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Let $R$ be an integral domain and a Noetherian U.F.D. with the following property:

for each couple $a,b\in R$ that are not both $0$, and that have no common prime divisor, there are elements $u,v\in R$ such that $au+bv=1$.

I want to show that $R$ is a P.I.D..

Could you give me some hints what we could do to show that $R$ is a P.I.D. ?

$$$$

EDIT:

Why is an ideal of the form $(x,y)$ principal?

Mary Star
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  • Let me point out that the hypothesis of your exercise contains some superfluous assumptions: you don't need $R$ noetherian, not even a UFD. It's enough to assume that $R$ is an integral domain which satisfies ACCP (and the condition about coprime elements, of course). Then $R$ is a PID. – user26857 Apr 16 '16 at 18:55

2 Answers2

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Every ideal generated by two elements is principal.

Let $a,b \in R$. Since $R$ is a UFD, we can write $a=da'$ and $b=db'$ with $d=\gcd(a,b)$ and $a',b'$ coprime. The hypothesis then imply that there are $u,v\in R$ such that $a'u+b'v=1$ and so $au+bv=d$. Therefore, $aR+bR=dR$, that is $(a,b)=(d)$.

Every finitely generated ideal is principal.

If $I=(r_1, r_2, \ldots, r_n)$ then by induction $J=(r_2, \ldots, r_n)$ is principal: $J=(s)$, and so $I=(r_1,s)$ is also principal by the previous fact.

Every ideal is principal.

Since $R$ is Noetherian, every ideal is finitely generated, and so is principal, by the previous fact.

All this can be summarized as:

The hypotheses imply that $R$ is a Bézout domain. Now a Bézout domain is a PID iff it is Noetherian.

lhf
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  • Isn't an ideal that is generated by one element principal? Why can we write $a=da'$ and $b=db'$ with $\gcd (a,b)$ and $a',b'$ coprime? – Mary Star Apr 15 '16 at 16:20
  • We have that $R$ is a U.F.D. iff $\forall r\in R\setminus {0}$, $r\notin U(R)$ the following hold: $$$$
    • $r=a_1 \cdots a_k$ with $a_i$ irreducible $$$$
    • If $r=a_1 \cdots a_k=b_1 \cdots b_t$ with $a_i, b_i$ irreducible then $k=t$ and $a_i=b_iu_i$ with $u_i\in U(R), \forall u=1, \dots , k$.
     – Mary Star Apr 15 '16 at 16:34
  • Ah ok... Could you explain to me the part: "By induction, every ideal is principal, because every ideal is finitely generated, since $R$ is Noetherian." ? When every ideal is finitely generated, then every ideal is a finite product of irreducible elements of $R$, or not? Why is every ideal then principal? – Mary Star Apr 15 '16 at 18:01
  • Why does it hold that $(r_1,r_2)=(s_1)$ ? Is that the definition of finitely generated? – Mary Star Apr 15 '16 at 18:24
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Suppose $R$ is not a PID. Let $I$ be a non-zero non-principal ideal, and $0\ne a_1\in I$. Clearly $(a_1)\subsetneq I$. Let $b_2\in I\setminus (a_1)$. We have $(a_1)\subsetneq (a_1,b_2)\subsetneq I$ since the hypothesis tells us that the ideal $(a_1,b_2)$ is also principal. Set $(a_2)=(a_1,b_2)$, pick an element $b_3\in I\setminus(a_2)$, consider the ideal $(a_2,b_3)$ (which is also principal), and so on. This way you find a strictly ascending chain of principal ideals in $R$, a contradiction.

