Is the number $-1$ prime?

From my understanding it's not prime because it's not greater than $0$. So my followup question is why did mathematicians exclude $-1$?

The definition of prime is having only two factors.

$-1 \cdot 1 = -1$

Solution 1:

By commonly used definitions, units (invertibles) such as $\,\pm 1\,$ are not considered primes, even though they satisfy the divisibility property $\,p\mid ab\,\Rightarrow\, p\mid a\,$ or $\,p\mid b\,$ that is key to proving the uniqueness of factorizations into irreducibles. Excluding units (and zero) as primes simplifies the statement of uniqueness of prime factorizations and many related results (cf. unit normalization)

But - as with any convention - what proves convenient in one context may prove inconvenient in another. For example, when studying quadratic forms and related results, some authors find it convenient to consider certain units to be prime, e.g. $\,-1\,$ is considered prime. This idea was first employed by H. Hasse and has more recently been popularized and extended by John H. Conway.

Below are excerpts of some of Conway's remarks explaining why he calls $\,-1$ prime. The context is explained at much greater length in his very beautiful book The Sensual (Quadratic) Form, which was cited in his prestigious 2000 Steele prize for mathematical exposition ("This is a book rich in ideas. They seem to burst forth from almost every page")

From p. vii of $ $ The Sensual (Quadratic) Form: $ $ This reminds me of the fact that some people always smile indulgently when I mention "the prime $\,-1$", $ $ and continue to use what they presume to be the grown-up name "$\infty$". But consider:

• Every nonzero rational number is uniquely a product of powers of prime numbers $\,p$.

• For distinct odd primes $\left(\frac{p}q\right)$ and $\left(\frac{q}p\right)$ differ just when $\,p \equiv q \equiv -1\pmod 4.$

• There is an invariant called the $p$-signature whose definition involves summing $p$-parts of numbers.

• If there are $p$-adically integral root vectors of norms $\,k\,$ and $\,kp,\,$ then $\,p\,$ is in the spinor kernel.

Each of these statements includes the case $\,p = {-}1,\,$ but none of them is even meaningful when we use the silly name "$\infty$". In the future, I shall smile indulgently back!

From a Sep. 29, 2000 post to usenet newsgroup geometry.research.independent

I call $\,-1$ a prime for a very real reason, not just for fun. Namely, the theory of quadratic forms largely consists of statements - let me call the typical one "Statement($p$)" - that are usually only asserted for the positive primes $\,p = 2,3,5,\ldots$. Now in my experience, in almost every case when Statement($p$) is true for all positive primes, it's automatically both meaningful and true for $\,p = -1\,$ too, and moreover, that additional assertion is just as useful.

Of course, it's also the case that $-1$ does behave differently to the positive primes, but the way this happens is usually that more is true when $\,p = -1,\,$ not less. If less were true, then it would indeed be silly to count $\,-1$ as a prime, but since it's almost always more, it's silly not too.

Let me discuss the simplest such statements in this light.

First, the form of the statement of unique factorization that I prefer is:

The multiplicative group of the non-zero rationals is the direct product of the cyclic subgroups generated by "my" primes, namely $\ {-}1,2,3,5,\,\ldots$.

The additional thing that's true here is that the subgroup generated by $\,-1$ has order $\,2$.

Of course you can still state this with "your" primes, but then have to say "generated by $\,-1$ and the primes", which makes it just a little bit longer.

Second, my preferred statement of quadratic reciprocity is:

If $\,p\,$ and $\,q\,$ are distinct odd primes (in my sense), then $\,\left(\frac{p}q\right)\,$ and $\,\left(\frac{q}p\right)\,$ differ precisely when $\,p\,$ and $\,q\,$ are both congruent to $\,-1$ modulo $4$.

If you want to make the same assertion without counting $\,-1$ as a prime, then you have to add an additional clause:

moreover, $\,\left(\frac{\!-1\,\!}p\right)\,$ is $\ {-}1$ precisely when $\,p\,$ is congruent to $\,-1$ modulo $4$.

Third, perhaps the best of my discoveries in that book is the fact that

for every odd "prime" $\,p\,$ (in my sense), the number

$$ p^a + [A\mid p^a] + p^b + [B\mid p^b] +\, \ldots \pmod 8$$

is an invariant of the quadratic form $\,{\rm diag}(p^a. A,\, p^b.B,\,\ldots )\,$ (with the understanding that $\,p^a\,$ is the $p$-part of $\,p^a.A,\,$ etc.), there being an analogous but slightly different statement for $\,p = 2.$

Once again, if you don't count $\,-1$ as a prime, the statement becomes longer.

From pp. 1-2 of $ $ The genus of a quadratic form: $ $ Our basic problem is to answer the question: when are two quadratic forms equivalent by a rational or integral change of basis? We shall temporarily suppose that all our forms are non-degenerate (i.e., have non-zero determinant), although degenerate forms really give no trouble.

The answer in the rational case is given by the celebrated Hasse-Minkowski theorem, which is usually stated in the form:

Two rational forms are equivalent over the rationals just if they are equivalent over the $\rm\color{#c00}{reals}$, and over the $p$-adic rationals for each positive prime number $p.$

This at first sounds like gobbledigook, because it seems to demand that you must first understand "$p$-adic", which really ain't so. Indeed, Minkowski, who understood the situation very well, did not have the notion of p-adic rational number.

The point really is, that there is a simple invariant for each $p$, and two rational quadratic forms of the same non-zero determinant and dimension are equivalent if and only if they are equivalent over the reals and have the same value of all these invariants.

It is conventional to avoid the exceptional treatment of the reals by regarding them as being the $p$-adic rationals for a rather special "prime number". If one is working over an algebraic extension of the rationals, there may be more than one of these "special primes", which Hasse (who introduced the idea) correctly called "unit primes." Unfortunately, this nomenclature didn't stick, and they are now usually called "infinite primes".

We're going to stay with the rational case, when there is only one such special prime, usually called "infinity." However, we'll go back to Hasse's way of thinking about things, and call it "$\color{#c00}{-1}$" which is really much more natural. The "$\color{#c00}{-1}\!-\!\!$adic" rational or integral numbers will be defined to be the $\color{#c00}{\rm real}$ numbers.

We can now restate the Hasse-Minkowski theorem so as to include $-1.$

Two forms of the same non-zero determinant are equivalent over the rationals just if they are equivalent over the $\,p$-adic rationals for $\,p = \color{#c00}{-1},2,3,5,\ldots$ (in other words, for all the prime numbers, where $\,\color{#c00}{-1}$ is counted as a prime).

This still includes the bogeyman word "$p$-adic," but I said that $p$-adic equivalence was determined by a simple invariant. Traditionally, it's been an invariant taking the two values $1$ and $-1$, and called THE "Hasse-Minkowski invariant." However, the definite article is undeserved, because there is no universal agreement on what "the Hasse-Minkowski invariant" means! There are two systems in use, and topologists further confuse the situation by still using an older version, the "Minkowski unit."

Solution 2:

The usual definition of "prime" is chosen so that get uniqueness of prime factorizations. At least from this perspective, negative numbers aren't prime, because if they were then we would have, for instance, the "prime factorizations" $4=2 \cdot 2$ and $4=(-2) \cdot (-2)$. It is also why $1$ is not prime, because if it were then we would have the "prime factorizations" $4=2 \cdot 2$ and $4=2 \cdot 2 \cdot 1$ and $4=2 \cdot 2 \cdot 1 \cdot 1$ and ...