# Can a complex number be prime?

I've been pondering over this question since a very long time. If a complex number can be prime then which parts of the complex number needs to be prime for the whole complex number to be prime.

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You should look into Gaussian integers mathworld.wolfram.com/GaussianPrime.html – Anurag A Feb 26 at 4:10
Look up Gaussian primes. – David Mar 18 at 5:31
One point I don't see explicitly in the answers is the definition of a prime depends on a specific set of numbers. The set of complex numbers has no primes for the same reason reals don't - that you can divide them up infinitely - a prime must be indivisible. But there are subsets of complex numbers that have primes, e.g. integers obviously & Gaussian integers as mentioned below. There may be other subsets with primes & it may be (beyond my knowledge) that primes in one subset will be distinct from the primes in another. This relates to ring theory & prime ideals as mentioned in the answers. – matty Mar 18 at 6:15
@user109256 I've already asked a very similar, if not identical question here. – samarbarrett Mar 18 at 7:32
Gaussian primes are fun. But also check out the primes in the Eisenstein integers. – PM 2Ring Mar 18 at 10:09

The notion of being "prime" is only meaningful relative to a base ring.

For instance, in the integers $\mathbb{Z}$ the number 5 is prime, whereas in the Gaussian integers $\mathbb{Z}[i]$ we have

$$5 = (2 + i)(2 - i) = 2^2 - i^2 = 4 - (-1) = 5$$

and in the ring $\mathbb{Z}[\sqrt{5}]$ we have

$$5 = (\sqrt{5})^2$$

so over these rings 5 is not a prime number.

The definition of prime you're probably familiar with -- a number is prime if it is divisible only by itself and one -- doesn't even really work over the integers: for instance, 5 is divisible not only by 1 and 5, but also by -1 and -5. So we need to formulate the definition of a prime differently, while still preserving the basic idea, to make sense of it in an arbitary ring.

Notice the following difference between primes and composites: since 5 is prime, if we have two numbers $a$ and $b$ such that $ab$ is a multiple of 5, then obviously one of $a$ or $b$ has to be a multiple of 5 just by unique factorization. On the other hand, if $ab$ is a multiple of 15, it may be the case that neither $a$ nor $b$ is a multiple of 15, because we might instead have $a$ a multiple of 3 but not 5 and $b$ a multiple of 5 and not 3. [*]

This gives us our definition of "prime" for a general ring: a ring element is prime if it is neither zero nor a unit, and moreover has the property that whenever it divides a product it must divide at least one of the factors.

There are a great many rings contained in the complex numbers, and in many of these rings there are non-real complex numbers that are primes in that ring. However, given any number that's prime in a given ring, there's a larger ring in which it's not prime, just as we saw above that 5 is prime over the integers, but not over the Gaussian integers $\mathbb{Z}[i]$ or over $\mathbb{Z}[\sqrt{5}]$.

In particular, since $\mathbb{C}$ is a field, every nonzero element is a unit, so nothing is prime over the complex numbers. (Similarly, nothing is prime over the real numbers, or the rational numbers.) However, I reiterate that many complex numbers are prime over smaller rings: for instance, it turns out that $2 + i$ is prime over $\mathbb{Z}[i]$.

[*] A slightly more straightforward generalization of the definition you're used to would be to look at nonzero, nonunit elements $r$ for which we only have $r = s t$ when either $s$ or $t$ is a unit. This actually gives a weaker notion called "irreducibility." In Unique Factorization Domains the two notions are the same (which explains why they're the same for the integers), but in rings like $\mathbb{Z}[\sqrt{-5}]$ where we do not have unique factorization you can have irreducible elements that are not prime. For instance, 3 is irreducible in $\mathbb{Z}[\sqrt{-5}]$, but we have $$3 \cdot 3 = 9 = (2 + \sqrt{-5})(2 - \sqrt{-5})$$ and 3 is not a divisor of $2 \pm \sqrt{-5}$, so it doesn't satisfy the definition of a prime.

