
– Eric Towers
Dec 5 '18 at 21:23
|
show 1 more comment
$begingroup$
No. $2 cdot 2^5 in (2^6)$ and $2 notin (2^6)$, but $2^6 not mid 2^5$. What you do know is there are $6$ powers of two among $2$ and $2^5$.
If $a b in (p^k)$, then $a b = m p^k$ for some $m in mathbb{Z}$. Since $a not in (p^k)$, $a = m' p^ell$ for $m,ell in mathbb{Z}$ where $m'|m$, $(m',p) = 1$, and $0 leq ell < p$. Then $b = frac{m}{m'}p^{k-ell}$. Since $k-ell geq 1$, $b^k in (p^k)$.
For the other direction, you should see how to use a composite, as indicated above, to show that a non-prime-power generator of the ideal can be split into parts preventing the ideal being primary.
answered Dec 5 '18 at 20:54
Eric TowersEric Towers
32.5k22369
$endgroup$
$begingroup$
Thanks for your ansewer. But, at this point, Can I formalize the fact that the primary ideals of $mathbb{Z}$ are $(p^k)$?
$endgroup$
– Jack J.
Dec 5 '18 at 20:56
$begingroup$
I don't see that you have shown any work to support that claim. Do you even have any reason to believe the primary ideals are principal ideals?
$endgroup$
– Eric Towers
Dec 5 '18 at 20:58
$begingroup$
So, I proved that $mathbb{Z}$ is a P.I.D and that the prime ideal of $mathbb{Z}$ are the ideals of the type $(p)$.
$endgroup$
– Jack J.
Dec 5 '18 at 21:01
$begingroup$
Every ideal of $mathbb{Z}$ is principal
$endgroup$
– Jack J.
Dec 5 '18 at 21:13
$begingroup$
I don't see that in any of the work you posted in your Question. We're not psychic -- you should tell us the progress you have already made.
$endgroup$
– Eric Towers
Dec 5 '18 at 21:23
|
show 1 more comment
$begingroup$
No. $2 cdot 2^5 in (2^6)$ and $2 notin (2^6)$, but $2^6 not mid 2^5$. What you do know is there are $6$ powers of two among $2$ and $2^5$.
If $a b in (p^k)$, then $a b = m p^k$ for some $m in mathbb{Z}$. Since $a not in (p^k)$, $a = m' p^ell$ for $m,ell in mathbb{Z}$ where $m'|m$, $(m',p) = 1$, and $0 leq ell < p$. Then $b = frac{m}{m'}p^{k-ell}$. Since $k-ell geq 1$, $b^k in (p^k)$.
For the other direction, you should see how to use a composite, as indicated above, to show that a non-prime-power generator of the ideal can be split into parts preventing the ideal being primary.
answered Dec 5 '18 at 20:54
Eric TowersEric Towers
32.5k22369
$endgroup$
No. $2 cdot 2^5 in (2^6)$ and $2 notin (2^6)$, but $2^6 not mid 2^5$. What you do know is there are $6$ powers of two among $2$ and $2^5$.
If $a b in (p^k)$, then $a b = m p^k$ for some $m in mathbb{Z}$. Since $a not in (p^k)$, $a = m' p^ell$ for $m,ell in mathbb{Z}$ where $m'|m$, $(m',p) = 1$, and $0 leq ell < p$. Then $b = frac{m}{m'}p^{k-ell}$. Since $k-ell geq 1$, $b^k in (p^k)$.
For the other direction, you should see how to use a composite, as indicated above, to show that a non-prime-power generator of the ideal can be split into parts preventing the ideal being primary.
answered Dec 5 '18 at 20:54
Eric TowersEric Towers
32.5k22369
edited Dec 6 '18 at 0:08
answered Dec 5 '18 at 20:54
Eric TowersEric Towers
32.5k22369
answered Dec 5 '18 at 20:54
Eric TowersEric Towers
32.5k22369
answered Dec 5 '18 at 20:54
Eric TowersEric Towers
32.5k22369
32.5k22369
$begingroup$
Thanks for your ansewer. But, at this point, Can I formalize the fact that the primary ideals of $mathbb{Z}$ are $(p^k)$?
