p-adic numbers vs real numbers

Could anyone give a concrete example of a p-adic number that is not a "real number"? that is, do we create "new numbers" (non real numbers) by completing Q with a non Archimedean norm? If so, what are they?

Solution 1:

Could anyone give a concrete example of a p-adic number that is not a "real number"?

1. If one is very very formal, one could say that no $p$-adic number is a real number. The sets $\mathbb Q_p$ and $\mathbb R$ can be thought of as disjoint.

However, we usually don't want to be that formal. Every field of characteristic $0$ contains a unique copy of $\mathbb Q$, the rational numbers. Hence we can, and usually do, say that both $\mathbb Q_p$ and $\mathbb R$ "contain" $\mathbb Q$. What that means in practice is that if you have any rational number, i.e. a fraction $m/n$ with $m,n \in \mathbb Z, n \neq 0$, you should be easily able to express it as a real number as well as as a $p$-adic number. For example $3/4$ "is" the real number $0.75$, and it also "is" the $5$-adic number $...111112.$ Likewise, $1/7$ "is" the real number $0.1428571428571428571....$ and it also "is" the $7$-adic number $...00.1$ (here, I expand a $p$-adic number in base $p$ to the left, and a real number in base $10$ (we're all so used to the decimal system) to the right). Note that for real numbers one could choose any other base and still the rational numbers would be the ones whose expansions are eventually periodic. There is a crucial fact here that any given rational number is uniquely determined as element of $\mathbb R$ as well as as element of $\mathbb Q_p$, i.e. e.g. $0.75$ "is" the rational number $3/4$ in the sense that no other real number solves $4x-3=0$.

2. That being said, usually questions like this get answered with examples like "$\sqrt{1-p^3} \in \mathbb Q_p$, but $\notin \mathbb R$" and "$\sqrt{p} \in \mathbb R$, but $\notin \mathbb Q_p$". That's fair and will make a beginner careful (and hopefully eager to learn how to come up with examples like these), but I think it is a bit informal, and the imprecision can cause serious trouble later. I will explain this in point 3. For now, I claim these examples are much better phrased as follows: "The equation $x^2-1+p^3=0$ has a solution in $\mathbb Q_p$, but none in $\mathbb R$; hence there is no field embedding $\mathbb Q_p \hookrightarrow \mathbb R$. And the equation $x^2-p=0$ has a solution in $\mathbb R$, but none in $\mathbb Q_p$; hence there is no field embedding $\mathbb R \hookrightarrow \mathbb Q_p$. (Note that for all this, we tacitly presuppose that $\mathbb Q$ is contained in both the $p$-adics and the reals as per 1.)

3. Now I will expand Mark Bennet's comment. I find this absolutely crucial because I remember I needed some time to get this right, and I have seen many students who don't get it. (Also, compare https://math.stackexchange.com/a/3260194/96384) Every beginner will want to say:

"But I learned that, for example, $\sqrt 2 \in \mathbb Q_7$. And everyone knows since high school that $\sqrt 2 \in \mathbb R$. So is not $\sqrt 2 \in \mathbb Q_7 \cap \mathbb R$ ?"

No, it is not. (Or: Only in a sense that is subtly but distinctly different from how a beginner would understand this. And the trouble with the usual phrasing of the examples in 2. is that it hides exactly this subtlety.) In a nutshell, the issue is that the smybol $\sqrt 2$ means one thing in the statement "$\sqrt 2 \in \mathbb R$", but another thing in the statement "$\sqrt 2 \in \mathbb Q_7$".

There are two elements in $\mathbb R$ whose square is $2$. As real numbers, they can be expressed to arbitrary precision, and in decimal expansion they start $x_1=1.41423562...$ and $x_2=-1.41423562...$ (without ever getting periodic), respectively.

There are two elements in $\mathbb Q_7$ whose square is $2$. As $7$-adic numbers, they can be expressed to arbitrary precision, and their $7$-adic expansions start $y_1=...16213.$ and $y_2=...50454.$ (without ever getting periodic), respectively.

