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I know if $x$ and $y$ are units, in say a commutative ring, then $xy$ is a unit with $(xy)^{-1}=y^{-1}x^{-1}$.

But if $xy$ is a unit, does it necessarily follow that $x$ and $y$ are units?

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How do you define unit? – Andrés E. Caicedo Jan 17 '12 at 21:54
@AndresCaicedo I define unit to be an invertible element. I'm assuming that the commutative ring is has $1$. – neel Jan 17 '12 at 21:55
Ok, so $xy$ has an inverse. Can you use that to define the inverse of $x$? The point is: To be invertible means that there is some element $z$ that when multiplied with $xy$ gives you 1. So you have $(xy)z=1$. Right? But then, this equation provides you with an element that multiplied with $x$ gives you 1. – Andrés E. Caicedo Jan 17 '12 at 21:57
@AndresCaicedo - you are using the associative law quite explicitly in your comment. I think it is another interesting question what happens if associativity fails ... of course the question as stated implies that multiplication is associative. – Mark Bennet Jan 17 '12 at 23:02
up vote 15 down vote accepted

Yes. Let $z=xy$. If $z$ is a unit with inverse $z^{-1}$, then $x$ is a unit with inverse $yz^{-1}$, and $y$ is a unit with inverse $xz^{-1}$, because $$x(yz^{-1})=(xy)z^{-1}=zz^{-1}=1$$ $$y(xz^{-1})=(yx)z^{-1}=(xy)z^{-1}=zz^{-1}=1$$

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Of course, that makes sense. Thanks Zev. – neel Jan 17 '12 at 21:57
No problem, glad to help! – Zev Chonoles Jan 17 '12 at 21:59

In a commutative ring, yes: let $u$ be such that $u(xy)=1 = (xy)u$. Then $$x(uy) = u(xy) = 1\quad\text{and}\quad (uy)x = u(xy) = 1,$$ so $x$ is invertible; and if $x$ is invertible, and $xy$ is a unit, then $y=x^{-1}xy$ is a product of units, hence a unit.

In a non-commutative ring, no; you can have a product be a unit yet neither factor be a unit. For example, let $$A = \prod_{i=1}^{\infty}\mathbb{Z}$$ and let $R$ be the ring of endomorphisms of $A$. Let $f\colon A\to A$ be the right-shift operator, and let $g\colon A\to A$ be the left-shift operator. Then $gf=1$ in $R$ and in particular the product is a unit, but neither $f$ nor $g$ are units ($g$ is not one-to-one, so it cannot be left-invertible, and $f$ is not onto, so it cannot be right-invertible).

Of course, if $xy$ is a unit in a non-commutative ring, then $x$ is right-invertible and $y$ is left-invertible, but that's the best you can say in general, as the example above shows.

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Thank you, I've not dealt much with noncommutative rings, so I appreciate this example. – neel Jan 17 '12 at 23:48
Can the downvoter explain why there is a downvote here? – Arturo Magidin Jan 20 '12 at 16:12
I've had my answer here and on another recent question downvoted without explanation too. – Zev Chonoles Jan 20 '12 at 16:20

HINT $\ $ Units are precisely the divisors of $1\:.\:$ Hence by transitivity of 'divides' we deduce

$$\rm xy\ unit\ \Rightarrow\ xy\ |\ 1\ \Rightarrow\ x\ |\ xy\ |\ 1\ \Rightarrow\ x\ unit\ \qquad QED$$

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

This can be viewed as the dual of the well-known "divides = contains" for principal ideals. Namely, let $\rm\: D(x) =\: $ the set of divisors of $\rm\:x\:$ and let $\rm\: M(x) =\: $ the set of multiples of $\rm\:x\:.\:$ Then we have

$\rm\qquad a\ |\ b\ \iff\ M(a) \supset M(b)\quad $ i.e. divides = contains $\qquad\ \ $ for multiple sets

$\rm\qquad a\ |\ b\ \iff\ D(a)\ \subset\: D(b)\quad $ i.e. divides = contained-in $\ $ for divisor sets

So $\rm\ u\ |\ 1\ \iff\ D(u)\:\subset D(1)\quad\:$ i.e. $(\Rightarrow)$ says: $\:$ if $\rm\:u\:$ is a unit then every divisor of $\rm\:u\:$ is a unit.

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NO if $R$ is not commutative

Definition: Suppose $R$ is any ring with unity $1$. An element $u$ is said to be unit in $R$ if there exits $v\in R$ such that $uv=1$ and $vu=1$.

Let $\mathbb{R}[x]$ be the (infinite dimensional) vector space of all polynomials over $\mathbb{R}$. Let $S$ denote the ring of all linear operators on $\mathbb{R}[x]$ with usual addition and composition of operators.

Let $D\colon\mathbb{R}[x]\rightarrow\mathbb{R}[x]$ demote the differential operator: $D(p(x))=p'(x)$.

Let $J\colon:\mathbb{R}[x]\rightarrow \mathbb{R}[x]$ denote the integral operator: $J(a_0+a_1x+\cdots + a_nx^n)=a_0x+a_1\frac{x^2}{2}+\cdots + a_n\frac{x^{n+1}}{n+1}$.

Then $D\circ J$ is identity operator which is obviously unit. But $D$ is not unit: it is not one-to-one (why?) hence it can not have a two-sided inverse.

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This was already noted before, see the answer by Arturo Magidin. And the OP assumed $R$ to be commutative. – Martin Brandenburg Apr 7 '15 at 12:35

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