## What is a prime, and who decides?

# What is a prime, and who decides?

*Some people view mathematics as a purely platonic realm of ideas independent of the humans who dream about those ideas. If that’s true, why can’t we agree on the definition of something as universal as a prime number?*

**Courtney R. Gibbons**

**Hamilton College**

### Introduction

*Scene:* It’s a dark and stormy night at SETI. You’re sitting alone, listening to static on the headphones, when all of the sudden you hear something: two distinct pulses in the static. Now three. Now five. Then seven, eleven, thirteen — it’s the sequence of prime numbers! A sequence unlikely to be generated by any astrophysical phenomenon (at least, so says Carl Sagan in *Contact*, the novel from which I’ve lifted this scene) — in short, proof of alien intelligence via the most fundamental mathematical objects in the universe…

Hi! I’m Courtney, and I’m new to this column. I’ve been enjoying reading my counterparts’ posts, including Joe Malkevitch’s column Decomposition and David Austin’s column Meet Me Up in Space. I’d like to riff on those columns a bit, both to get to some fun algebra (atoms and ideals!) and to poke at the idea that math is independent of our humanity.

### Introduction, Take 2

Scene: It’s a dark and stormy afternoon in Clinton, NY. I’m sitting alone at my desk with two undergraduate abstract algebra books in front of me, both propped open to their definitions of a prime number…

- Book A says that an integer $p$ (of absolute value at least 2) is
*prime*provided it has exactly two positive integer factors. Otherwise, Book A says $p$ is*composite*. - Book B says that an integer $p$ (of absolute value at least 2) is
*prime*provided whenever it divides a product of integers, it divides one of the factors (in any possible factorization). Otherwise, Book B says $p$ is*composite*.

* Note: *Book A is Bob Redfield’s

*Abstract Algebra: A Concrete Introduction*(Bob is my predecessor at Hamilton College). Book B

*Abstract Algebra: Rings, Groups, and Fields*by Marlow Anderson (Colorado College; Marlow was my undergraduate algebra professor) and Todd Feil (Denison University).

I reached for the nearest algebra textbook to use as a tie-breaker, which happened to be Dummit and Foote’s *Abstract Algebra*, only to find that the authors hedge their bets by providing Book A’s definition and then saying, well, actually, Book B’s definition can be used to define prime, actually. Yes, it’s a nice exercise to show these definitions are equivalent. I can’t help but wonder, though: which is what it really * is* to be prime, and which is merely a consequence of that definition?

### Who Decides?

Some folks take the view that math is a true and beautiful thing and we humans merely discover it. This seems to me to be a way of saying that math is independent of our humanity. Who we are, what communities we belong to — these don’t have any effect on Mathematics, Platonic Realm of Pure Ideas. To quantify this position as one might for an intro to proofs class: For each mathematical idea $x$, $x$ has a truth independent of humanity. And yet, two textbooks fundamental to the undergraduate math curriculum are sitting here on my desk with the audacity to disagree about the very definition of arguably the **most** pure, **most** platonic, **most** absolutely mathematical phenomenon you could hope to encounter: prime numbers!

This isn’t a perfect counterexample to the universally quantified statement above (maybe one of these books is wrong?). But in my informal survey of undergraduate algebra textbooks (the librarians at Hamilton really love me and the havoc I wreak in the stacks!), there’s not exactly a consensus on the definition of a prime!

As far as I can tell, the only consensus is that we shouldn’t consider $-1$, $0$, or $1$ to be prime numbers.

But, uh, why not?! In the case of $0$, it breaks both definitions. You can’t divide by zero (footnote: well, you shouldn’t divide by if you want numbers to be meaningful, which is, of course, a decision that someone made and that we continue to make when we assert “you can’t divide by zero”), and zero has infinitely many positive integer factors. But when $\pm 1$ divides a product, it divides one (all!) of the factors. And what’s so special about exactly two positive divisors anyway? Why not “at most two” positive divisors?

