0

Prove every integral ideal $J$ is identical with $\Bbb{J}_m$ for some $m$.

Suppose $J \neq \{0\} = \Bbb{J}_0$. By the least integer principle, there exists an $m \in J$ such that $rm \in J$ in $r \in \Bbb{Z}$. But then if $J \neq \Bbb{J}_{m_0}$ for some $m_0 \in \Bbb{Z}$, $J$ cannot be an integral ideal. But this violates LIP.

True or true mates?

Don Larynx
  • 4,703

1 Answers1

1

Let's define $\mathbb{J_m}$ to be the set of all integers that are integral multiples of $m$. In set notation, $\mathbb{J_m}=\{t\in \mathbb{Z}|t=bm, b,m\in \mathbb{Z}\}$, where $b$ is not fixed.

We want to first show that $J$ must contain positive integers. If $n\in \mathbb{Z^-}$ is | $n\in J$, then pick any $r\in \mathbb{Z^-}$. By definition of an integral ideal, $m=rn$ must be in $J$ and $m\in \mathbb{Z^+}$. Therefore, $J$ contains positive integers.

Let $m$ be the least positive integer in $J$. We want to show that $J=$ $\mathbb{J_m}$.

Let $j\in J$. Assume $j$ is not an integral multiple of $m$. Then, by Euclid's Division Lemma $j$ can be written as $j=bm+r$ where $0< r<m$ and $b\in \mathbb{Z}$.

$bm\in J$ by definition of integral ideal. ∴ $r=(j-bm) \in J$. But, $0< r<m$ and $r\in J$. This would mean that $r$ is the least positive integer in $J$ contradicting our assertion. Therefore, $j$ is a multiple of $m$ and is included in $\mathbb{J_m}$.

Now let $j\in \mathbb{J_m}$, then $j$ can be written in the form $j=rm$ and is by definition a member of $J$.

QED

Vladimir Louis
  • 157
  • Your set notation at the beginning implies there is only one element in $\Bbb {J}_m $, which is $ t $ – Don Larynx Jan 27 '15 at 03:39
  • It reads "The set of all t included in the set of integers such that t is equal to bm where b and m are integers". I think more than one t fits the bill. – Vladimir Louis Jan 27 '15 at 03:50
  • perhaps am mistaken. If you don't mind, please rewrite the set. But, I think the proof is solid past that point. – Vladimir Louis Jan 27 '15 at 04:05
  • "$bm∈J$ by definition of integral ideal." But we never assumed J was an integral ideal. – Don Larynx Jan 27 '15 at 13:11
  • I was proving your statement in which you declared $J$ as an integral ideal. I only defined $\mathbb{J_m}$ because you did not define it. We appear to be learning number theory from the same book so I knew what $\mathbb{J_m}$ is. Every time I search for a question it appears you have posted it so I just try to answer them in order to improve my understanding of the subject. We can grow together. – Vladimir Louis Jan 27 '15 at 13:18
  • I understand now. – Don Larynx Jan 27 '15 at 15:33
  • Also, let $b$ not be fixed at the beginning. That should do it. – Don Larynx Jan 27 '15 at 15:35
  • I wanted to put the restriction, but it looked weird. I am still learning the website's platform. Thanks for the vote. – Vladimir Louis Jan 27 '15 at 17:53