Note: several answers (including this one) have brought up the Gaussian integers specifically. They're indeed an example of a subring of the complex numbers containing non-real complex numbers, but just to be clear they're in no way "the" natural example here -- they're on the same footing as all the others.

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I like your answer, as it covers the fact that the notion can be generalised without going into too many abstract algebra concepts which the OP may not be familiar with, and too many specific equations. It's possible a reader is not familiar with the notation of ℤ[i] and such but I'm sure the meaning comes through anyway. – matty Mar 18 at 8:58
The Gaussian integers is "the" natural example in the sense that it is the smallest ring extending the integers in which -1 has a square root. – kasperd Mar 18 at 10:28
Even as someone with absolutly no background in number theory i could learn a lot from this. Great answer! (+1) – tired Mar 18 at 10:55
@kasperd: but why should -1 have a square root rather than -5? Or why not the smallest ring where 1 has a seventh root other than itself? – Tom Church Mar 18 at 12:18

As with real numbers, you can factor any complex number $z$ as $2\times\dfrac z 2$. For example, $31$ can be factored as $2\times\dfrac{31}2$. However, in defining prime numbers one is not working within the set of all real numbers, but within the set $\{1,2,3,\ldots\}$. (And when Euclid wrote about prime numbers in the third century BC, he did not consider $1$ to be a number.)

Mathematicians sometimes work within the set $\mathbb Z= \{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$, and then $(-3)\times(-2)=6$ is not considered to be a different factorization from $3\times2=6$, because you can change $3$ to $-3$ by multiplying it by a "unit", in this case by $-1$. There are two "units" in $\mathbb Z$, namely $\pm1$, and the thing that qualifies them as "units" is that they are divisors of $1$.

Mathematicians also work with "Gaussian integers", which are complex numbers of the form $a+bi$ where $a,b\in\mathbb Z$, i.e. $a,b$ are ordinary integers. Within the Gaussian integers, numbers that cannot be factored are "Gaussian primes". The number $5$ is not a Gaussian prime because $5=(2+i)(2-i)$. It is also $(1+2i)(1-2i)$, but that is not a different factorization because $1+2i$ can be reached from $2-i$ by multiplying it by a unit, namely $-i$, since $(1+2i)(-i) = 2+i$.

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I've looked up the properties of Gaussian primes here, mathworld.wolfram.com/GaussianPrime.html , but it's not clear to me why numbers that satisfy those properties cannot be factored into new Gaussian integers. Do you know why that is? – samarbarrett Feb 27 at 23:54
@samarbarrett : I wouldn't take the present statements on the Mathworld page to be a definition, although it's phrased as if that's what it was intended to be. A Gaussian prime is simply a Gaussian integer that cannot be factored except by pulling out one the four units, $\pm1,\pm i$ as a factor. ("Units" are divisors of $1$.) That's the definition. $\qquad$ – Michael Hardy Feb 28 at 16:18

The concept of primality is extended to other systems, including the Gaussian integers, as irreducibility. An irreducible element is one which cannot be written as the product of non-unit elements; compare this with the definition of a prime.

This is similar in principle, but can have (initially) unexpected consequences. For example, the decomposition of natural numbers into primes is unique, but the decomposition of some of those same numbers into irreducible complex numbers is not!

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Actually, the integer concept of "prime" splits to two different concepts - one concept being irreducible, and the other, confusingly, called prime. – Thomas Andrews Feb 26 at 5:22
If you're saying factorization in the Gaussian integers is non-unique, neither is factorization in the integers. $10=5\times 2 = -5\times -2$. – Milo Brandt Feb 26 at 21:52
The natural numbers are not the integers. Negative numbers are not primes. – Nij Feb 26 at 23:39
Well, but it's important to highlight the difference between the trivial sort of non-uniqueness that happens in UFDs like $\mathbb{Z}$ or $\mathbb{Z}[i]$ and the nontrivial sort of non-uniqueness that happens in non-UFDs like $\mathbb{Z}[\sqrt{5}i]$. – Daniel McLaury Mar 18 at 8:42

If you are thinking of Gaussian primes, $a+bi$ can be prime with neither $a$ nor $b$ prime. The simplest example is $1+i$. More complicated is $4+15i$. Here I used the fact that if $a^2+b^2$ is an ordinary prime, then $a+bi$ is a Gaussian prime.

