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If $u_1u_2\cdots u_n=1$ in a commutative ring, then all of $u_i$ are units.

Does the proof follow some logic like the following:

$u_1(u_2\cdots u_n)=1\implies u_1,u_2\cdots u_n$ are both units, so $u_1$ is a unit,

$u_2(u_1u_3u_4\cdots u_n\implies u_2,u_1u_3\cdots u_n)$ are both units, so $u_2$ is a unit,

etc

Or is there some other way to show this?

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    $\begingroup$ That looks good. $\endgroup$ – Empy2 Nov 3 '15 at 12:22
  • $\begingroup$ @Michael I was afraid of that :P. Now I must work out a way to formalise this ! $\endgroup$ – Isthisnorm al Nov 3 '15 at 12:25
  • $\begingroup$ As the ring is commutative, you have $u_i^{-1}=u_1u_2\dots \hat u_i \dots u_n$ (where the "hat" means that this term is deleted from the product). $\endgroup$ – lulu Nov 3 '15 at 12:27
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Here is a proof by induction.

The cases $n=1$ and $n=2$ are clear.

If $n>2$, let $v=u_1u_2$. Then $vu_3\cdots u_n=1$ and by induction $v, u_3, \ldots, u_n$ are units. Use the case $n=2$ on $v$ to conclude that $u_1, u_2$ are units.

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  • $\begingroup$ What about induction for $v_{n}=u_1\cdots u_n$ and $1=vu_{n+1}$ $\endgroup$ – Isthisnorm al Nov 3 '15 at 12:31
  • $\begingroup$ @Isthisnormal, well, $1=vu_{n+1}$ does not imply $v=1$, and so I don't see how induction would work in that case. $\endgroup$ – lhf Nov 3 '15 at 15:16
  • $\begingroup$ I meant to show $v_n$ and $u_{n+1}$ are units in each iteration. Well I didn't write what I wanted to right, so I'll try to get it going myself $\endgroup$ – Isthisnorm al Nov 3 '15 at 15:19
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No induction required.

If there exists $v$ such that $(u_1\dotsm u_n)v=1$, then $u_2\dotsm u_nv $ is the inverse of $u_1$. Similarly for all other $u_i$s.

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    $\begingroup$ I know that, it's what I said in the question. But it doesn't seem very rigorous. I think you are implicitly inducting to apply the definition of a unit $\endgroup$ – Isthisnorm al Nov 3 '15 at 13:28
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    $\begingroup$ The last comment is correct: what is written here establishes the truth of the claim "for one $n$ at a time," but what is necessary is to prove it "for all $n\in \Bbb N$ at once." That is heuristically what induction does for you. Once you understand how a proof goes for a particular $n$, induction lets you apply it to all of $\Bbb N$ simultaneously. So this solution is not an 'induction free' proof of the statement. $\endgroup$ – rschwieb Nov 3 '15 at 14:13
  • $\begingroup$ I don't agree with you. The induction is merely in the rewriting of $\displaystyle\prod_{i=1}^nu_i=k\prod_{\substack{i=1\\i\neq k}}^nu_i$, and this has nothing to see with units. I might have written that way. In my opinion, it' much lighter (and convincing) to write the way I did $\endgroup$ – Bernard Nov 3 '15 at 14:38

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