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On the way to explain a $p$-adic expansion, we consider, when dealing with natural numbers, if we take $p$ to be a fixed prime number, then any positive integer expansion in the form can be written as a base $p$ expansion

$$\sum_{i=0}^n a_i\; p^i$$

where the $a_i$ are integers in $\{0, …, p−1\}$.

Is there a way commonly knwon to relate this expansion to the canonical form of the fundamental theorem of arithmetic?

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In some sense, but it's not very exciting. See . – Qiaochu Yuan Jun 3 '13 at 12:03
@QiaochuYuan thanks. But unfortunately I can not find on that link any concrete relation between the two subjects. May you explain deeper, what you see and I cannot see? – al-Hwarizmi Jun 3 '13 at 12:13
The $p$-adic order of a number $n$ is both the exponent of $p$ in the prime factorization of $n$ and an important property of its $p$-adic expansion. – Qiaochu Yuan Jun 3 '13 at 12:15
@QiaochuYuan thanks. This sends me to valuation theory! Yes good hint. Thanks. – al-Hwarizmi Jun 3 '13 at 12:17
up vote 2 down vote accepted

There is no obvious direct connection between a number's base $p$ expansion and its factorization; after all, if you could go from $n$'s binary representation to its factorization with feasible time and resources, you would have broken a fundamental pillar of modern computer security! However if computation is no concern and we have data associated to all primes $p$, the answer changes.

If $|x|_p=p^{-v_p(x)}$ is the $p$-adic absolute value, we have the product formula

$$\prod_v |x|_v=1$$

for all nonzero $x\in\bf Q$, where $v$ ranges over finite primes and the "infinite prime" (i.e. the usual archimedean, Euclidean absolute value $|\cdot|$). This is roughly equivalent to FTA. The product formula generalizes to arbitrary global fields.

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thanks for the inspiring answer. – al-Hwarizmi Jun 15 '13 at 21:21

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