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

A: 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).
