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Let $A$ be a ring and $u,v \in A^\times$. When do we have that $u + v \in A^\times$?

I think that A is needed to be an integral domain. For example consider $\mathbb{Z/6}$. Both $1$ and $5$ is a unit but their sum $1+5=0$ is not a unit.

Ninja
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    In any ring with one, both $1$ and it's additive inverse will be units, but their sum will be zero. – Mike Pierce Dec 21 '16 at 21:23
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    I'm doubtful that there's a nice answer for this other than "whenever it happens by chance." Obviously, $1+(-1)$ is not a unit in any ring and even in integral domains like $\mathbb Z$, we have $1+1$ is not a unit. – Milo Brandt Dec 21 '16 at 21:23
  • But then in $\mathbb{Z}/25\mathbb{Z}$, $1$, $2$ and $1+2$ are units. – Pierre-Guy Plamondon Dec 21 '16 at 21:23
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    Assuming $A$ commutative, $u+v$ is a unit if and only if $uv^{-1}+1$ is a unit. So you could ask when $1+u$ is a unit when $u$ is a unit and the answer is “almost never”. Much more interesting is the case when the sum of two non-units is a non-unit, which happens if and only if the ring is local. – egreg Dec 21 '16 at 21:38

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I do not think there is such a condition. For any unital ring such that $1 \neq 0,$ $1$ and $-1$ are units, but $1+(-1)=0,$ which is not a unit.

Pawel
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I don't sure about an answer for your question. However, there is a nice result of F. Beukers and Schlickewei about the number of solutions of the unit equation in a finitely generated subgroup.

ToThichToan
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In any field, the sum of two units is a unit, unless they're additive inverses. For a non-field example, let $\zeta$ be a primitive sixth-root of unity and let $A$ be the ring of integers in $\mathbb{Q}[\zeta]$. Then note that $\zeta^2+1 = \zeta$. (Because $0 = \zeta^6 - 1 = (\zeta^3-1)(\zeta^3+1) = (\zeta-1)(\zeta^2+\zeta+1)(\zeta+1)(\zeta^2-\zeta+1)$ and we can set this last factor equal to $0$.)

B. Goddard
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  • As mentioned above, we have that in any field, both $-1$, and $1$ are units. We have that $1+(-1) = 0$, which is not a unit. – Mark Schultz-Wu Dec 21 '16 at 21:49
  • Not hard to fix. – B. Goddard Dec 21 '16 at 22:53
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    Also thoroughly uninteresting, as you might as well have said "in a field, everything nonzero is a unit". The sum part here is unimportant. – Mark Schultz-Wu Dec 21 '16 at 22:54
  • @Mark Yet, that may be the answer the instructor is looking for. Also note that most of my answer is not about the field example. It was just an introductory, throw-away comment. – B. Goddard Dec 21 '16 at 22:57
  • How is a field extension a non-field example? – Oiler Dec 22 '16 at 04:18
  • @Oiler Because it works in the ring of integers. – B. Goddard Dec 22 '16 at 10:01
  • @B.Goddard then you might want to say 'in the ring of integers' – Oiler Dec 23 '16 at 18:51
  • @Oiler Yeah, yeah, there's always something that could be added or subtracted or reversed or some comma added. You could always write your own answer, and then have it say just what you want. Win-Win. – B. Goddard Dec 23 '16 at 19:28
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    $\mathbb{Q}[\zeta]$ is a field. You use it as a 'non-field example' in your answer. Making my own answer with that phrasing adds nothing new to the discussion but your answer can be improved. – Oiler Dec 24 '16 at 05:36