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Statement:

Let $ (R,+,\cdot) $ be a ring that has at least two invertible elements. Let $a,b$ be two invertible elements such that $$ord(a)=ord(b)=+\infty$$ Let consider the following sequence $$c_n=a^n-b^n$$ Let $S=\{c_1,c_2,...,c_m,...\}$ a set of elements from $R$.

Prove that if $S=\{c_1,c_2,...\}$ is finite then $$0 \in S$$

Attempt:

I obtained that:

$$a^n-b^n=c_n$$ then $$a^{n+1}-ab^n=ac_n$$ and $$a^{n+1}-b^{n+1}=c_{n+1}$$ then $$(b-a)b^{n}=ac_n-c_{n+1}$$ Since $S$ is finite then $$M=\{ac_n-c_{n+1},n \in \mathbb Z_+\}$$ is finite.

So $$\{(b-a)b^n,n \in \mathbb Z_+\}$$ is finite. Then there exists $i$ and $j$ such that $$(b-a)b^i=(b-a)b^j$$

How should I continue with this. I would highly appreciate somebody's help.

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    $\begingroup$ Is there a connection to this question you had before? $\endgroup$ May 14 at 18:14
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    $\begingroup$ Is the ring assumed to be commutative? $\endgroup$ May 14 at 21:46
  • $\begingroup$ No, the ring is not commutative. $\endgroup$
    – shangq_tou
    May 15 at 6:50

1 Answer 1

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Modified to NOT assume $R$ is commutative

Continuing from where you left off, you could use your line of reasoning to show the following: For any positive integer $l$ [not just $l=1$], there is a positive integer $k_l$ such that $$(a^{k_l}-1)c_l = 0.$$ [Indeed for each such $l$ there exists distinct integers $i$ and $j$ such that $a^ic_l = a^jc_l$, and assume $j >i$. Then $k_l=j-i$ works.] Furthermore, $(a^k-1)c_l =0$ for any $k$ that is a multiple of $k_l$, because even in a noncommutative ring, $(a^{C_1}-1)$ and $(a^{C_2}-1)$ commute for any two integers $C_1$,$C_2$. So let $T$ be a finite set of integers $l$ such that for all $m \in \mathbb{N}$ there is an $l \in T$ such that $c_m=c_l$, and set $$k = \prod_{l \in T} k_l.$$ Then $$(a^k-1)c_m = 0 \ \forall m \in \mathbb{N}.$$

In fact, for any multiple $K$ of $k$, it follows that $$(a^K-1)c_m =0 \ \forall m \in \mathbb{N}.$$ Thus we conclude with the following:

Claim 1 There is a positive integer $k \in \mathbb{N}$ such that the equation $$(a^K-1)c_m=0 \ \forall m \in \mathbb{N}$$ holds for all multiples $K$ of $k$.

We also make the following claim:

Claim 2 There exists a multiple $K$ of $k$ and a positive integer $n$ such that the equation $c_{n+K}=c_n$ holds.

Proof: Let $N$ be the number of distinct $c_l$; $l \in \mathbb{N}$. Then for any positive integer $m$, at least two of $c_m, c_{m+jk}$; $j=1,2,\ldots, N$; must be the same by the Pigeonhole Principle. ■

So, using Claims 1 and 2, let $K$ be a multiple of $k$, where $k$ is as in Claim 1, and let $n$ be an integer such that the equation $c_n=c_{n+K}$ holds.

Then simple algebra gives $$a^Kc_n +a^Kb^n -b^{n+K} = c_{n+K}.$$ plugging into this the equation $c_n=c_{n+K}$ and rearranging gives $$(a^K-1)c_{n+K} + c_Kb^n = 0.$$ But, as $K$ is a multiple of $k$ as in Claim 1, the equation $(a^K-1)c_{n+K}=0$ holds, so we are left with the equation $$c_Kb^n = 0.$$ As $b^n$ is invertible, it follows that the equation $$c_K=0$$ must hold. ■

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  • $\begingroup$ For the equation $$ a^Kc_n +a^Kb^n -b^{n+K} = c_{n+K} $$ can you provide a little more detail? $\endgroup$
    – quasi
    May 16 at 8:51
  • $\begingroup$ $$a^Kc_n +a^Kb^n-b^{n+K}$$ $$= a^K(a^n-b^n)+a^Kb^n-b^{n+k}$$ $$= (a^{K+n}-a^Kb^n)+a^kb^n-b^{n+K}$$ $$=a^{K+n} -a^Kb^n+a^kb^n -b^{n+K}$$ $$=a^{n+K}-b^{n+K}$$ $$= c_{n+K}$$. $\endgroup$
    – Mike
    May 16 at 15:27
  • $\begingroup$ @quasi here it is in the above comment $\endgroup$
    – Mike
    May 16 at 15:41
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    $\begingroup$ Got it. Thanks.${}{}{}{}{}{}{}$ $\endgroup$
    – quasi
    May 17 at 2:46

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