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An irreducible element of a ring is a non-unit such that it cannot be written as a product of two non-units.

A UFD is a domain such that each non-unit $x\in R \backslash \{0\}$ can be written as a product of irreducible elements.

So if we take an irreducible element $r$ in a UFD, then it's not a unit and so admits a unique factorization into irreducible elements $r_1\dots r_n$ (which are also non-units). As it's irreducible it cannot be written as a product of two non-units so if $r = ab$ then $a$ or $b$ is a unit.

Does this mean that irreducible elements can be multiplied together to form units? And therefore non-units can be multiplied together to form units?

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    Everything you've stated is correct, but I don't see how you come to the conclusion that irreducible elements can be multiplied together to form units. – Ted May 20 '12 at 17:45
  • It's sort of irrelevant for the question in the title, but in your setup note that since you are in a UFD it must be the case that $ n=1$ and $r_1 = ur$ for some unit $u$. – Dylan Moreland May 20 '12 at 17:54

4 Answers4

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In a factorization, note that $n$ can be 1 (or zero!).

In general if $ab$ is a unit, then there exists $c$ so that $(ab)c=1$. Now, what can you say about $a$?

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    Noting that multiplication is associative ... – Mark Bennet May 20 '12 at 17:50
  • It implies that 'a' has a right inverse but not necessarily (at least I don't see it) a left inverse, and 'b' would have a left inverse. So is what you're saying not, in general, true for non commutative rings? – Anshuman Agrawal Jan 30 '24 at 17:35
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Hint $ $ Units = divisors of $1,$ so divisors of units are units $\rm\:a\:|\:b\:|\:1\:\Rightarrow\:a\:|\:1\:$ by transitivity of $ $ '|'.

Thus the unit group $\rm\,U\,$ is a saturated monoid, i.e $\rm\:ab\in U\iff a,b\in U.$

Hence nonunit $\rm\:\!a\:\Rightarrow\: ab\:\!$ nonunit, another example of a complementary view of a subgroup.

Bill Dubuque
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I don't understand your "Does that mean..."

In a UFD, the factorization into irreducibles of an irreducible is just itself. Just like, in $\mathbb{Z}$, the way to express $2$ as a product of primes is by writing $2=2$.

In a commutative ring, any divisor of a unit is itself a unit: If $x$ divides $u$ and $u$ is a unit, then there exists $y$ such that $xy=u$, and there exists $v$ such that $uv=1$. Then $x(yv) = (xy)v = uv = 1$, so $x$ is a unit.

In particular, in a commutative ring, a product that contains at least one nonunit factor is a nonunit; and a product of units is a unit (easy to verify).

In a noncommutative ring, however, it is possible for a non-unit to divide a unit (on one side). Recall that in a noncommutative ring, a unit is an element $u$ that has a two-sided inverse, that is an element $v$ such that $uv=vu=1$. An element may have a one-sided inverse and not be a unit.

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.

Arturo Magidin
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The product of two non-zero elements is a unit iff the elements themselves are units. If $a$ and $b$ are irreducible this means that $a$ and $b$ are non-units, and hence $ab$ is non-unit; so the product of two or more irreducibles cannot be a unit.

311411
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Muhammad Usman
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