# The Fundamental Theorem of Arithmetic and Unique Factorization

We know, from basic arithmetic, that any natural number greater than $1$ is either a prime number or can be factored uniquely into a product of prime numbers. For example, consider the number $150$:

$\displaystyle 150=2\cdot3\cdot5\cdot5$

This is only unique, of course, up to the order in which multiplication is performed (since multiplication in the natural numbers is commutative), and up to a unit, i.e.

$\displaystyle 150=2\cdot3\cdot5\cdot5=1\cdot2\cdot3\cdot5\cdot5=1\cdot1\cdot2\cdot3\cdot5\cdot5=...$

so we just consider all these different factorizations to be the same. This fact is so important that it is referred to as the fundamental theorem of arithmetic.

Now let us consider other kinds of numbers. We start with the integers, which are almost like the natural numbers except that there are negative numbers. The most important thing we should take note of here is that aside from the unit $1$, we also have another unit, namely $-1$.

Let’s consider the number $-28$:

$\displaystyle -28=(-1)\cdot2\cdot2\cdot7$

This is also

$\displaystyle -28=(-1)\cdot2\cdot2\cdot7=(-1)\cdot(-1)\cdot(-1)\cdot2\cdot2\cdot7=...$

Once again, we consider these factorizations to be the same. Aside from these differences involving units, and also the order of multiplication, the factorization is unique.

Now we move on to less familiar numbers. Consider the Gaussian integers, complex numbers whose real and imaginary parts are both integers. In more technical terms, we write

$\displaystyle \mathbb{Z}[i]=\{a+bi|a,b\in\mathbb{Z}\}$.

The units here are, in addition to $1$ and $-1$, $i$ and $-i$.

For Gaussian integers, the number $5$ is not a prime number anymore, but instead factorizes as follows:

$5=(2+i)(2-i)$.

Note that this is also the same as

$5=(1+2i)(1-2i)$.

Since these two factorizations differ only by units:

$(2+i)=i(1-2i)$

$(2-i)=(-i)(1-2i)$

$(2+i)(2-i)=i(1-2i)(-i)(1-2i)$

Meanwhile, the number $3$ is still a prime number.

It is a theorem, which I shall not prove in this post, that the “ordinary” prime numbers that leave a remainder of $3$ when divided by $4$ “remain prime” for Gaussian integers. Those that leave a remainder of  $1$ when divided by  $4$ are no longer prime for Gaussian integers. We say that they “split”. The number 2 is a special case, we say that it “ramifies”; this means it factorizes, again, up to units, as a power:

$2=(-i)(1+i)(1+i)$

But through all this, even though factorization in the Gaussian integers is very different from factorization in the natural numbers or the integers, the factorization in the Gaussian integers remains unique, of course up to the order of the factors and up to units.

We now show an example of factorization no longer being unique. Consider the (complex) numbers

$\displaystyle \mathbb{Z}[\sqrt{-5}]=\{a+b\sqrt{-5}|a,b\in\mathbb{Z}\}$.

We now have, for the number $6$, the following two factorizations which are truly different, not just in order or up to units:

$6=2\cdot3$

$6=(1+\sqrt{-5})(1-\sqrt{-5})$

Unique factorization has failed. And in numbers like these, the failure of unique factorization can be measured by something called the class number. When unique factorization holds, the class number is $1$. The class number is one of the most important quantities in all of algebraic number theory and will be tackled in later posts.

References:

Fundamental Theorem of Arithmetic on Wikipedia

Algebra by Michael Artin