We do have $x_1=-x_2$ and $y_1=-y_2$. (This is a purely algebraic fact, as the negative of a square root is also a square root. For example, there are also two elements in $\mathbb F_{17}$ which are square roots of $\bar 2$, namely, $z_1=\bar 6$ and $z_2=\bar{11}$, and they are also each others negatives, $z_1 = -z_2$.)

We can look at the subfields $K:= \mathbb Q(x_1) = \mathbb Q(x_2) \subset \mathbb R$ as well as $E:= \mathbb Q(y_1) = \mathbb Q(y_2) \subset \mathbb Q_7$.

We can also define purely algebraically, without knowing anything about real or $p$-adic numbers, the field $F:= \mathbb Q[X]/(X^2-2)$.

We can then see that the fields $K, E, F$ are all isomorphic. We desperately want to just call all of them $\mathbb Q(\sqrt 2)$. Well, as long as we promise to only do algebra, of course we can do that.

(By "doing only algebra", I mean things like $(1+\sqrt2)^2 = 3+2\sqrt2$, or $(1+\sqrt2)^{-1} = \sqrt2-1$, which are true just by the property that the symbol "$\sqrt2$" signifies a number whose square is $2$. The symbol can stand for each of $x_1,x_2,y_1,y_2$. (This should look somewhat surprising when a beginner thinks about it. Yes, if $\sqrt2$ would mean $\color{red}{-}1.4142...$, these two equations, and many others, still hold.)

But as soon as we want to invoke anything else (like an "order" in the field, talk about somethig being "positive" vs. "negative", say some numbers are big or small, close to each other, ...; i.e. metric or topologies), we should absolutely not do that.

And even if we do only do algebra, we should notice that: Because although the fields $K, E, F$ are all isomorphic, these isomorphisms are not quite canonic. Rather, for each pair of them there are two isomorphisms.

In particular there are two isomorphisms $K \simeq E$. One of them sends $x_1$ to $y_1$, and the other sends $x_1$ to $y_2$. Now we are used to giving the real number starting $1.4142...$, the one I called $x_1$ above, the name and notation "$\sqrt 2$". As mathematicians we have to unlearn this: There is a purely conventional choice here, namely, we just picked the positive (and this notion is already beyond algebra, a property of $\mathbb R$) solution of $x^2-2=0$ as "the" square root. As long as we only do algebra, this is arbitrary, and we could have chosen the other one as well. Everything which distinguishes $x_1$ from $x_2$ is a property of them as real numbers, not just algebra. And everything which distinguishes $y_1$ from $y_2$ is a property of them as $7$-adic numbers, not just algebra.

$x_1$ is positive, $x_2$ is negative. But neither $y_1$ nor $y_2$ are positive or negative. Which one should we match to $x_1$?

$x_1$ is closer to $1$ than $x_2$ is (as real numbers). The powers of $x_1-1$ get smaller and smaller, the powers of $x_2-1$ do not (in $\mathbb R$).

On the other hand, $y_1$ and $y_2$ as $7$-adic numbers both have the same distance from $1$. The powers of $y_1-1$ and $y_2-1$ get neither smaller nor bigger, but remain of the same size. However, $y_1$ is close to $3$, and $y_2$ is close to $4$. So which one of $y_1$ and $y_2$ do you want to call $\sqrt 2$ and which one $-\sqrt 2$, the one that is close to $3$ or the one that is close to $4$?

Or: The polynomial $p(X)= X^2 - 8X + 14$ has a pair of solutions in each of the fields $K, E, F$. We could just call each of those pairs $4 \pm \sqrt 2$. Then we would only do algebra and be fine (and keep the ambiguity $\pm$). But if somebody tried to nail us down, well which of the solutions is ... bigger (further away from $0$ in absolute value)? Hm. If $\pm \sqrt 2$ mean $x_1, x_2$, then the solution $4+x_1$ is bigger than $4+x_2= 4-x_1$ as real numbers (has bigger real absolute value).

But if $\pm \sqrt 2$ mean $y_1, y_2$, then the solution $4+y_2$ is bigger than $4+y_1= 4-y_2$ (has bigger $7$-adic absolute value).

