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What is the motivation behind the concept of definition of Gaussian prime numbers?

I am interested about the part of the definition when both real and imaginary coefficients are non-zero. Then a complex number is called a Gaussian prime number iff $a^2+b^2$ is an ordinary prime number. Why?

Bill Dubuque
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Janik
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    It is very characteristic of this site that such a beautiful question, which leads right into one of the most amazing rabbit holes of math, has a couple of close votes. – Mariano Suárez-Álvarez Dec 28 '22 at 07:54
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    Welcome to the site. Please add some more context that motivated your question. Are you currently studying the ring of Gaussian integers? Are you trying to solve a problem? Are you self-studying? This will improve your question and make it more suitable for out site. See How to ask a good question? – Pedro Dec 28 '22 at 12:50
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    To give a helpful answer it is essential to know your knowledge level, e.g. do you have any knowledge of ring theory or elementary number theory? – Bill Dubuque Dec 28 '22 at 19:54
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    The statement regarding Gaussian prime in your post is not a definition but rather a theorem. The concept of prime (as well as irreducible) can be generalised to rings and you need to get familiar with some ring theory to understand and appreciate answers to your question. Looking forward for updates to your post with more context (perhaps including responses to queries posted in comments). – Paramanand Singh Dec 29 '22 at 00:55

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The notions of "integer" and "prime" are simply too good to be restricted to the familiar context of $\mathbb Z$ - the integers with which we are familiar. It turns out that there are natural generalisations of these ideas to wider contexts. What you see here is just the beginning of a whole new world for mathematicians to explore - a world in which we can study, for example, the solutions of polynomial equations, or treat the graphs (solution sets) of polynomial equations as geometric objects.

So if we have a linear polynomial with integer coefficients $p(x)=ax+b: a\neq 0$ then the (rational) root $x=-\frac ba$ is an integer precisely when $a=\pm 1$ (or we might say that $a$ is a unit). If $a=-1$ we can change all the signs in $p(x)$ and get a "monic" polynomial with the same roots (ie leading coefficient is 1). The generalisation which works as we expand to new contexts (by adding the roots of polynomial equations like $p(x)=x^2+1$ to the numbers we already have) is to say that an algebraic integer is the root of a monic polynomial with integer coefficients (allowing algebraic integer coefficients does not affect what is counted as an integer).

Then we say that a (non-unit, non-zero) algebraic integer is prime precisely when $p|ab$ ($p, a, b$ algebraic integers) implies that either $p|a$ or $p|b$ (or both). This is different from being irreducible: an example of how extending the context distinguishes between ideas which seem naturally to belong together. In fact the notion of "prime" can usefully be extended beyond this and following Kummer we find the notion of a prime ideal - an ideal is not a single number, but a set of numbers, and with a prime ideal the whole set behaves collectively like a prime number. The concept of an ideal was developed in the search for a proof for Fermat's Last Theorem, and in particular to deal with some unexpected technicalities, like the failure of unique factorisation when numbers are extended. The Gaussian Integers do still have unique factorisation, but if we were to add $\sqrt {-5}$ to the mix, for example, unique factorisation would fail. Early attempts at FLT could easily miss the fact that unique factorisation was being assumed.


Gaussian Primes help to answer the question "which integers can be expressed as the sum of two integer squares?" and also the number of such representations, where they exist.

Mark Bennet
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    You need to excude units (and possibly $0$) in your definition of a prime. The claim that "the concept of an ideal was developed..." is a bit misleading. The actual history is much richer than that. – Bill Dubuque Dec 28 '22 at 19:45
  • @BillDubuque Thank you - I have changed it - but please leave your comment as it gives a perfect example of how it is so easy to make an unnoticed assumption. – Mark Bennet Dec 28 '22 at 19:48
  • To add a little without editing the post, the additional primes introduced when we add $i$ to our number system, with $i^2=-1$ are of various kinds. We get $\pm ip, p\equiv 3 \bmod 4$ - but $i$ is a unit, so these primes are the familiar ones. When $p\equiv 1 \bmod 4$, we get (precisely) two "Gaussian Primes" which are complex conjugates of each other - we say these primes split. The prime $2$ gives $2=i(1-i)^2$ where $1-i=-i(1+i)$ is prime: the presence of a multiple factor is called "ramification". The three possibilities are general and theoretically significant. – Mark Bennet Dec 28 '22 at 19:58
  • And @BillDubuque it is of course really important to make a distinction between primes and units: or at least to understand units as a different thing - there was a Lehmer who insisted on counting 1 as prime, and that was common once. The generalisations work more smoothly if they are distinguished. – Mark Bennet Dec 28 '22 at 20:05
  • There are some (rare) contexts where it proves convenient to include some units as primes, e.g. here are good arguments (by Hasse & Conway) for considering $-1$ to be prime in certain contexts. – Bill Dubuque Dec 28 '22 at 20:10
  • @BillDubuque You remind me that JH Conway used to regard -1 as a prime eg in his book on quadratic forms. – Mark Bennet Dec 28 '22 at 20:12
  • @BillDubuque Conway was one of my lecturers in Cambridge way back when - he was one of those remarkable expositors who did his best to make apparently complicated things seem obvious. I meant thanks for reminding me, and the linked post is of significant interest. – Mark Bennet Dec 28 '22 at 20:17
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The Gaussian integers are a subset of complex numbers of the form $a+ib$ where $a$ and $b$ are integers. They are also denoted $\mathbb{Z}[i]$.

The Gaussian primes can be thought of as the intersection of two sets: a) the prime elements of $\mathbb{Z}[i]$ and b) the positive integers.

There's a theorem that says that any prime number is a prime element in $\mathbb{Z}[i]$ if and only if it is congruent to 3 mod 4.

For example, $2$ factors as $(1+i)(1-i) = 1 - (-1) = 1+1 = 2$.

Additionally, $5$ factors as $(2+i)(2-i)$.

The result that you're stating about $a^2 + b^2$ is incomplete:

An element of the Gaussian integers is a prime element if and only if its norm (i.e. $a^2 + b^2$ is prime) or it is a prime number of the form $4n + 3$ (up to multiplication by a unit).

For example, the norm of $3$, which is a Gaussian prime, is $9$, which is not prime.

Greg Nisbet
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