The impact of the FTA is so great, I cannot overstate its significance. This section collates a few examples, but you will see them throughout the text, as well as in the next section, when we connect it back to congruences.
Most importantly, lots of theorems now have reasons, not just proofs. This is an important point about mathematics! You will (re)prove a few things in the Exercises to see this. This ability boils down to the fact that \(\gcd(a,b)=1\) now means that \(a\) and \(b\) do not share any common prime factors. Here is a first example to give the feel.
If \(a\mid c\), \(b\mid c\), and \(\gcd(a,b)=1\), then \(a=\prod p_i\) and \(b=\prod q_j\) but none of the \(p_i\) can be any of the \(q_j\) (or the gcd would include that prime).
Since by the FTA \(c=\prod r_k^{e_k}\) where the \(r_k\) are distinct, the \(p_i\) must be some of the \(r_k\) and the \(q_j\) must be different ones, so that \(\prod p_i q_j\) still divides \(c\).
So if \(a\mid c\), \(b\mid c\), and \(\gcd(a,b)=1\), then \(ab\mid c\), which is part of Proposition 2.4.6.
As another example, the proofs from Section 3.7 become far simpler. We can prove the first here, and save the second for Exercise 6.6.11.
Let's show that \(a^2\mid z^2\) implies \(a\mid z\).
The reader should note that for such proofs, the implicit use of Corollary 6.3.6 is crucial along with the FTA.
Nearly as important is that computing all kinds of things becomes easier. If we let \(a=\prod_{i=1}^n p_i^{e_i}\) and \(b=\prod_{i=1}^n p_i^{f_i}\), where it's possible that \(e_i\) or \(f_i\) is zero at times, then we can often get formulas for various combinations of \(a\) and \(b\).
Given two numbers \(x\leq y\), we let the maximum and minimum be defined by \begin{equation*}\max(x,y)=y\text{ and }\min(x,y)=x\end{equation*} with the clear extension to a min or max of a set consisting of more than two numbers.
Then we have formulas of this kind.
\begin{equation*}ab=\prod_{i=1}^n p_i^{e_i+f_i}\end{equation*}
\begin{equation*}\gcd(a,b)=\prod_{i=1}^n p_i^{\min(e_i,f_i)}\end{equation*}
Assuming \(b \mid a\), \begin{equation*}a/b=\prod_{i=1}^n p_i^{???}\end{equation*} (See Exercise 6.6.7.)
Another use of the FTA is to help us do in a systematic way results that were probably first obtained by extremely ad-hoc methods. As an example, it is likely that you have seen a proof that \(\sqrt{2}\) is irrational, and it probably used mostly the concept of “evenness”. But we can prove that \(\sqrt{m}\notin \mathbb{Q}\) (for \(m\) not a perfect square) in a very similar fashion.
Most deeply, it gives us a canonical way to describe every integer in terms of simpler integers, and gives a measure of simplicity. We'll exploit this some much later, such as in Chapter 24.
Here are some ways to calculate these things in Sage. Simply replace the numbers you are interested in.
Note that the first of these functions gives just a list of the prime divisors, while the second one gives the full prime power factorization.
Finally, let's note that depending on the context, the computational, proof, or other tools won't all be necessary. To demonstrate that, let's introduce some useful additional notation.
For \(p\) prime, we say that \(p^k\parallel n\) precisely when \(p^k\mid n\) but \(p^{k+1}\) does not divide \(n\).
We write \(n!\) for the product of the integers from \(1\) to \(n\), called \(n\) factorial.
As an example, \(5^2\parallel 75\). The prime factorization of a number clearly gives you information about this.
Since \(2^{18} \parallel 20!\) and \(5^4\parallel 20!\), we can conclude that \(20!\) ends with exactly 4 zeros merely from the prime factorization, which we could certainly get without multiplying it out (though in this case Sage does that first). We can check this:
