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I wish to enquire about the properties of units in abstract algebra.

In a ring $R$, a unit $u$ is an invertible element. Let $u=ab$. Is it possible that $a$ and $b$ are not units? Is it possible that they're prime?

Motivation: If $u=ab$, then $(ab)^{-1}$ exists. We know that if the inverses of $a$ and $b$ exist, then $(ab)^{-1}=b^{-1}a^{-1}$. However, if the inverses of $a$ and $b$ don't exist, I feel $ab$ can still be a unit. However, I'm not sure of this.

Thanks in advance!

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    $\begingroup$ If you're interested in commutative rings, you should say so. $\endgroup$
    – egreg
    Commented Oct 17, 2013 at 9:30

3 Answers 3

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Let $R = L(\ell^2)$ the linear continuous operators on $\ell^2$. Define $a,b \colon \ell^2 \to \ell^2$ by $$ a(x_1, x_2, \ldots) = (x_2, x_3, \ldots) $$ and $$ b(x_1, x_2, \ldots) =(0, x_1, x_2, \ldots) $$ Then $ab = 1$ is the identity, hence invertible, but neither $a$ nor $b$ are units as $a$ is not one to one, where $b$ is not onto.

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  • $\begingroup$ The endomorphism ring of an infinite dimensional vector space is a common source for counterexamples showing that something true in commutative rings is not necessary true in the noncommutative case. $\endgroup$
    – egreg
    Commented Oct 17, 2013 at 9:29
  • $\begingroup$ @egreg In this case, we can forget about rings and just talk about semigroups. Then every semigroup is a subsemigroup of a semigroup of endomorphisms under function composition, as above. So all counter-examples lie in this world! (Of course, the question is about rings, but if you want to put in in terms of rings then every semigroup is a subsemigroup of the semigroup underlying this ring...) $\endgroup$
    – user1729
    Commented Oct 17, 2013 at 10:39
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It is not possible in commutative rings. In general, if $ab$ is a unit then there exists, $v\in R$ such that $(ab)v=1$. Thus $a(bv)=1$. So, $a$ is a right unit. S Similarly it can be shown that $b$ is a left unit.

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    $\begingroup$ But only in commutative rings! Let $R = L(\ell^2)$, and $b : e_n \mapsto e_{n+1}$ the right shift, $a \colon e_n \mapsto e_{n-1}$, $ae_1 = 0$ the left shift. Then $ab = 1$ is a unit, but neither $a$ or $b$ are. $\endgroup$
    – martini
    Commented Oct 17, 2013 at 9:09
  • $\begingroup$ @ Martini: Yeah, you are right. I am sorry, let me rectify. If $ab$ is a unit, then $a$ is a right unit and $b$ is a left unit. $\endgroup$
    – Anupam
    Commented Oct 17, 2013 at 9:28
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If $ab$ is invertible, then there is an $x\in R$ with $x\cdot ab = ab\cdot x = 1$. From $(xa)b = 1$ we get that $b$ is left-invertible, and from $a(bx) = 1$ we get that $a$ is right-invertible.

The example in the answer of martini shows that a left-invertible element is not necessarily right-invertible.

Rings with the property "left-invertible iff right-invertible" are called Dedekind finite. So in Dedekind finite rings, your property is true. Special cases of Dedekind finite rings are commutative rings (answer of Anupam) and finite rings.

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