Let the ring of the integer $p$-adic numbers $\mathbb{Z}_p$.

Could you explain me the following sentences?

  • It is a principal ideal domain. $$$$
  • The function $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ is an embedding. (So, $\mathbb{Z}$ is considered $\subseteq \mathbb{Z}_p$) $$$$
  • The units of the ring $\mathbb{Z}_p$:

$$\mathbb{Z}^*=\mathbb{Z} \setminus p \mathbb{Z}$$

so the units are

$$= \{ \sum_{n=0}^{\infty} a_n p^n | a_0 \neq 0\}$$

  • Each element $x$ of $\mathbb{Z}_p \setminus \{ 0 \}$ has a unique expression of the form $x=p^m u | m \in \mathbb{N}_0$

  • $\mathbb{Z}_p$ has exactly these ideals:

$$0, p^n \mathbb{Z}_p (n \in \mathbb{N}_0)$$

Furthermore, $\cap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_0=\{0\}$ and $\frac{\mathbb{Z}_p}{p^n \mathbb{Z}_p} \cong \frac{\mathbb{Z}}{p^n \mathbb{Z}}$

Last but not least, the unique maximal ideal of $\mathbb{Z}_p$ is $p \mathbb{Z}_p$.

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    $\begingroup$ What do you mean by "explain"? Define the terms? Provide intuition? Give proofs? What do you know already? In particular, what definition of $\mathbb{Z}_p$ are you working with? $\endgroup$ – Slade Oct 13 '14 at 17:16
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    $\begingroup$ Provide intuition or give proof, both of them would be fine! $$$$ $$\mathbb{Z}_p=\{ (\overline{x_n})_{n \in \mathbb{N}_0} \in \Pi_{n=0}^{\infty} \frac{\mathbb{Z}}{p^{n+1}\mathbb{Z}} | x_{n+1} \equiv x_n \pmod {p^{n+1}}$$ $\endgroup$ – evinda Oct 13 '14 at 17:50
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    $\begingroup$ This is a tall order. I would give the general advice that, if you're looking for such a deep and involved explanation of a subject, that you should spend a lot more time writing your question. In particular, you should really try to explain what the subject looks like from your point of view, because this really helps posters know where to start from, and what to focus on—but it also humanizes the experience, which makes people more likely to want to answer your question. $\endgroup$ – Slade Oct 13 '14 at 19:04
  • $\begingroup$ All that said, I find this subject really interesting, and I'm pretending to work on my thesis (which is on a related topic), so I'll put something together anyway. But next time, you'd better put some heart into it! $\endgroup$ – Slade Oct 13 '14 at 19:05
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    $\begingroup$ In general, one should ask questions only after they have attempted them. It appears as if no attempt is made here. You can't figure out where your stumbling points are if you never allow yourself to stumble in the first place. $\endgroup$ – RghtHndSd Oct 13 '14 at 20:15

Let's look at a different ring first: $R=k[[X]]$, the ring of formal power series over a field. We can also think of $R$ as the ring of sequences $f_0, f_1, f_2,\ldots$, where each $f_i$ is a polynomial of degree $\leq i$, and $f_i = f_{i+1} \pmod {X^{i+1}}$ for each $i$.

In other words, we construct a power series by choosing polynomials of higher and higher degree, and doing so in a consistent way. Yet another way to write this is $\displaystyle R = \varprojlim_{i\geq 0} k[X]/(X^i)$

(This last way emphasizes that $k[[X]]$ is the completion of $k[X]$ with respect to the $(X)$-adic topology, if you are interested in the general form of this construction. Completions are very important throughout algebra!)

Easy fact: $(X^i)$ is an ideal of $R$, and $\bigcap_{i\geq 0} (X^i) = (0)$. Intuitively, this tells us that the sequence $(X),(X^2),(X^3),\ldots$ "approaches zero", that is, we can evaluate how "close" two power series $f$ and $g$ are by seeing how many of their leading terms are the same. If $f-g \in (X^6)$, that's "closer" than $f-g\in (X^5)$. If $f-g\in (X^i)$ for all $i$, then $f=g$.

Another easy fact: Every polynomial is also a power series! In other words, there is a natural injective homomorphism $k[X] \to k[[X]]$. It is not hard to check that this map gives an isomorphism $k[X]/(X^i) \cong k[[X]]/(X^i)$ for every $i\geq 0$. (Intuitively: a truncated power series is no different from a truncated polynomial.)

Harder fact: Any power series with nonzero constant term is invertible. To see this, suppose that $f=a+gX$, with $a\neq 0$. By scaling, we may assume that $a=1$, so $f=1+gX$.

We have $(1+gX)(1-gX) = 1-g^2 X^2$, which is "close" to $1$. And $(1+gX)(1-gX+g^2X^2) = 1-g^3 X^3$, which is even closer to $1$. And just to make the point, $(1+gX)(1-gX+g^2X^2 - g^3 X^3) = 1-g^4 X^4$, closer still.

So we define $h$ to be the power series $1-gX+g^2X^2-g^3X^3 + \cdots$. Take a minute to verify that we are allowed to do this. The key is that, for any $i$, we have an explicit formula for $h \pmod{X^i}$, and these formulas are consistent with each other.

Then $fh = 1-g^iX^i = 1 \pmod{X^i}$. Because this holds for all $i$, $fh=1$. So $f$ is invertible.

