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Let $\Bbb{Z}_p$ be the $p$-adic integers given by formal series $\sum_{i\geq 0} a_i p^i$. I'm having trouble proving that it's an integral domain.

Daniel Donnelly
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    To be an integral domain, you have to prove the axioms for a commutative ring (for multiplication and addition: closure, commutativity, associativity, identity, distributivity, and additive inverses) as well as cancellation for multiplication. What are you having trouble with? – Newb Dec 21 '13 at 20:33
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    @Newb I'm having trouble with proving that there are no zero divisors other than zero - isn't that the definition of an integral domain? I already know that $\Bbb{Z}_p$ is comm ring with $1$. – Daniel Donnelly Dec 21 '13 at 20:34
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    So if there were a zero divisor other than $0$, what would that mean for $p$? – abiessu Dec 21 '13 at 20:37
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    Hint: if $\sum_{i\ge 0}a_ip^i$ is nonzero, there is some minimal $i$ where $a_i\neq 0$. – vadim123 Dec 21 '13 at 20:38

4 Answers4

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The ring of $p$-adic integers $\mathbb{Z}_p$ is the endomorphism ring of $\mathbb{Z}(p^\infty)$ (the Prüfer $p$-group).

Since $\mathbb{Z}(p^\infty)$ is divisible, the image of an endomorphism is divisible, but the only divisible subgroups of $\mathbb{Z}(p^\infty)$ are $\{0\}$ and $\mathbb{Z}(p^\infty)$ itself, because all proper subgroups are finite. Therefore every nonzero endomorphism is surjective. Thus $fg=0$, with $g\ne0$ implies $f=0$.

egreg
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The p-adic integers are a discrete valuation ring, i.e., with an absolute valuation $\nu$. Hence $0=ab$ means $0=\nu(ab)=\nu(a)\nu(b)$ in the real numbers, which have no zero divisors, hence either $\nu(a)=0$ or $\nu(b)=0$, i.e. $a=0$ or $b=0$.

Dietrich Burde
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Although the description of $\mathbb Z_p$ as "formal series" resembling power series in $p$ has some appeal, and was part of Hensel's original motivation, this is not a good viewpoint from which to ascertain many of the properties, as it turns out.

A relatively elementary description is that $\mathbb Z_p$ is the metric completion of $\mathbb Z$ with respect to the $p$-adic metric $|a-b|_p$, where the $p$-adic norm $|a|_p$ is $|p^n\cdot c|=p^{-n}$ for integer $c$ relatively prime to $p$. Yes, the exponent has a sign flip, so that highly-divisible-by-$p$ integers are tiny. (One can also, perhaps better-in-the-long-run, describe $\mathbb Z_p$ as the projective limit of $\mathbb Z/p^n$... which also shows that projective limits occur in real life, not just as artifacts... but usually this is too much "load" at the point $p$-adic things are introduced.)

From this viewpoint, it is much more feasible to prove that there are no zero-divisors... (using continuity).

paul garrett
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  • I know this is the preferable view, but I'm going through Alain. P. Robert's book and it starts out with that def then later gets more mathy / abstract definition. I haven't gotten there yet! – Daniel Donnelly Dec 21 '13 at 20:41
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    Ok, but/and then as a warm-up you might see what it takes to prove that the real numbers, described as decimal expansions, form an integral domain. Seriously! The awkwardness obviously is in talking precisely about the artifact of the "carries"... :) – paul garrett Dec 21 '13 at 20:46
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As $\Bbb Z_p/p^n\Bbb Z_p\cong\Bbb Z/p^n\Bbb Z$, to show $a,b\ne0\Rightarrow ab\ne0$ it suffices to show that $ab\not\equiv0$ mod $p^n$ for a high enough power $p^n$ given $a,b\ne0$ (see vadim's hint in the comments).

anon
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