$endgroup$
– Jack J.
Dec 5 '18 at 20:56
$begingroup$
I don't see that you have shown any work to support that claim. Do you even have any reason to believe the primary ideals are principal ideals?
$endgroup$
– Eric Towers
Dec 5 '18 at 20:58
$begingroup$
So, I proved that $mathbb{Z}$ is a P.I.D and that the prime ideal of $mathbb{Z}$ are the ideals of the type $(p)$.
$endgroup$
– Jack J.
Dec 5 '18 at 21:01
$begingroup$
Every ideal of $mathbb{Z}$ is principal
$endgroup$
– Jack J.
Dec 5 '18 at 21:13
$begingroup$
I don't see that in any of the work you posted in your Question. We're not psychic -- you should tell us the progress you have already made.
$endgroup$
– Eric Towers
Dec 5 '18 at 21:23
|
show 1 more comment
$begingroup$
Thanks for your ansewer. But, at this point, Can I formalize the fact that the primary ideals of $mathbb{Z}$ are $(p^k)$?
$endgroup$
– Jack J.
Dec 5 '18 at 20:56
$begingroup$
I don't see that you have shown any work to support that claim. Do you even have any reason to believe the primary ideals are principal ideals?
$endgroup$
– Eric Towers
Dec 5 '18 at 20:58
$begingroup$
So, I proved that $mathbb{Z}$ is a P.I.D and that the prime ideal of $mathbb{Z}$ are the ideals of the type $(p)$.
$endgroup$
– Jack J.
Dec 5 '18 at 21:01
$begingroup$
Every ideal of $mathbb{Z}$ is principal
$endgroup$
– Jack J.
Dec 5 '18 at 21:13
$begingroup$
I don't see that in any of the work you posted in your Question. We're not psychic -- you should tell us the progress you have already made.
$endgroup$
– Eric Towers
Dec 5 '18 at 21:23
$begingroup$
Thanks for your ansewer. But, at this point, Can I formalize the fact that the primary ideals of $mathbb{Z}$ are $(p^k)$?
$endgroup$
– Jack J.
Dec 5 '18 at 20:56
$begingroup$
Thanks for your ansewer. But, at this point, Can I formalize the fact that the primary ideals of $mathbb{Z}$ are $(p^k)$?
$endgroup$
– Jack J.
Dec 5 '18 at 20:56
$begingroup$
I don't see that you have shown any work to support that claim. Do you even have any reason to believe the primary ideals are principal ideals?
$endgroup$
– Eric Towers
Dec 5 '18 at 20:58
$begingroup$
I don't see that you have shown any work to support that claim. Do you even have any reason to believe the primary ideals are principal ideals?
$endgroup$
– Eric Towers
Dec 5 '18 at 20:58
$begingroup$
So, I proved that $mathbb{Z}$ is a P.I.D and that the prime ideal of $mathbb{Z}$ are the ideals of the type $(p)$.
$endgroup$
– Jack J.
Dec 5 '18 at 21:01
$begingroup$
So, I proved that $mathbb{Z}$ is a P.I.D and that the prime ideal of $mathbb{Z}$ are the ideals of the type $(p)$.
$endgroup$
– Jack J.
Dec 5 '18 at 21:01
$begingroup$
Every ideal of $mathbb{Z}$ is principal
$endgroup$
– Jack J.
Dec 5 '18 at 21:13
$begingroup$
Every ideal of $mathbb{Z}$ is principal
$endgroup$
– Jack J.
Dec 5 '18 at 21:13
$begingroup$
I don't see that in any of the work you posted in your Question. We're not psychic -- you should tell us the progress you have already made.
$endgroup$
– Eric Towers
Dec 5 '18 at 21:23
$begingroup$
I don't see that in any of the work you posted in your Question. We're not psychic -- you should tell us the progress you have already made.
$endgroup$
– Eric Towers
Dec 5 '18 at 21:23
|
show 1 more comment
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$begingroup$
You take each pie a day. Can I say that you take two pie at one time?
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 5 '18 at 20:54