1

Recently, I learned the following knowledge.

(1) All $\mathbb{Q}_p$ are non-archimedean local fields and $\mathbb{R}$ is an archimedean local field. They are extensions of $\mathbb{Q}$, but their topologies are different. They correspond to a place of $\mathbb{Q}$ respectively.

(2) The algebraic closure of $\mathbb{Q}_p$, denoted by $\mathbb{C}_p$, is isomorphic to $\mathbb{C}$. (Only as a field.) I suspect that for any prime number $p$, $\mathbb{Q}_p$ is not isomorphic to $\mathbb{R}$, and for any prime numbers $p_1$ and $p_2$, $\mathbb{Q}_{p_1}$ is not isomorphic to $\mathbb{Q}_{p_2}$. But I can't prove it.

Then, my question is whether my guess is true and how to prove that $\mathbb{Q}_p$ and $\mathbb{R}$ are not isomorphic. Furthermore, whether any two different local fields are not isomorphic (only as fields)?

I have read some books on algebraic number theory and I can't find the answer, and I won't prove it myself. Is there any literature to introduce this knowledge?

J W
  • 2,351
XUSEN
  • 39
  • 2
    For literature: Robert's book "A Course in p-adic Analysis" covers this material. For how to show that $Q_p$ is not isomorphic to $R$: try counting roots of unity. The field $F_p$ has a primitive $(p-1)$th root of unity, so by Hensel's Lemma, the p-adic integers has a primitive $(p-1)$th root of unity, so $Q_p$ does as well. Meanwhile $R$ does not have a primitive $(p-1)$ root of unity for any $p>3$. This quick argument shows you that $Q_p$ cannot be isomorphic to $R$ as long as $p>3$. Also Sam Hopkins is right that these questions are better for MSE. –  Aug 04 '22 at 05:21
  • 1
    @A.S. Thank your very much! – XUSEN Aug 04 '22 at 05:24
  • A kind of a duplicate is here. – Jyrki Lahtonen Aug 04 '22 at 06:48
  • 1
    See also https://math.stackexchange.com/questions/93633/is-mathbb-q-r-algebraically-isomorphic-to-mathbb-q-s-while-r-and-s-denote – Sam Hopkins Aug 06 '22 at 01:16
  • For the record, I learned to denote by $\mathbb C_p$ not an algebraic closure of $\mathbb Q_p$, but the completion of such. These two options again might be isomorphic as fields, but certainly not as topological fields, and one should not call $\mathbb C_p$ a field which is not complete. – Torsten Schoeneberg Aug 16 '22 at 05:49

2 Answers2

2

They can all be distinguished by the question of which square roots of integers exist. Namely, over $\mathbb{R}$ all square roots of positive integers exist. On the other hand, over $\mathbb{Q}_p, p \ge 3$ we have that by Hensel's lemma that if $\gcd(p, n) = 1$ then $\sqrt[2]{n}$ exists iff $n$ is a quadratic residue $\bmod p$, and over $\mathbb{Q}_2$ the condition is that if $\gcd(2, n) = 1$ then $\sqrt[2]{n}$ exists iff $n \equiv 1 \bmod 8$; these conditions are not always satisfied for positive $n$ but are sometimes satisfied for negative $n$, which is enough to show that $\mathbb{R}$ is not isomorphic to $\mathbb{Q}_p$ for any $p$.

To distinguish $\mathbb{Q}_p$ from $\mathbb{Q}_q$ for distinct primes $p, q$ it suffices to observe that by the Chinese remainder theorem we can always find $n$ such that $\gcd(n, pq) = 1$ and $n$ is a quadratic residue $\bmod p$ but not a quadratic residue $\bmod q$ (with $p$ or $q$ respectively replaced by $8$ for the $\mathbb{Q}_2$ case).

Qiaochu Yuan
  • 468,795
  • 2
    I like to do it by counting the roots of unity in the fields $\Bbb Q_p$, with a special argument for the pair ${2,3}$. – Lubin Aug 04 '22 at 19:49
0

According to your guidance, I came up with a method myself, but I'm not sure whether it is correct. Any $n$-th root of unity in $\mathbb{Q}_p$ belongs to $\mathbb{Z}_p$. By Hensel's Lemma, the field $\mathbb{F}_p$ has a primitive $n$-th root of unity if and only if $\mathbb{Z}_p$ has a primitive $n$-th root of unity. For any prime numbers $p_1<p_2$, the field $\mathbb{F}_{p_1}$ does not has a primitive $(p_2-1)$-th root of unity. We deduce that $\mathbb{Q}_{p_1}$ can not be isomorphic to $\mathbb{Q}_{p_2}$.

I think the key step in this kind of problem is to use Hensel's lemma and polynomials.

XUSEN
  • 39
  • Thanks @Qiaochu yuan Excuse me, is my idea correct? – XUSEN Aug 05 '22 at 06:27
  • You need to be a little careful about the case where $p \mid n$, and this argument doesn't distinguish $\mathbb{Q}_2$ from $\mathbb{Q}_3$ (note that $\mathbb{Q}_2$ has a primitive second root of unity even though $\mathbb{F}_2$ doesn't). – Qiaochu Yuan Aug 05 '22 at 08:19
  • With the exception of $p=2$, there are as many roots of unity in $\Bbb Z_p$ as in $\Bbb F_p$, so your idea was perfectly sound. – Lubin Aug 08 '22 at 02:11
  • @Lubin Excuse me, can you tell me how to prove "With the exception of p=2, there are as many roots of unity in Z_p as in F_p". By Hensel's Lemma, I can't get it. My statement "By Hensel's Lemma, the field Fp has a primitive n-th root of unity if and only if Zp has a primitive n-th root of unity. " is wrong because the lift of amodp may not equal to a. – XUSEN Aug 09 '22 at 04:00
  • Let’s see, @XUSEN — By “$a\mod p$”, I suppose you mean an integer between $1$ and $p-1$. These are almost never roots of unity in $\Bbb Z_p$. Look at $\Bbb Z_5$, which has a perfectly good (primitive) fourth root of unity. Is this what’s confusing you? In any event, the fact is that every nonzero element of $\Bbb F_p$ lifts to a root of unity in $\Bbb Z_p$. That’s a direct consequence of (strong) Hensel. Just take $X^{p-1}-1$ in characteristic zero, and use Hensel on it: Since it factors into linears modulo $p$, it factors into linears over $\Bbb Z_p$, and there you are. – Lubin Aug 09 '22 at 16:50