# Why is two prime?

1. Apr 13, 2014

### pondzo

just a quick question. why is two prime if its has factors, (1+i) and (1-i)?

2. Apr 13, 2014

### mrnike992

The definition of a prime number is (according to Merriam-Webster) "Any positive integer greater than 1 and exactly divisible only by 1 and itself."

Key words being "positive integer," meaning that imaginary numbers are not considered when defining prime numbers.

3. Apr 13, 2014

### pondzo

ahhh i see, why do you think this was the orginal definition? wouldn't it makes sense to include the complex numbers? so gaussian primes are then considered the 'primes'.

4. Apr 13, 2014

### micromass

Staff Emeritus
Very good question. The notion of a prime number is highly dependent of its underlying ring.

So if we talk about primes in $\mathbb{Z}$, then we only allow factors which are in $\mathbb{Z}$.
However, the Gaussian integers $\mathbb{Z}$ also forms a very nice ring. In that ring, if we talk about a prime in $\mathbb{Z}$, then we only allow factors in $\mathbb{Z}$. In particular, the factors $1+i$ and $1-i$ are allowed, thus $2$ is not a prime in $\mathbb{Z}$.

So if we talk about prime numbers, the underlying ring should always be given. If it is not given, then it is safe to assume that it is $\mathbb{Z}$.

The reason that we usually work in $\mathbb{Z}$ and not in $\mathbb{Z}$ is because imaginary numbers were not accepted for a very long time. In particular, when the notion of prime number was invented by the Greeks, nobody even had the slightest idea that such things as imaginary numbers existed. So the standard has become to work in $\mathbb{Z}$ since other rings were inconceivable for a very long time.

We can ask ourselves which numbers of $\mathbb{Z}$ are actually primes in $\mathbb{Z}$. These numbers are exactly the primes of the form $4n+3$ or $-(4n+3)$ where $n$ is a positive integer. The positive primes of these form are exactly the primes which can not be expressed as the sum of two squares, which is not a coincidence.

5. Apr 13, 2014

### mrnike992

I'd also like to point out that if we include prime numbers, then we could (correct me if I'm mistaken) also include $(1 + i^5)(1- i^5)$ and so on, leaving us with an infinite number of factors, which is impractical, so we eliminate the obvious and infinite possibilities, to leave us with only natural numbers.

6. Apr 13, 2014

### pondzo

Then shouldn't the 'ultimate goal' be to find a formula that produces all primes in Z(i) and not Z? and furthermore, would a search for such a formula for all primes in Z be futile, as the sequence of primes 2,3,5,7,11,13... should be the sequence Z$\cap$Z(i), namely 3, 7, 11, 19... (but then i guess this gets back to what you said about whether you are talking of primes in Z or in Z(i) or which ever ring).

7. Apr 13, 2014

### micromass

Staff Emeritus
No. A version of the fundamental theorem of arithmetic does hold for the Guassian integers. In particular, we define a unit in $\mathbb{Z}$ to be an invertible element. Thus one such that $u^{-1}$ is also a Gaussian integer. Clearly, the units are $1,~-1,~i,~-i$.
Then we can define a Gaussian prime $p$ as a nonzero element of $\mathbb{Z}$ such that if we can write $p = ab$ for $a,b\in \mathbb{Z}$, then either $a$ or $b$ is a unit.
The fundamental theorem then states that for any $x\in \mathbb{Z}$ there exist a finite number of Gaussian primes $p_1,~....,p_n$ such that $x=p_1\cdot...\cdot p_n$. Furthermore, if $x=q_1\cdot.... q_m$ is another such decomposition, then $n=m$ and it is possible to reorder the sequence of the $q_i$ such that $p_i = u_i q_i$ for some unit $u_i$.

This theorem can be generalized even to further rings. If a ring additionally satisfies $a\neq 0$ and $b\neq 0$ implies $ab\neq 0$, then such a ring is called a Unique Factorization Domain or a UFD: http://en.wikipedia.org/wiki/Unique_Factorization_Domain
Thus both $\mathbb{Z}$ and $\mathbb{Z}$ are UFD's.

Finally, I would like to say that my definition of a prime as $p=ab$ implies $a$ or $b$ is a unit, is actually called an irreducible element. The actual definiton of a prime is the following: $p$ is a prime if for any $a$ and $b$ such that $p$ divides $ab$ (thus there exists some $c$ such that $pc=ab$) we either got that $p$ divides $a$ or $p$ divides $b$ (thus there exists some $d$ such that $dp = a$ or $dp = b$. Euclid has proven in his Elements that these two definitions are equivalent in $\mathbb{Z}$. Furthermore, we can actually prove the two definitions are equivalent in each UFD. In an arbitrary integral domain, we always have that each prime is irreducibe. The converse is false, for example, the ring $\mathbb{Z}[5i]$.

Both rings $\mathbb{Z}$ and $\mathbb{Z}$ are interesting to study. There are many interesting results which hold for primes in $\mathbb{Z}$ which do not hold for the Gaussian primes. I don't think we should argue about whether to consider primes in $\mathbb{Z}$ or $\mathbb{Z}$. Both situations should be studied. In the same way, we shouldn't be arguing on whether to do analysis on $\mathbb{R}$ or on $\mathbb{C}$ as both are interesting and both yield important results.

8. Apr 13, 2014

### 1MileCrash

One ring is not more important or true than the other.

9. Apr 13, 2014

### SteamKing

Staff Emeritus
Which came first, the chicken or the egg?

People have played around with prime numbers since at least the Greeks, if not earlier. These people thought that fractions were advanced math and were struggling to grasp irrational numbers. There were no negative numbers and the idea of zero was still in the future. Complex numbers were not developed until many centuries later.

As was stated earlier, some things were developed solely around positive integers.