3

If $R$ is the ring of integers, let $U$ be the ideal consisting of all multiples of $17.$ Prove that if $V$ is an ideal of $R$ and $U\subset V\subset R$ then either $V = R$ or $V = U.$

I tried, solving this problem as follows:

First of all, it's given $R$ is the ring of integers. We assume the operations defined on $R$ are the usual addition and multiplication of integers.

If $V=U$ then we are done. If $V\neq U$ then, $\exists $ an $a\in V$ such that $17$ does not divide $a$ i.e $(a,17)=1.$

If $a$ leaves a remainder $r$ when divided by $17$ then, $a=17q+r$ for some $q\in\Bbb Z.$ Now, $a=17q+r\in V\implies 2a=17q'+2r\in V$ as $V$ is an ideal of $R.$ This also implies, that $a,2a,3a,...,17a$ are in $V.$ But all these $17$ numbers are congruent to $r,2r,3r,...,17r$ respectively modulo $17.$ Now, $1,2,3,...,17$ is a set of complete residue of $17.$ As, $(a,17)=1$ so, is, $(r,17)=1,$ (this is because, if $x\equiv y\pmod m$ then, $(x,m)=(y,m)$ and in here, $a\equiv r\pmod 17$) and this further implies $r,2r,3r,...,17r$ is a set of complete set of residues of $17.$

(In the above assertion, we have used the lemma, that if $r_1,r_2,r_3,\cdots,r_m$ is a complete residue system congruent modulo $m$ and if $\exists a$ such that $(a,m)=1$ then, $ar_1,ar_2,ar_3,\cdots,ar_m$ is also a complete residue system congruent modulo $m$ )

So, we have numbers of all the possible reminders (on division by $17$) occuring on $V.$

Now, if $b\in V$ and $b=17q+s$ such that $s\neq 0,$ is an arbitrary remainder on division by $17$, then, $b=17(q+k)+s$ is also in $V,$ where $k\in \Bbb Z$ is an arbitrary integer. So, all the numbers having a remainder of $s$ on division by $17$ occurs in $V,$ (because, $(q+k)$ covers all possible quotients and hence all possible numbers ). As, $s$ is an arbitrary residue/remainder and all possible residues of $17$ occurs in $V$ , so $V$ has all the integers possible, $\implies$ $V=R.$

I think my solution's alright? But can I do anything, to improve the readability of this solution?

Should I add/remove anything to make this solution perfect? I want to know if tlany changes are demanded in here that will make the solution more comprehensible? Any suggested changes in the proof-writing will be highly appreciated.

Finally, if you find any errors in my solution, please do let me know.

Thomas Finley
  • 2,734
  • $\Bbb Z$ is Euclidean so a PID by the 2nd dupe, and in PIDs it is easy to prove that prime ideals are maximal (since contains = divides for principal ideals), see the first dupe. – Bill Dubuque Jul 29 '23 at 12:47

1 Answers1

4

This can be a lot cleaner than what you have written.

As you said, if $U=V$ we are done. If not, let $v\in V$ and $v\not\in U$. Since $17$ is a prime number we know that the $\text{gcd($17$, $v$)}=1$ so there exists $a,b\in\mathbb Z$ such that $$a17 + bv=1$$.

Since $V$ is an ideal and $17\in V$ we know that $a17\in V$. Likewise, $bv\in V$. Ideals are closed under addition so we know that $a17+bv=1\in\ V$ and since $1\in V$, for any $z\in\mathbb Z$, $1z=z\in V$.

Therefore, $V=\mathbb Z$

John Douma
  • 11,426
  • 2
  • 23
  • 24
  • Yes, but my solution is valid, right? Thanks for your valuable suggestion – Thomas Finley Jul 29 '23 at 07:40
  • @ThomasFinley To be honest, your solution is hard to read. I didn't quite follow the logic that all remainders being present in $V$ implied that all integers must be in $V$. If you can show that given $r\in{0,\dots, 16}$, that $17k+r\in V$ for all $k$ then it seems that your proof is correct. – John Douma Jul 29 '23 at 07:48
  • For example, suppose the $a$ in your proof was $a=19$. Then the residue $r$ would be $2$ and $r, 2r,\dots 17r$ would all be even and would not comprise a complete set of residues. – John Douma Jul 29 '23 at 07:54
  • I have edited the solution to make it clearer. Do you see any problems now? – Thomas Finley Jul 29 '23 at 09:04
  • 2
    Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 29 '23 at 12:42
  • @BillDubuque Sorry, but till now we are not taught about PIDs. – Thomas Finley Jul 29 '23 at 13:05
  • @ThomasFinley A PID is an integral domain where every ideal $I$ is principal $I = (a)$. It is trivial to show that $\Bbb Z$ is a PID using the Euclidean algorithm - see the 2nd dupe. That's all you need. It is very simple (and really the essence of the matter). – Bill Dubuque Jul 29 '23 at 13:08