Context: In this exact moment of this online lecture the following statement is made (consider misunderstandings):

Let $R$ be a ring. An element $a \in R$ is called irreducible if it cannot be written as a product of two elements in $R$ unless one of those elements is a unit.

That is, $a$ is irreducible if $a=bc\implies \left(b\in R^\times \text{ or } c\in R^\times \right).$

Example in the polynomial ring over the rationals:

$p=18 - 12t=(6)(3-2t)$ is irreducible despite factoring out $6.$

Here is the quote:

$(6)$ doesn't count, because it is a unit in $\mathbb Q,$ although it would count over the integers, $\mathbb Z[t].$

My take is that the understanding the speaker is making reference to is the concept of the polynomial ring over the rationals as a vector space with basis $\color{red}1, x, x^2,\dots.$ And I don't know how this translates to the subring of polynomials over the integers, $\mathbb Z[t],$ except the lead in Wikipedia regarding the basis of such polynomials being


for $k=0,1,2,\dots$ An unfamiliar expression thus far...

  • 3
    $\begingroup$ Polynomials over $Z$ form a commutative ring by the usual operations. You don't need the vector space and basis concepts to work with it. The point here is that $18-12t$ is not irreducible as a polynomial over $Z$ since you can write it as the product two non-units (i.e. 6 is not a unit in $Z$), whereas it is irreducible over $Q$ (as is any linear polynomial) since every constant non-zero poly (i.e. a non-zero rational number) is a unit over $Q$. $\endgroup$ – Ned Jan 20 '18 at 16:37
  • $\begingroup$ @Ned Two questions: 1. When you say "6 is not a unit in $Z$" are you referring to the integers with + and x operations? If so, I thought that $\mathbb Z[t]$ in the lecture made reference to the polynomials with integer coefficients, instead. And, 2. is "every constant non-zero poly" a unit because of the vector space concept? $\endgroup$ – Antoni Parellada Jan 20 '18 at 16:43
  • $\begingroup$ 1. Yes, but the units in a polynomial ring are precisely the units in the coefficient ring, since non-constant polys don't have multiplicative inverses in the polynomial ring. 2. When the coefficient ring is a field such as $Q$, then every non-zero constant poly has a multiplicative inverse, namely it's inverse in the field (i.e. the constant polys in a polynomial ring are an exact copy of the coefficient ring [isomorphic is the technical term]) $\endgroup$ – Ned Jan 20 '18 at 17:39

As others have said, vector spaces are not required here.

The issue is how to generalise the concept of irreducible or prime to more general rings. Often, but not always, they are the same. I will ignore the distinction here since I am just trying to explain the motivation of the definition of unit. One interesting question in a ring is whether prime factorisation is unique; sometimes it is and sometimes it is not.

Is prime factorisation unique in $\mathbb{Z}$? We normally say it is but how about $6 = 2 \times 3 = (-2) \times (-3)$. Does that violate unique factorisation? No, because we usually ignore the negatives integers. Also, $6 = 2 \times 3 = 1 \times 2 \times 3$. Does that violate unique factorisation? No, because it is no longer usual to consider $1$ prime. It is not composite either, so what is it? It's another category: "unit" - an element with an inverse in the ring.

Now let's look at a more interesting example: the Gaussian Integers, $\mathbb{Z}[i]$. These are the complex numbers with integer real and imaginary components. At first, it seems that prime factorisation is not unique since $10 = 2 \times 5 = (3 + i) \times (3 - i)$. However, the catch is that $2$ is not a prime in this ring since $2 = (1 + i) \times (1 - i)$. $(1 + i)$ and $(1 - i)$ are prime. Is the factorisation unique? Well, $2 = (-1 + i) \times (-1 - i)$ so it seems not. However, these are just the same factors multiplied by $i$ and $-i$ and these are units, they have inverses in the ring. These units allow you to create multiple factorisations. So, it is usual to ignore them; a prime factorisation is considered unique if any other can be obtained from it by rearranging the terms or multiplying them by units.

Back to the normal integers, the problem can be easily hidden by picking the positive integers as the canonical factors. This is not so easy in the Gaussian integers or arbitrary rings.

Now, to your examples. In $\mathbb{Q}[t]$, all degree zero polynomials are units, they have inverses, and hence are not interesting in factorisations. But in $\mathbb{Z}[t]$, only the two degree zero polynomials $1$ and $-1$ have inverses, the others might or might not be prime or irreducible.

Finally, consider a field such as $\mathbb{Q}$ or $\mathbb{R}$. All elements, except $0$, have an inverse and hence are units. There are no primes and hence the question of unique factorisation is uninteresting.

  • $\begingroup$ Thank you (+1). Very lucid and well written. $\endgroup$ – Antoni Parellada Jan 20 '18 at 18:05

It's always the missing steps...

Here is the answer (Wikipedia so often to the rescue):

In mathematics, an invertible element or a unit in a ring $R$ is any element $u$ that has an inverse element in the multiplicative monoid of $R$, i.e. an element $v$ such that

$uv = vu = 1R,$ where $1R$ is the multiplicative identity.

The multiplicative identity in $\mathbb Q[t]$ is $1;$ hence $6$ is a unit since $1/6 \in \mathbb Q[t].$ Yet, $1/6 \notin \mathbb Z[t].$


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