Is it possible that $(ab)^{-1}$ is defined although $a^{-1},b^{-1}$ are not?

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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.