So you see there is a lot of ambiguity and traps here as soon as we leave the field $\mathbb Q$. And here I only wanted to add a square root, which have the nice extra feature that they always come in pairs and we are tempted to just forget about the $\pm$ ambiguity! Now what will happen if we want to identify solutions of cubic polynomials in $\mathbb R$ with ones in $\mathbb Q_p$? For example, the equation $X^3-X+\frac{7}{20}=0$ has three different solutions $X_1,X_2,X_3$ in $\mathbb R$ (graph them with calculus: one of them is "close to $-1$, the other two are "to the left and right of $1/2$"). It also has three different solutions in $\mathbb Q_7$: One of them is close to $1$, one is close to $0$, and one is close to $-1$. If you want to identify them (and that is what you want to do say that $X_1, X_2, X_3$ are "also" $7$-adic numbers): Which one is which? And why?

And now imagine higher degree polynomials ... and maybe several at once, or various layers of field extensions ...

Upshot: If one wants to (and for purely algebraic reasons one might want to) identify certain subfields of $\mathbb R$ and of $\mathbb Q_p$ (or, for that matter, of some $\mathbb Q_p$ and a different $\mathbb Q_\ell$), one can do that. But one has to be very careful then to not invoke properties of them which belong to one of them and not to the other.

(Added:) Also, e.g. here is a recent question where the underlying problem is the false belief that a symbol string like "$\,\mathbb Q(\sqrt[3]{2})$" would well-define a subfield of $\mathbb Q_p$, or, for that matter, e.g. "$\,\mathbb Q_5(\sqrt[3]{2})$" would well-define a field extension of $\mathbb Q_5$. These root symbols and the way we are taught to just accept and use them in high school, as if they are well-defined in themselves, will forever be obstacles on beginners' paths to algebra and number theory.

4. If one believes in the axiom of choice, then there are embeddings $\mathbb Q_p \hookrightarrow \mathbb C$ (without axiom of choice, there are none; with choice, there are many, many more than anyone can imagine, and none of them is explicit). See e.g. Is there an explicit embedding from the various fields of p-adic numbers $\mathbb{Q}_p$ into $\mathbb{C}$?.

I want to say: Unfortunately. Because this might re-enforce the false belief that I just tried to argue against in number 3.

For example, in your response to Lubin's comment about $i:= \sqrt{-1} \in \mathbb Q_5$, you say something to the effect of "oh but $i \in \mathbb C$, I know that". Well. As long as "$i$" is just a placeholder for any number whose square is $-1$, yes you can just use that. For example you have $(1+i)^2 = 2i$, or $i^{-1} = -i$, or $(4-i)(3+i)=13+i$ for all such numbers $i$: it doesn't matter if you mean one of the complex or one of the $5$-adic roots of the equation $x^2+1=0$ with $i$ here. But I bet you imagine the complex numbers $\pm i$ one unit away from $0$, perpendicular to the real number line? That's right in the complex numbers as Gaussian plane. But those two numbers we also call $\pm i$ in $\mathbb Q_5$: One of them is close to $2$, and the other is close to $3$. Actually, the first is close to $7$, and the other close to $18$ (yes, the natrual numbers $7$ and $18$!). Actually, the first is pretty close to $57$, the other is very close to $68$ ... Actually, I can find natural numbers that are closer to them than any given $\epsilon$. They are not far away from $\mathbb Q$ at all (nothing in $\mathbb Q_5$ is)!

Also, you know of course that $\mathbb C$ has an automorphism, complex conjugation, which flips $i$ and $-i$. But $\mathbb Q_5$ has no automorphism which flips those two numbers we call $i$ and $-i$ in it. (Indeed, $\mathbb Q_5$ has no non-trivial automorphism at all.)

5. One more example: Take the polynomial $x^3-6 =0$.

This polynomial has three roots in $\mathbb C$. I assume you are familiar with writing them conventionally as $$\sqrt[3]{6}, \qquad e^{\frac{2\pi i}{3}}\cdot \sqrt[3]{6}, \qquad e^{\frac{4\pi i}{3}}\cdot \sqrt[3]{6}.$$ You see that one of them is real, while the other two are truly complex and conjugates of each other.

