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Let $R$ be a commutative ring with identity with $b\in R$. Let $T$ be the subring of all multiples of $b$, $T=\{r\cdot b : r \in R\}$. If $u$ is a unit in $R$ with $u \in T$, prove that $T=R$.

Could you help me some suggestions?

I really have no clues to do this questions, I can only show $1\cdot R$ belongs to $T$. I even don't know the general way to prove two rings are equal.

Brian Fitzpatrick
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ZHJ
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  • Hint: Here "equal" means equality as sets, so you need to show that every element of $R$ is contained in $T$. – tabstop Feb 10 '14 at 02:14
  • Do you know how to show that two sets are equal? – Brian Fitzpatrick Feb 10 '14 at 02:14
  • Yeah, just show T contains R and R contains T.Here, it is easy to show T is subring of R, I try to show R is subring of T, but I failed. – ZHJ Feb 10 '14 at 02:16
  • Hint: T contain a unit, and is closed under multiples. But every element is the multiple of a unit so ... – Bill Dubuque Feb 10 '14 at 02:28
  • I do not understand your comment. – ZHJ Feb 10 '14 at 02:33
  • What part don't you understand? Note: you need to write @username in order for the user to be notified of your comment. You can complete usernames using the Tab key, e.g. @Bill – Bill Dubuque Feb 10 '14 at 02:38
  • You don't necessarily know that $b$ is a unit; but you do know that $u$ is a unit, and therefore all multiples of $u$ are also in $T$ (since if $x=q\cdot u$, then $x=q\cdot (r\cdot b)=(q\cdot r)\cdot b\in T$). Now what do you know about multiples of a unit.... – tabstop Feb 10 '14 at 02:39
  • @Bill Dubque Thank you. I got you, and what next? Sorry, I am still confused – ZHJ Feb 10 '14 at 02:45
  • For example, if $,2\in T,$ then, since $T$ is closed under multiples, it contains all multiples of $2$. Similarly, if $,u\in T,$ then $T$ contains all multiples of $u$. Thus, since every element is a multiple of a unit $u$, we infer that $T$ contains every element of $R$. – Bill Dubuque Feb 10 '14 at 02:48