Well, if you’re reading this, you probably have had a course in algebra, and so you know (or can be easily persuaded, I hope!) that the integers have a natural (what’s natural is a matter of opinion, of course) algebraic analog in a ring of polynomials in a single variable with coefficients from a field $F$. The resemblance is so strong, algebraically, that we call $F[x]$ an integral domain (“a place where things are like integers” is my personal translation). The idea of prime, or “un-break-down-able”, comes back in the realm of polynomials, and Book A and Book B provide definitions as follow:

- Book B says that a nonconstant polynomial $p(x)$ is
*irreducible*provided the only way it factors is into a product in which one of the factors must have degree 0 (and the other necessarily has the same degree as $p(x)$). Otherwise, Book B says $p(x)$ is*reducible*. - Book A says that a nonconstant polynomial $p(x)$ is
*irreducible*provided whenever $p(x)$ divides a product of polynomials in $F[x]$, it divides one of the factors. Otherwise, Book A says $p(x)$ is*reducible*.

Both books agree, however, that a polynomial is reducible if and only if it has a factorization that includes more than one irreducible factor (and thus a polynomial cannot be both reducible and irreducible).

Notice here that we have a similar restriction: the zero polynomial is excluded from the reducible/irreducible conversation, just as the integer 0 was excluded from the prime/composite conversation. But what about the other constant polynomials? They satisfy both definitions aside from the seemingly artificial caveat that they’re not allowed to be irreducible!

Well, folks, it turns out that in the integers and in $F[x]$, if you’re hoping to have meaningful theorems (like the Fundamental Theorem of Arithmetic or an analog for polynomials, both of which say that factorization into primes/irreducibles is unique up to a mild condition), you don’t want to allow things with multiplicative inverses to be among your un-break-down-ables! We call elements with multiplicative inverses *units*, and in the integers, $(-1)\cdot(-1) = 1$ and $1\cdot 1 = 1$, so both $-1$ and $1$ are units (they’re the only units in the integers).

In the integers, we want $6$ to factor uniquely into $2\cdot 3$, or, perhaps (if we’re being generous and allowing negative numbers to be prime, too) into $(-2)\cdot(-3)$. This generosity is pretty mild: $2$ and $-2$ are *associates*, meaning that they are the same up to multiplication by a unit. One statement of the Fundamental Theorem of Arithmetic is that every integer (of absolute value at least two) is prime or factors uniquely into a product of primes *up to the order of the factors and up to associates*. That means that the list prime factors (up to associates) that appear in the factorization of $6$ is an invariant of $6$, and the number of prime factors (allowing for repetition) in any factorization of $6$ is another invariant (and it’s well-defined). Let’s call it the *length* of $6$.

But if we were to let $1$ or $-1$ be prime? Goodbye, fundamental theorem! We could write $6 = 2\cdot 3$, or $6 = 1\cdot 1\cdot 1 \cdots 1 \cdot 2 \cdot 3$, or $6 = (-1)\cdot (-2) \cdot 3$. We have cursed ourselves with the bounty of infinitely many distinct possible factorizations of $6$ into a product primes (even accounting for the order of the factors or associates), and we can’t even agree on the length of $6$. Or $2$. Or $1$.

The skeptical, critical-thinking reader has already been working on workarounds. Take the minimum number of factors as the length. Write down the list of prime factors without their powers. Keep the associates in the list (or throw them out, but at that point, just agree that $1$ and $-1$ shouldn’t be prime!). But in the polynomial ring $F[x]$, dear reader, **every** nonzero constant polynomial is a unit: given $p(x) = a$ for some nonzero $a \in F$, the polynomial $d(x) = a^{(-1)}$ is also in $F[x]$ since $a^{(-1)}$ is in the field $F$, and $p(x)d(x) = 1$, the multiplicative identity in $F[x]$. So, if you allow units to be irreducible in $F[x]$, now even an innocent (and formerly irreducible) polynomial like $x$ has infinitely many factorizations into things like ($a)(1/a)(b)(1/b)\cdots x$. So much for those workarounds!

So, since we like our Fundamental Theorems to be neat, tidy, and useful, we agree to exclude units from our definitions of prime and composite (or irreducible and reducible, or indecomposable and decomposable, or…).

