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In any ring, is the product of a non-unit with any other element necessarily a non-unit?

Alexander Gruber
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goblin GONE
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2 Answers2

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This is clearly true in a commutative ring. Suppose $r$ is a non-unit and $s$ is some other element in the ring. Suppose $rs$ is a unit, so $rst=1$ for some element $t$. Then $r(st)=1$ and $r$ is a unit, a contradiction.

This is less clearly false in a general non-commutative ring. I refer you to Arturo's answer here.

Here's a standard example:

Let $V$ be the vector space of all real sequences, with pointwise addition. Let $R$ be the ring of all linear transformation $V\to V$, with multiplication being composition of functions. Then we can let $\lambda\colon V\to V$ be the left-shift operator that maps the sequence $(a_0,a_1,a_2,\ldots)$ to $(a_1,a_2,a_3,\ldots)$, and $\rho\colon V\to V$ be the right-shift operator that maps the sequence $(a_0,a_1,a_2,\ldots,)$ to $(0,a_0,a_1,a_2,\ldots)$.

Then $\lambda\rho=1$, so $\lambda$ is a left divisor of a unit, and $\rho$ is a right divisor of a unit. However, neither $\lambda$ nor $\rho$ are units: a unit in $R$ must be a bijective linear transformation on $V$, but $\lambda$ is not one-to-one, and $\rho$ is not onto, so neither element is a unit.

Community
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Potato
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  • Thanks for catching my cognitive misfire below. I blame daylight savings time. I somehow bent it into "product of nonunit with unit". – rschwieb Mar 11 '13 at 19:17
  • @rschwieb No problem. I'm wondering if this is true in a non-commutative ring now... – Potato Mar 11 '13 at 19:18
  • Also true in any Dedekind-finite ring. This includes one-sided Noetherian rings, any domain, any local ring, and two-sided self injective rings. – rschwieb Mar 11 '13 at 19:20
  • @rschwieb See my edit. – Potato Mar 11 '13 at 19:25
  • Yeah I was just thinking that any example of a non Dedekind finite ring would suffice :) If $xy=1$ and $yx\neq 1$, then $x$ is clearly a nonunit, and $xy$ is clearly a unit. So, the OP's original condition holds iff the ring is DF. – rschwieb Mar 11 '13 at 19:27
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Hint $\rm\ U\,$ closed under divisors $\,\Rightarrow\,$ its complement $\rm\:U'$ closed under multiples.

Proof $\ $ If not, there is $\rm\:n\in U',\:$ with $\rm\:nx\in U,\:$ contra $\rm\,U\,$ closed under divisors.

Note $\ $ The set of units $\rm\,U\,$ is closed under divisors, i.e. if $\rm\,u\,$ is a unit then so is every divisor of $\rm\,u.\,$ Indeed, units are precisely the divisors of $\,1,\:$ hence by transitivity of 'divides' we deduce $$\rm xy\ unit\ \Rightarrow\ xy\ |\ 1\ \Rightarrow\ x\ |\ xy\ |\ 1\ \Rightarrow\ x\ unit\ $$

i.e. the set of all divisors of a fixed element is closed under taking divisors (by transitivity).

Math Gems
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  • This is wrong when the ring is not assumed to be commutative. – Potato Mar 11 '13 at 19:26
  • @Potato I see no mention in the question that the ring is noncommutative. The standard convention is that ring means commutative unless otherwise qualified. If not then almost all MSE questions on rings would be "wrong". If this is the source of the downvote, then I think that is a poor reason to downvote. – Math Gems Mar 11 '13 at 20:02
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    I really don't think that is the standard convention. The question clearly mentions "any" ring. – Potato Mar 11 '13 at 21:20
  • @Potato I'm a ring theorist with many years experience. My experience is that noncommutativity is normally not implied unless there is some explicit mention (or ambient context) that implies such. As such, "any ring" means usually means "any commutative ring". – Math Gems Mar 11 '13 at 21:34
  • I'll defer to you then. – Potato Mar 12 '13 at 05:43
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    I'm a ring theorist with several years of experience (quite probably not as many as you) but my impression is that the majority of ring theorists I've interacted with would not assume commutativity as part of the defintion of a ring. In any case, it never hurts to qualify answers when the convention is in doubt. – rschwieb Mar 12 '13 at 13:33
  • @rschwieb Probably 99% of ring theory questions here do not concern noncommutative rings. Even less so for nonassociative rings such as as Lie algebras or Jordan algebras. As such, the default denotation for "ring" is normally commutative and associative. That's the way it should be in a general level math forum. – Math Gems Mar 12 '13 at 14:05
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    @MathGems Probably also 99% of commutative ring theory questions are labeled as commutative. It's perhaps a little unrealistic to compare associativity and commutativity. Anyhow, I've achieved my purpose of raising the awareness of differences of assumptions, here. I don't want to moralize about "how it should be", I just think more information is better. – rschwieb Mar 12 '13 at 14:22
  • @rschwieb I don't agree. Nor do I find constructive arguing about conventions. So let's agree to disagree. – Math Gems Mar 12 '13 at 14:23
  • @MathGems Agreed! – rschwieb Mar 12 '13 at 15:13