user26857
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  • How do we know that $(a_1,b_2)$ is also principal? – Mary Star Apr 16 '16 at 09:40
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    @MaryStar If $a_1$ and $b_2$ are coprime (that is, have no common prime divisor), then $(a_1,b_2)=(1)$ by hypothesis. If they are not coprime, then set $d=\gcd(a_1,b_2)$, write $a_1=da_1'$, $b_2=db_2'$ with $a_1',b_2'$ coprime, so $(a_1',b_2')=(1)$ and finally $(a_1,b_2)=(d)$. – user26857 Apr 16 '16 at 13:56
  • If $a_1$ and $b_2$ are coprime then from the property of $R$, we have that there are $u,v\in R$ such that $a_1u+b_2v=1$. Is this equivalent with $(a_1,b_2)=(1)$ ? – Mary Star Apr 16 '16 at 14:00
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    @MaryStar Yes, it is! What means $(a_1,b_2)=(1)$? Recall that the elements of the ideal $(a_1,b_2)$ are of the form $a_1x+b_2y$ with $x,y\in R$. Then this ideal equals $R$ (or, equivalently $(1)$) iff $1=a_1z+b_2t$ for some $z,t\in R$. But this is exactly the hypothesis. – user26857 Apr 16 '16 at 14:09
  • Ah ok... Why do we have to suppose that $I$ is a non-principal ideal? Does it have to stand because $R$ is not a P.I.D. ? – Mary Star Apr 16 '16 at 14:26
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    @MaryStar Yes, you are right! I've supposed that $R$ is not a PID and then I want to arrive to a contradiction. That's why I started with a non-principal ideal. – user26857 Apr 16 '16 at 14:28
  • So, we have found a strictly ascending chain of principal ideals in $R$. Since $R$ is Noetherian this chain has to stop after a finite number of steps. Does this mean that $I$ must be also principal? And that is the contradiction? – Mary Star Apr 16 '16 at 15:24
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    @MaryStar Let me recall how the proof works: suppose $R$ is not a PID. Then there must exist ideals of $R$ which are not principal. Let $I$ be such an ideal. Then $I\ne(0)$, and then one can chose a non-zero element $a_1\in I$. Now the proof shows that one can find $a_2,\dots,a_n,\ldots\in I$ such that $(a_1)\subsetneq(a_2)\subsetneq\cdots\subsetneq(a_n)\subsetneq\cdots$. This contradicts the hypothesis that $R$ is noetherian. – user26857 Apr 16 '16 at 15:34
  • To find that $(a_1)\subsetneq(a_2)\subsetneq\cdots\subsetneq(a_n)\subsetneq\cdots$ should $I$ have infinitely many elements? – Mary Star Apr 16 '16 at 16:03
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    @MaryStar I think so, but why do you care? The exercise doesn't ask us such things. – user26857 Apr 16 '16 at 16:04
  • Ah ok... I want to understand the proof... In general, are the ideals finite or infinite? – Mary Star Apr 16 '16 at 16:09
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    @MaryStar Do you refer to finite or infinite as sets? Or to finitely or infinitely generated? – user26857 Apr 16 '16 at 16:12
  • As sets.... Can both (finite/infinite) happen? – Mary Star Apr 16 '16 at 16:13
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    @MaryStar Well, it depends: the ideals of a finite ring like $\mathbb Z/n\mathbb Z$ are all finite. However, if the ring is an infinite integral domain (like $\mathbb Z$) then all its non-zero ideals are infinite: once an ideal contains a non-zero element $a$ then it contains all multiples $ra$, $r\in R$, and these are infinitely many ($r_1a=r_2a$ $\implies$ $r_1=r_2$). – user26857 Apr 16 '16 at 16:13
  • Ah ok... I see... Thank you very much!! :-) – Mary Star Apr 16 '16 at 16:15
  • I have a question... If $R$ is an integral domain and U.F.D., for example $\mathbb{Z}[x]$, does it follow that it is a P.I.D. ? – Mary Star Apr 18 '16 at 20:22
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    @MaryStar Of course not! $\mathbb Z[X]$ is not a PID; see http://math.stackexchange.com/questions/36169/show-that-langle-2-x-rangle-is-not-a-principal-ideal-in-mathbb-z-x – user26857 Apr 18 '16 at 20:28
  • Ah ok... I see... Thanks!! :-) – Mary Star Apr 19 '16 at 20:34