By the way, this implies that any nonzero power series can be written uniquely in the form $uX^n$, with $u\in R^\times$. (Proof: Divide by the leading term!) This observation will be important later.

Intuition: The above has an interpretation in the domain of complex analysis: If $f$ is an analytic function on an open set containing $0\in\mathbb{C}$, then either $f\equiv 0$, or $f$ vanishes at $0$ with order $n\geq 0$, and $f/X^n$ is invertible (i.e. nonvanishing) in a neighborhood of $0$.

Here is a very useful general fact about rings: if $P\subsetneq R$ is a proper ideal, and every element outside of $P$ is invertible, then $P$ is the unique maximal ideal of $R$. In fact, the statement $R\setminus P = R^\times$ is equivalent to the statement that $P$ is the unique maximal ideal.

Why is this? Well, if $I\subsetneq R$ is any proper ideal, then $I\cap R^\times = \emptyset$. This implies that $R\setminus P = R^\times \subset R\setminus I$. In other words $I\subset P$. So every ideal of $R$, besides $R$ itself, is contined in $P$!

Applied to the setting of power series, we see that $f\notin (X) \implies f\in R^\times$. Therefore, $(X)$ is the unique maximal ideal of $R$.

Next, let's determine all the ideals of $R$. Specifically, let's show that the only nonzero proper ideals are the ones we already know about: $(X), (X^2), \cdots$.

Let $I$ be a nonzero proper ideal, so that $I\subset (X)$. If $I\subset (X^i)$ for all $i$, then $I=(0)$, so choose the smallest $n$ such that $I\subset (X^n)$, but $I\not\subset (X^{n+1})$.

Pick some $f\in I$ with $f\notin (X^{n+1})$. We have $f\in (X^n)$ however, so we can write $f=g X^n$, where $g\notin (X)$. But then $g$ is invertible, so $fg^{-1} = X^n \in I$. This implies that $(X^n)\subset I$. By assumption, $I\subset (X^n)$, so $I=(X^n)$.

Note that we can now conclude very easily that $R$ is a principal ideal domain, because we know all the ideals, and they're all principal!

Going down your list: We have just established, for $k[[X]]$, every property that you were interested in for $\mathbb{Z}_p$: We characterized the units, we established that every ideal is a power of the maximal ideal, we described factorization, we pointed out that it's a PID, and we mentioned the inclusion $k[X] \to k[[X]]$ and its nice properties.

Every single thing about the above explanation still holds if we replace $k[X]$ by $\mathbb{Z}$, $(X)$ by $(p)$, and $k[[X]]$ by $\mathbb{Z}_p$. All the proofs still work, though they may be a little less intuitive in some places.

In fact, the integers are "almost" a polynomial ring $\mathbb{Z}/p [p]$, in the sense that every nonnegative integer has a representation $k = a_0 + a_1 p + \cdots +a_n p^n$, and this representation is unique if we add the restriction $a_i \in \{0,1,\cdots , p-1\}$.

In other words, the point is to think of the base-$p$ representation as something like a polynomial in $p$. If we pursue this metaphor, $\mathbb{Z}_p$ is something like a power series in $p$—specifically, it's just the collection of base-$p$ representations, but where we allow infinitely many digits.

The only thing lacking from the previous section is: What happens to negative integers? While nonnegative integers look quite ordinary in $\mathbb{Z}_p$, we should be careful to see what happens to the negative ones.

$-1$ is, of course, not a positive integer. But watch this: $-1 \equiv p-1 \pmod{p}$, $-1 \equiv (p-1)(1+p)\pmod{p^2}$, $-1 \equiv (p-1)(1+p+p^2)\pmod{p^2}$, et cetera.

So, $-1$ has a base $p$ representation (albeit one with infinitely many terms) given by $-1 = (p-1) + (p-1)p + (p-1)p^2 + (p-1)p^3 + \cdots$. While your teachers would normally yell at you for summing a divergent series like this, it is a perfectly valid statement in $\mathbb{Z}_p$, and it completes our picture of the embedding $\mathbb{Z}\to\mathbb{Z}_p$.

This may be the least part of the explanation, but in principle it is exactly the same as the identity $0.99999\cdots = 1$, which we can verify by calculating $1-0.99999\cdots$. Similarly, we can verify that $-1 = (p-1) + (p-1)p + (p-1)p^2 + (p-1)p^3 + \cdots$ by calculating $1+ (p-1) + (p-1)p + (p-1)p^2 + (p-1)p^3 + \cdots = p + (p-1)p + (p-1)p^2 + (p-1)p^3 + \cdots = p^2 + (p-1)p^2 + (p-1)p^3 = p^3 + (p-1)p^3 + \cdots = \cdots$, which visibly lies in $\bigcap_{i\geq 0} (p^i) = (0)$!

By the way, $\mathbb{Z}_p$ is often described as the metric completion of $\mathbb{Z}$ with respect to the $p$-adic norm $\| a\| = p^{-\nu_p (a)}$ (e.g. $\| p\| = 1/p$, $\|k\| = 1$ when $(k,p)=1$).

If you squint closely, these two definitions are the same: we are saying that two integers $a$ and $b$ are "close" when $a-b$ is divisible by lots of powers of $p$, which is somehow the same as the property $\bigcap_{i\geq 0} (p^i) = (0)$.


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