2 Answers2

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The critical thing to realize here is that if $u$ is a unit, and $u = ab$, then $a$ and $b$ are both units. For if $u$ is a unit, then $uv = 1$ for some $v \in R$, so that $1 = (ab)v = a(bv) = b(av)$, where we have used the commutativity of $R$. So $u \in T$ a unit implies $b \in T$ is a unit is well, since $u = ab$ for some $a \in R$. Now since $b \in T = Rb$ is a unit, for any $s \in R$ take $r = sb^{-1} = s(av)$. Then $s = s1 = sb^{-1}b = rb \in Rb = T$, so in fact $R \subset T$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
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  • There's no need to deduce that $b$ is a unit since we are already given a unit $,u\in T,,$ see the comments to the question. – Bill Dubuque Feb 10 '14 at 02:52
  • @BillDubuque: yes, I do believe you are correct, but I thought, since $T = Rb$, it might be a little more clear to make it explicit. Clearer to me, in any event. Thanks for the input. Best. – Robert Lewis Feb 10 '14 at 02:56
  • I am really appreciative with both of your comments, sincerely thank you. – ZHJ Feb 10 '14 at 02:59
  • @user127534: Thank you very much, and glad I could help out. – Robert Lewis Feb 10 '14 at 03:03
  • @RobertLewis Just a little confusing, why do you take r=sb^-1 ? – ZHJ Feb 10 '14 at 03:35
  • @user127534: we have $T \subset R$, and we want to show $R \subset T$ so we can affirm $R = T$. Now $T$ is the set $Rb$ of all multiples of $b$ in $R$. So taking $r = sb^{-1}$ for $s \in R$ gives an element of $R$ which when multiplied by $b$, yields $s$, so we can see explicitly that $s \in T = bR$. Just a little trick, really. Does that clarify? If not, please leave a comment and I'll get back to you. OR if you have any other questions . . . – Robert Lewis Feb 10 '14 at 03:40
  • @RobertLewis My question is we want to prove s=br, but you just let r=sb^-1, which I feel is just another way to say s=br. What you suppose is what we want to prove. – ZHJ Feb 10 '14 at 03:47
  • @user127534: we need to show every $s \in R$ is of the form $rb$ for some $r \in R$. Given that $b \in T$ is a unit, I need some element of $r \in R$ with $s = rb$. I know $s$, presumably, so I can exhibit $r$ by saying $r = sb^{-1}$. You start with $s$, then find $r$. That's the order of the logic, though it is in effect another way of saying $s = br$. But it starts from what we have, and establishes what we need. Do these remarks help? – Robert Lewis Feb 10 '14 at 03:53
  • @RobertLewis Now I feel this is the trickiest part of this problem. – ZHJ Feb 10 '14 at 03:56
  • @user127534: It is a little tricky, but not very, and it's a common trick that occurs in many places. All I want do to is show every $s \in R$ can be written $rb$ for some $r \in R$. So I just "solve" the equation $s = rb$, using the fact that $b$ is a unit. I know it seems tricky at first, but you'll see this trick again and again in your mathematical studies and career, I'll betcha! – Robert Lewis Feb 10 '14 at 04:07
  • @RobertLewis Again, really thanks for your help! This is my first time to study abstract algebra. I definitely need to practice more. – ZHJ Feb 10 '14 at 04:10
  • @user127534: well, we all need to practice more, and it really does help. These things become second nature with time, though they seem sort of odd at first. Best of luck, and feel free to ask if you have further questions! Regards. – Robert Lewis Feb 10 '14 at 04:12
  • @user127534 It's not a trick - see my answer where I explain the conceptual foundations. – Bill Dubuque Feb 11 '14 at 18:14
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I'd like to dispel the notion in the comments that the key idea is a "trick". Instead, I will show that it boils down to the transitivity of "contains" or "divides" (i.e. $\ b\mid c,\,\ c\mid d\ \Rightarrow\ b\mid d),\ $ using the following simple but fundamental correspondence (ubiquitous in divisibility theory)

Theorem $\ \ bR \supseteq aR\iff b\mid a\,$ in $\,R\quad$ [Contains = Divides for principal ideals]

Proof $\ (\Rightarrow)\,\ a = a\cdot 1\in aR\subset bR\,\Rightarrow\, a = br,\ $ for some $\,r\in R,\,$ so $\,b\mid a\,$ in $R.$
$(\Leftarrow)\ \ b\mid a\,$ in $R\,\Rightarrow\, a = br,\ $ for some $\,r\in R,\ $ so $\ aR = brR\subseteq bR,\,$ by $\,rR\subseteq R.\ $ QED

Corollary $\ $ If $\,c\in bR\,$ and $\,c\mid d\,$ then $\,bR \supset dR$.

Proof $\ $ Translating containment relations to corresponding divisibilty relations we obtain $\ c\in bR\,\Rightarrow\, b\mid c,\,$ so $\, c\mid d\,\Rightarrow\, b\mid d\,\Rightarrow\, bR\supseteq dR,\,$ by transitivity of "divides" and Theorem. $\ $ QED

Your exercise is the special case of the Corollary when $\,c\,$ is a unit, i.e. $\,cr = 1\,$ for some $\,r\in R,\,$ therefore $\,c\mid 1,\,$ so with $\,d=1\,$ in the corollary, we conclude that $\,bR\,\supseteq\, 1\cdot R = R.\,$ Hopefully this clarifies the motivation behind my hint in the comments to your question.

While imposing this conceptual structure costs slightly more effort, the rewards are great, since this structure plays a fundamental conceptual role in divisibility theory and related algebra. Indeed, if you later study algebraic number theory you will learn that ideals satisfying "contains = divides" play a key role in Dedekind's beautiful ideal-theoretic restoration of unique factorization in algebraic number rings (Dedekind domains are those domains whose proper ideals have prime factorizations (necessarily unique); they are the domains whose ideals satisfy "contains = divides", and a suitable finiteness condition, e.g. Noetherian).

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  • Excellent expiation! I really learn a lot from your comments. I will definitely learn in-detailed the concepts you mentioned. – ZHJ Feb 12 '14 at 19:07