### More Consequences (or Precursors)

Lately I’ve been working on problems related to semigroups, by which I mean nonempty sets equipped with an associative binary operation — and I also insist that my semigroups be commutative and have a unit element. In the study of factorization in semigroups, the Fundamental Theorem of Arithmetic leads to the idea of counting the distinct factors an element can have in any factorization into *atoms* (the semigroup equivalent of irreducible/prime elements; these are elements $p$ that factor only into products involving units and associates of $p$).

One of my favorite (multiplicative) semigroups is $\mathbb{Z}[\sqrt{-5}] = \{a + b \sqrt{-5} \, : \, a,b \in \mathbb{Z}\}$, favored because the element $6$ factors distinctly into two different products of irreducibles! In this semigroup, $6 = 2\cdot 3$ and $6 = (1+\sqrt{-5})(1-\sqrt{-5})$. It’s a nice exercise to show that $1\pm \sqrt{-5}$ are not associates of $2$ or $3$, yielding two distinct factorizations into atoms!

While we aren’t lucky enough to have unique factorization, at least we have that the number of irreducible factors in any factorization of $6$ is always two. That is, excluding units from our list of atoms leads to an invariant of $6$ in the semigroup $\mathbb{Z}[\sqrt{-5}]$.

Anyway, without the context of more general situations like this semigroup (and I don’t know, is $\mathbb{Z}[\sqrt{-5}]$ one of those platonically true things, or were Gauss et al. just really imaginative weirdos?), would we feel so strongly that $1$ is not a prime integer?

### Still More Consequences (or precursors)

Reminding ourselves yet again that the integers form a ring under addition and multiplication, we might be interested in the ideals generated by prime numbers. (What’s an ideal? It’s a nonempty subset of the ring closed under addition, additive inverses, and scalar multiplication from the ring.) We might even call those ideals prime ideals, and then generalize to other rings! The thing is, if we do that, we end up with this definition:

(Book A and B agree here:) An ideal $P$ is prime provided $xy \in P$ implies $x$ or $y$ belongs to $P$.

But in the case of the integers — a principal ideal domain! — that means that a product $ab$ belongs to the principal ideal generated by the prime $p$ precisely when $p$ divides one of the factors.

From the perspective of rings, every (nonzero) ring has two trivial ideals: the ring $R$ itself (and if $R$ has unity, then that ideal is generated by $1$, or any other unit in $R$) and the zero ideal (generated by $0$). If we want the study of prime ideals to be the study of interesting ideals, then we want to exclude units from our list of potential primes. And once we do, we recover nice results like an ideal $P$ is prime in a commutative ring with unity if and only if $R/P$ is an integral domain.

### Conclusions

I still have two books propped open on my desk, and after thinking about semigroups and ideals, I’m no closer to answering the question “But what is a prime, really?” than I was at the start of this column! All I have is some pretty good evidence that we, as mathematicians, might find it useful to exclude units from the prime-or-composite dichotomy (I haven’t consulted with the mathematicians on other planets, though). To me, that evidence is a reminder that we are constantly updating our mathematics framework in reference to what we learn as we do more math. We look back at these ideas that seemed so solid when we started — something fundamentally indivisible in some way — and realize that we’re making it up as we go along. (And ignoring a lot of what other humans consider math, too, as we insist on our axioms and law of the excluded middle and the rest of the apparatus of “modern mathematics” while we’re making it up…)

And the math that gets done, the math that allows us to update our framework… Well, that depends on *what* is trendy/fundable/publishable, *who* is trendy/fundable/publishable, and *who* is making all of those decisions. Perhaps, on planet Blarglesnort, math looks very different.

### References

Anderson, Marlow; Feil, Todd. *A first course in abstract algebra. Rings, groups, and fields.* Third edition. ISBN: 9781482245523.

Dummit, David S.; Foote, Richard M. *Abstract algebra.* Third edition. ISBN: 0471433349.

Redfield, Robert. Abstract algebra. A concrete introduction. First edition. ISBN: 9780201437218.

Geroldinger, Alfred; Halter-Koch, Franz. *Non-unique factorizations. Algebraic, combinatorial and analytic theory.* ISBN: 9781584885764.