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I came across the following problems about p-adic norms:

Problem. Show that $$\prod_{p} |x|_p = \frac{1}{|x|}$$ where the product is taken over all primes $p = 2,3,5, \dots$ and $x \in \mathbb{Q}$.

We have the following: $|x|_2 = 2^{-\max \{r: 2^{r}|x \}}$, $|x|_3 = 3^{-\max \{r: 3^{r}|x \}}, \ \dots$ so that $$\prod_{p} |x|_p = \frac{1}{2^{\max \{r: 2^{r}|x \}}} \cdot \frac{1}{3^{\max \{r: 3^{r}|x \}}} \cdots$$

Then it seems that by the Fundamental Theorem of Arithmetic the result follows. Is this the right idea?

Problem. If $x \in \mathbb{Q}$ and $|x|_p \leq 1$ for every prime $p$, show that $x \in \mathbb{Z}$.

We know that $$\prod_{p} |x|_p = \frac{1}{|x|} \leq 1$$

Then suppose for contradiction that $x \notin \mathbb{Z}$? Or maybe the product is a null sequence?

Added. Suppose $x \notin \mathbb{Z}, \ x \in \mathbb{Q}$. Then $x = \frac{r}{s}$ where at least one prime $p$ divides $s$. Then $\text{ord}_{p} x = \text{ord}_{p} r- \text{ord}_{p} s$. So $$|x|_{p} = p^{-\text{ord} _{p} x} = p^{ \text{ord}_{p} s- \text{ord}_{p} r}$$

$$= \frac{p^{\text{ord}_{p} s}}{p^{\text{ord}_{p} r}} > 1$$

which is a contradiction?

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$@$damien: I just noticed you edited the question. Yes, I think you have it now, although your solution could be worded better: *why* is the final quantity bigger than $1$? (I think you know, but you haven't explicitly said.) – Pete L. Clark Jul 11 '11 at 2:14
@Pete L. Clark: Because $\text{ord}_{p} s > \text{ord}_p r$. – Damien Jul 11 '11 at 2:15
$@$damien: right, but that doesn't follow from "where at least one prime $p$ divides $s$" unless we assume that $r$ and $s$ are relatively prime. It's just a matter of spelling things out... – Pete L. Clark Jul 11 '11 at 2:45
up vote 5 down vote accepted

Your solution to the First Problem is correct (you could perhaps be more explicit by writing out the prime-power factorization of $x$ and observing that it is the reciprocal of the expression you have).

I don't think that the product formula (that is what the first problem is called) is so helpful for the Second Problem: just because a rational number has absolute value at least one doesn't mean it's an integer!

Hint for this part: if a rational number is not an integer, then when put in lowest terms its denominator is divisible by at least one prime $p$. Now re-express this using the $p$-adic norm. (Another way to say this is that you are given infinitely many inequalities: $|x|_p \leq 1$ for all primes $p$. Thus you have infinitely many pieces of information about $x$. By multiplying all these inequalities together, you see from the First Problem that this amounts to one piece of information about $x$: $|x| \geq 1$. But that's not enough for what you want, so evidently we lost a lot of information by multiplying together all our inequalities, which is of course what happens in general when you add/multiply inequalities: you lose some information.)

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Suppose $x \notin \mathbb{Z}, \ x \in \mathbb{Q}$. Then $x = \frac{r}{s}$ where at least one prime $p$ divides $s$. Then $\text{ord}_{p} x = \text{ord}_{p} r- \text{ord}_{p} s$. – Damien Jul 11 '11 at 1:39
See my added section in the question. – Damien Jul 11 '11 at 2:13

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