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The analogous concept would be to consider the Gaussian integers $g = m + ni : m,n \in \Bbb{Z}$ and say that $g$ is prime if there is no pair of Gaussian integers such that $hk = g; |h| \neq 1; |k|\neq 1$.

So for example, in the field of Gaussian integers, $5 = (2+i)(2-i)$ is not prime.

Two caveats here:

(1) There is no simple way to tell of $g$ is prime, based on whether its real and imaginary parts are prime.

(2) The Gaussian Integers are not a unique factorization domain; that is, a number can have two non-trivially distinct factorizations. The latter property is very likely the place where Fermat's "too big for the margin" proof was flawed, because there is indeed a very clever (maybe remarkable) proof of his last theorem that uses numbers of this form and assumes unique factorization.

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Pretty sure the Gaussian integers are a unique factorization domain. Maybe you're thinking of other field extensions of $\mathbb{Z}$, like $\mathbb{Z}[\sqrt{-5}]$? – mjqxxxx Mar 18 at 5:57
@DanBrumleve: It's called "Lame's proof" (after the mathematician Gabriel Lame) and you find it easily on Google. Some famous mathematician once stated that Fermat probably had something like Lame's proof, but in retrospect this seems unlikely -- it's more likely that he mistakenly assumed a proof based on infinite descent generalized. – Daniel McLaury Mar 18 at 6:23
The mathematician evoked by Daniel is Lamé (with an acute accent on the e), but this is probably a lame comment... – Georges Elencwajg Mar 18 at 9:15
Some remarks on this answer: first of all, $\mathbb Z[i]$ is most definitely a PID (and if it weren't, your definition of prime would be incorrect, you currently have the definition of irreducible element), secondly Fermat definitely did not try to prove FLT using factorization over number fields, as these concepts wouldn't be developed for another century or two, and thirdly one can tell if a Gaussian integer $a + bi$ with $a, b \neq 0$ is prime simply by checking if its norm $a^2 + b^2$ is prime. – Marc Paul Mar 18 at 10:53
$\mathbb Z[i]$ has a division algorithm, so it is certainly a UFD. – Spooky Mar 20 at 6:08

Yes, a complex number can be prime (in the traditional sense of the word). Recall that $\mathbb R \subseteq \mathbb C.$ Therefore, all numbers that you would traditionally think of as being prime are themselves complex (though not non-real). So in this case, we require of $a+bi$ that $a$ be prime (in the traditional sense) and $b=0.$

However, there is the notion of being Gaussian prime. A Gaussian prime is a Gaussian integer (a complex number $a+bi$ such that $a,b\in\mathbb Z$) satisfying one of the following:

• If $a,b\neq 0,$ then $a+bi$ is Gaussian prime iff $a^2+b^2$ is (traditionally) prime;

• If $a=0,$ then $bi$ is Gaussian prime iff $|b|$ is (traditionally) prime and $|b|\equiv 3 \pmod 4;$

• If $b=0,$ then $a$ is Gaussian prime iff $|a|$ is (traditionally) prime and $|a|\equiv 3 \pmod 4.$

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A prime is an integer $p>1$ that cannot be written as $p=ab$ where $a,b>1$ are integers. Clearly, any prime in this sense is also a complex number (since $\mathbb{N}\subset\mathbb{C}$).

More abstractly, an element $p$ of a commutative ring is prime if it is nonzero, noninvertible (i.e. not a unit), and satisfies the condition $$p\mid ab\implies p\mid a\text{ or }p\mid b.$$ You can check that this is consistent with our original definition in $\mathbb{N}$.

David points out an example in $\mathbb{Z}[i]$.

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