Well, the polynomial also has three roots in $\mathbb Q_7$. One of them is $$\text{close to } 24, \text{ another one close to } 40, \text{ and a third one close to } 34.$$

If you want to write down an embedding $\mathbb Q_7 \hookrightarrow \mathbb C$, you have to decide: Which one of those three numbers in $\mathbb Q_7$ do you want to send to the real number $1.817 ...$ we call $\sqrt[3]{6}$?

The answer, which is actually a problem, is that you could send any of them there. If you like the number $24$ better than $40$, go for it and identify that third root of $6$ with the real number $\sqrt[3]{6}$. The axiom of choice says, if I throw more and more polynomials (and then also a bunch of transcendentals) at you, you will always be able to identify them with a bunch of complex numbers in a compatible way. However, your neighbour can also say the spell "axiom of choice", but in her compatible choice $\mathbb Q_7 \hookrightarrow \mathbb C$, she identified a different root (say, the one close to $34$) with the real number $\sqrt[3]{6}$. She might also have matched a different number to $\sqrt{17e}+4.51\pi$ than we did; there's many numbers; we might never even know. Of course for some purposes, one might not care about those. Sometimes one gets away with that. Sometimes, not.

6. (Added in response to comment by Lubin:) Above I mentioned a (presumably) transcendental number at the end. If I were to expand on that, I would actually continue my theme and say that with the transcendental numbers, almost more so than with algebraic numbers, the problem is not so much that "no $p$-adic number can be identified with $\pi$" but (with axiom of choice) "literally every $p$-adic number $t$ which is transcendental over $\mathbb Q$ [of which there are ... plenty] could be identified with $\pi$", in the sense that by axiom of choice there are field isomorphisms $\mathbb C \simeq \mathbb C_p$ sending $\pi$ to $t$.

What a beginner might not get is that for the goal of calling some element of $\mathbb C_p$ by the name "$p$-adic $\pi$", it would be totally useless to say the spell "choice" and claim there is an iso as above, by which one would have merely fixed one's favourite transcendental element $t$ of $\mathbb C_p$ and given it a fancy name.

Rather, one would have to investigate what "roles" the classical $\pi$ "plays" in the real numbers (e.g.: area and / or half the circumference of the unit circle; first positive zero of the sine function; four times the value of arctan at $1$; certain rational multiple of $\zeta$-values; certain multiple of the period of the complex logarithm; ...), and then investigate if the $p$-adic analogues of any of these properties single out a certain $p$-adic number. I understand that's exactly what Professor Lubin is doing in his answer to Can $\pi$ be defined in a p-adic context?, and it turns out that many of those properties either do not have a meaningful analogue in $p$-adics to begin with, or the numbers singled out by them "over there" are something trivial (like $0$ or $1$) and certainly should not be called $p$-adic $\pi$'s.

That does not exclude that some more involved "roles" that the classical $\pi$ (or a related number like $2\pi i$, or another classical special number like $e$, $\gamma$ [?], ...) "plays" in the archimedean world might have well-defined analogues in the $p$-adic world which are played by well-defined unique elements of some $p$-adic field or ring. I do find it fascinating that e.g. in Fontaine's $B_{dR}$ (an overring of $\mathbb C_p$) there is an element which in a certain sense "plays the role" of the complex period $2\pi i$.

Solution 2:

But if you go into these things more deeply, you’ll see that the $p$-adic worlds (nonArchimedean systems) are entirely different from the real-complex world (the Archimedean system).

My favorite way of illustrating this is that within the complex numbers, any root of unity lies on the unit circle, as I hope you know. And they’re dense on the circle: every neighborhood of a root of unity contains infinitely many other roots of unity.

Entirely different for the $p$-adic universe: there, the roots of unity are a discrete set. Indeed, within a neighborhood of radius $1/3$ of a root $\zeta$ of unity, there are no other roots of unity.