# Prime factorization of 1

Fundamental Theorem of Arithmetic says every positive number has a unique prime factorisation. Question: If 1 is neither prime nor composite, then how does it fit into this theorem?

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look at the first sentence of en.wikipedia.org/wiki/Fundamental_Theorem_of_Arithmetic – miracle173 Jun 23 '11 at 14:32
@miracle173: That is just one way of stating the theorem. – TonyK Jun 23 '11 at 14:44
From a wider point of view, $1,-1$ are units in $\Bbb Z$. More generally, a ring $A$ is called a UFD or said to be factorial if every nonzero nonunit element admits a unique factorization into primes save order and unit multipliers. Since $1,-1$ are units, we don't care about them. =) – Pedro Tamaroff Apr 19 '14 at 21:24

Let us remember that an empty product is always 1. Hence, 1 has the empty product as its prime factorization. This product is vacuously a unique product of primes.

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Or, as exponent vectors, e.g. $\:2^3\cdot 3^0\cdot 5^1\mapsto\: <3,0,1,0,0,\cdots>,\ \ 1\mapsto\: <0,0,0,\cdots>\:$. – Bill Dubuque Jun 23 '11 at 15:37

You need to change the theorem because anything that works for this contradicts the theorem. Any natural number greater than 1 can be written as a product of prime factors.

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You could say that i is prime because it's only factors are $i$ and $1$. It has no prime factors on the non-complex list of numbers therefore it is prime correct? $i^4=1$ (nothing in the theorem bans complex numbers) so you can say that $i^4$ is it's prime factorization. Being empty could be a solution but i^4 is also a solution meaning there are $2$. But there can't be. Also $i^8$ and $i^12$ work meaning there are infinite. So that means that $i$ is not prime but then what are it's other factors? I think that $1=i^4$ is better than $1= .$You could argue that you cannot count $i^4$ because $i^8$ also works but if $1=$ then $12= *2^2*3$ which contradicts the theorem more than $i^4$. So, what is the prime factorization of $1$?

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Sorry to raise more questions as an answer but there is no other way to make my point. – Locke May 25 '14 at 3:52
First of all, FTArithmetic talks about $\mathbb Z$, so imaginary numbers do not come into the picture. But even if you extended the ring to $\mathbb Z[i]$, the ring of lattice points on the complex plane, which is also a UFD, factors of $i$ don't really count since $i$ is a unit (it divides $1$). This is also why, in the FTArithemetic, factors of $-1$ are not really considered. – Nishant May 25 '14 at 4:03

The OP hasn't misinterpreted the theorem. Every nonzero integer can be written as a product of primes.(GTM84 P.3) Just the exponents are all zeros...

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How fascinating. – The Chaz 2.0 Oct 15 '11 at 2:34
This is rather lousy referencing style. OP will very likely be unaware of what GTM means and thus be unable to track it down. Why not just link to it? For the record: you're referring to Ireland and Rosen, A classical introduction to modern number theory, Springer Graduate Texts in Mathematics, volume 84. Since you have that book in front of you, what does it cost you to give that information? While it costs the reader some minutes to figure out and locate what you're referring to. – t.b. Oct 15 '11 at 4:46
This is a question of style, not content. You don't have to exclude 1, if you interpret an empty set of primes as a factorisation; but you might want to for pedagogical purposes, so that it doesn't distract from the important issues. – TonyK Jun 23 '11 at 14:42