If $I$ is an ideal of $R=\{\frac{a}{b}\in \mathbb{Q}\mid p\not\mid b\}$, for some fixed prime $p$, show that $(p^t)\subset I.$

Question

Background:

Exercise 14:

Let $p$ be a fixed prime integer and let $R$ be the set of all rational numbers that can be written in the form $\frac{a}{b}$ with $b$ not divisible by $p$. Prove that

(a) $R$ is an integral domain containing $\mathbb{Z}$. [Note $n=\frac{n}{1}$].

(b) If $\frac{a}{b}\in R$ and $p\nmid a$, then $\frac{a}{b}$ is a unit in $R.$

(c) If $I$ is a nonzero ideal in $R$ and $I\neq R$, then $I$ contains $p^t$ for some $t>0.$.

Questions:

I have a question about part (c) above. I understand that I have to show if $I$ is an ideal of $R$, then for some fixed prime $p,$ there exists a smallest $t>0$ such that $p^t\in I$. First to show that $(p^t)\subset I.$ We show that $p\in I$ We have $\frac{p}{1}\in R,$ because $p\not\mid 1.$ Also, let $\frac{m}{n}\in R,$ where $p\not\mid n,$ then $p\mid m.$ Because if $p\not\mid m,$ would imply $\frac{1}{m}\in R,$ and so $m\cdot\frac{1}{m}=1\in I,$ and $I=R,$ a contradition. Hence $p\mid m,$ and we can let the smallest integer $t>0,$ so that $p^t\mid m$ and $m=p^tw,$ where $p\not\mid w,$ then $\frac{1}{w}\in R,$ and $\frac{m}{w}=p^t\in I.$

I think I might need to do some more work to show from the conclusion that $p\mid m$, i can let the smallest integer $t>0$ so that $p^t\mid m$. Also, the solution states to let $x=p^t\cdot \frac{a}{b},$ since $\frac{a}{b},$ is a unit by part (b) above, then $p^t=x\cdot\frac{a}{b}.$ I am not clear how the solution in the book simply could just let $x=p^t\cdot \frac{a}{b}.$

Answer 1

There are several errors in the assertion:

"I understand that I have to show that if $I$ is an ideal of $R$, then for some fixed prime $p$, there exists a smallest $t\gt 0$ such that $p^t\in I$."

First, $p$ is fixes throughout. It is part of the definition of $R$. Saying "for some fixed prime $p$" after saying "If .... then" implies that $p$ depends on the premise of the implication; here, that the choice of $p$ may depend on the choice of $I$. That is incorrect. Again, $p$ is fixed throughout the entirety of these statements.

Second, it is not true that this happens for any ideal $I$. The statement is very clear that it is restricted to nonzero proper ideals $I$.

Third, nothing in (c) talks about a "smallest" $t$; it only talks about a $t\gt 0$. Of course, if there are any such $t$ then there is a smallest one, but that is immaterial and irrelevant as far as the statement (c) is concerned.

What you need to show is exactly what (c) says. That if $I$ is a nonzero proper ideal of $R$, then there is a $t\gt 0$ such that $p^t\in I$. The prime $p$ is part of the definition of $R$.

To your work: showing that $p\in R$ follows simply from invoking part (b): you show in (b) that every integer is an element of $R$, so there is no reason whatsoever for proving it again.

Your argument that if $\frac{m}{n}\in I$ and $p\nmid m$ then $1\in I$ is not correct as written. You claim that $m\left(\frac{1}{m}\right)\in I$; but you do not know that $m$ is in $I$ nor that $n$ is in $I$ yet. You only know that $\frac{m}{n}\in I$. Before you assert that $m\left(\frac{1}{m}\right)\in I$, you need to argue that at least one of the factors is in $I$, and you have failed to do so.

Likewise, the claim that $\frac{m}{w}\in I$ is incomplete. How do you know it lies in $I$? All you know about $I$ is that it is nonzero and it is not all of $R$. Why does it contain that particular element?

Note that you have failed to use that $I\neq (0)$, which suggests there are other assumptions in your argument that you are making but not explaining, rendering it at best incomplete.

The correct argument: since $I$ is nonzero, it contains a nonzero element $\frac{m}{n}$. We may assume that $\gcd(m,n)=1$. If $p\nmid m$, then $\frac{n}{m}\in R$, so $\left(\frac{n}{m}\right)\left(\frac{m}{n}\right)=1\in I$ (product of an element of $R$ by an element of $I$ lies in $I$), and therefore $I=R$. Thus, if $I\neq R$, then $p\mid m$.

Now write $m=\frac{p^tw}{n}$, with $p\nmid w$. Then $\frac{n}{w}\in R$, and therefore $\frac{n}{w}\left(\frac{m}{n}\right)\in I$ (product of an element of $R$ by an element of $I$). Thus, $\frac{n}{w}\left(\frac{m}{n}\right) = \frac{np^tw}{wn} = \frac{p^t}{1}\in I$, as desired.

The solution is taking $x$ to be your $\frac{m}{n}$, $b$ is your $n$, and $a$ is your $w$. So they are saying to write $x$ as $\frac{p^ta}{b}= p^t\left(\frac{a}{b}\right)$ with $p\nmid a$. So they are saying: since $\frac{p^ta}{b}=p^t\left(\frac{a}{b}\right)\in I$, and $\frac{a}{b}$ is a unit, then $p^t\in I$. It is a trivial fact that if $I$ is an ideal, and $ux\in I$ for a unit $u$, then $x\in I$.