Prime Numbers

We call positive whole numbers the natural numbers and denote them by $\mathbf{N}$. Thus

$$\mathbf{N}= \{1,2,3,4,\ldots\}.$$

We call all the whole numbers, both positive and negative, the integers, and write

$$\mathbf{Z}= \{\ldots, -2, -1, 0, 1,2,\ldots\}.$$

They are denoted by $\mathbf{Z}$ because the German word for the integers is ``Zahlen'' (and $19$th century German number theorists rocked).

Definition 1.1   If $a, b\in \mathbf{Z}$ then ``$a$ divides $b$'' if $ac=b$ for some $c\in\mathbf{Z}$.

To save time, we write

$$a \mid b.$$

For example, $2 \mid 6$ and $389\mid 97734562907$. Also, everything divides 0.

Definition 1.2   A natural number $p>1$ is a prime if $1$ and $p$ are the only divisors of $p$ in $\mathbf{N}$. I.e., if $a\mid p$ implies $a=1$ or $a=p$.

Primes:

$$2,3,5,7,11,\ldots,389,\ldots,2003,\ldots$$

Composites:

$$1,4,6,8,9,10,12,\ldots,666=2\cdot 3^2\cdot 37, \ldots, 2001=3\cdot 23\cdot 29,\ldots$$

Primes are ``primal''--every natural number is built out of prime numbers.

Theorem 1.3 (The Fundamental Theorem of Arithmetic)   Every positive integer can be written as a product of primes, and this expression is unique (up to order).

Warning: This theorem is harder to prove than I first thought it should be. Why?

First, we are lucky that there are any primes at all: if the natural numbers are replaced by the positive rational numbers then there are no primes; e.g., $2 = \frac{1}{2}\cdot 4$, so $\frac{1}{2}\mid 2$.

Second, we are fortunate to have unique factorization in  $\mathbf{Z}$. In other ``rings'', such as $\mathbf{Z}[\sqrt{-5}] = \{a + b\sqrt{-5} : a, b\in\mathbf{Z}\}$, unique factorization can fail. In $\mathbf{Z}[\sqrt{-5}]$, the number $6$ factors in two different ways:

$$2\cdot 3 = 6 = (1+\sqrt{-5})\cdot (1-\sqrt{-5}).$$

If you are worried about whether or not $2$ and $3$ are ``prime'', read this: If $2 = (a+b\sqrt{-5})\cdot (c+d\sqrt{-5})$ with neither factor equal to $\pm 1$, then taking norms implies that

$$4 = (a^2 + 5b^2)\cdot (c^2 + 5d^2),$$

with neither factor $1$. Theorem 1.3 implies that $2=a^2 + 5b^2$, which is impossible. Thus $2$ is ``prime'' in the (nonstandard!) sense that it has no divisors besides $\pm 1$ and $\pm 2$. A similar argument shows that $3$ has no divisors besides $\pm 1$ and $\pm 3$. On the other hand, as you will learn later, $2$ should not be considered prime, because the ideal generated by $2$ in $\mathbf{Z}[\sqrt{-5}]$ is not prime. We have $(1+\sqrt{-5})\cdot (1-\sqrt{-5}) = 6\in (2)$, but neither $1+\sqrt{-5}$ nor $1-\sqrt{-5}$ is in $(2)$. We also note that $(1+\sqrt{-5})$ does not factor. If $(1+\sqrt{-5}) = (a+b\sqrt{-5})\cdot (c+d\sqrt{-5})$, then, upon taking norms,

$$2\cdot 3 = (a^2 + 5b^2)\cdot (c^2 + 5d^2),$$

which is impossible.