fundamental theorem of arithmetic, proof of the

To prove the fundamental theorem of arithmetic, we must show that eachpositive integer has a prime decomposition and that each suchdecomposition is unique up to the order (http://planetmath.org/OrderingRelation) of the factors. Beforeproceeding with the proof, we note that in any integral domain, everyprime is an irreducible element. Furthermore, by Euclid’s lemma, every irreducible element in $\mathbb{Z}$ is prime. As a result, prime elements and irreducible elements are equivalent in $\mathbb{Z}$. We will use this fact to prove the theorem.

Note: all our arithmetic will be carried out in the natural numbers, not the integers.

Existence

Now we prove the existence of a prime decomposition. Since 1 has aprime decomposition and any prime has a prime decomposition, itsuffices to show that any composite number has a prime decomposition.We claim that any composite number is divisible by some prime number.To see this, assume n is a composite positive integer. Thenthere exist positive integers a and b, necessarily strictlysmaller than n, such that n = ab. If a is notprime we can write it as a product of smaller positive integers in thesame way. Continuing this process, we obtain a strictly decreasingsequence of natural numbers. By the well-ordering principle, thissequence must terminate, and by construction, it must terminate in aprime number. Hence n is divisible by a prime number, asdesired.

To complete the proof of existence, we apply the well-orderingprinciple again. If n is composite, then it is divisible by someprime. Moreover, the quotient is strictly smaller than n. Ifthe quotient is not prime then as before we find a prime that dividesit; continuing this process, we produce a strictly decreasing sequenceof natural numbers. By the well-ordering principle, this sequenceterminates in a prime number. The product of the prime numbers wefound in the construction of this sequence is simply n. Thusevery composite number has a prime decomposition.

Uniqueness

Now we show uniqueness up to order of factors. Suppose we are giventwo prime decompositions

$$n=p_1\mathrm{\cdots }{p}_k=q_1\mathrm{\cdots }{q}_{\mathrm{\ell }}$$

of n, where $k\le \mathrm{\ell }$ and the factors in each prime decompositionare arranged in nondecreasing order. Since $p_1$ divides n, itdivides the product $q_1\mathrm{\cdots }{q}_{\mathrm{\ell }}$, so by primality must dividesome $q_i$. Thus there is a natural number b such that $q_i=b\cdot p_1$. Since $q_i$ is an irreducible element and $p_1$ is not aunit, it must be that b is a unit, that is, b = 1. Hence$p_1=q_i\ge q_1$. Similarly, there is some j such that $q_1=p_j\ge p_1$. Since $p_1\ge q_1$ and $q_1\ge p_1$, it follows that$p_1=q_1$. Cancelling these factors yields the simpler equation

$$p_2\mathrm{\cdots }{p}_k=q_2\mathrm{\cdots }{q}_{\mathrm{\ell }}.$$

By repeating the above procedure we can show that $p_i=q_i$ for alli from 0 to k. Cancelling these factors gives the equation

$$1=q_{k+1}\mathrm{\cdots }{q}_{\mathrm{\ell }}.$$

Since a nontrivial product of primes is greater than 1, the right-handside of this equation is the empty product. We conclude that k =l. Hence all prime decompositions of n use the same numberof factors, and the factors which appear are unique up to the order inwhich they appear. This completes the proof of uniqueness and therebythe proof of the theorem.

Remark. Note that Euclid’s Lemma is necessary in order to prove the uniqueness portion of the theorem. Also, an alternative way of proving the existence portion of the theorem is to use induction: if $n$ is composite, then $n=ab$ for some integers (this is true, for if one of $a,b$ is $n$, then the other integer must be $1$). By induction, both $a$ and $b$ can be written as product of primes, which implies that $n$ is a product of primes.