11

Let $p$ be an odd prime number and $a\in \mathbb {Z}$ with $\sqrt[p]{a}\notin \mathbb{Z} $. Prove that $p$ is ramified in the number field $\mathbb{Q}(\sqrt[p]{a})$.

My idea is to apply Dedekind's Theorem: A rational prime number is ramified in a number field $\mathbb K$ if and only if $p \mid d(\mathbb K)$ where $d(\mathbb K)$ stands for the discriminant of $\mathbb K$ (over $\mathbb Q$). So the question boils down to computing $d(\mathbb K)$. I tried to find the minimal polynomial of $\sqrt[p]{a}$ over $\mathbb Q$ but failed.

There is a lemma which I ignored at first: For $\mathbb K=\mathbb Q(w)$ where $w$ is a root of an irreducible polynomial which satisfies Eisenstein's Theorem ($p$ is used when applying it), $p$ is totally ramified in $\mathbb K$. I neglected it because $a$ may not have a prime factor of degree $1$.

Eric Wofsey
  • 330,363
user341609
  • 111
  • Isn't the minimal polynomial just $x^p-a$? – JasonM Jun 23 '16 at 03:08
  • @JasonM How to prove this? – user341609 Jun 23 '16 at 03:11
  • @JasonM Irreducible? Use Eisenstein' Theorem? – user341609 Jun 23 '16 at 03:27
  • I think I've solved the problem: decompose $a$ into a product of prime numbers. We may assume each factor's degree is less than p and hence coprime to p. Let $\alpha=\sqrt[p]{a}$, and we can find some $t$ with $(t,p)=1$ such that some factor's degree of $a^t$ is coprime to p. And $\mathbb Q(\alpha)=\mathbb Q(\alpha^t)$. Then we can prove $x^p-\beta$ is irreducible since some factor of $\beta$ is of degree one (Eisenstein' Theorem) . @JasonM Is this correct? – user341609 Jun 23 '16 at 04:33
  • I think it's correct @user341609 – user12580 Jun 23 '16 at 04:43
  • 1
    This question helps you with irreducibility. Finding an integral basis may still be taxing... but may not be necessary to settle the question(?) – Jyrki Lahtonen Jun 23 '16 at 08:26
  • @JyrkiLahtonen For $\mathbb K=\mathbb Q(w)$ where $w$ is a root of an irreducible polynomial which satisfies Eisenstein's Theorem ($p$ is used when applying it), $p$ is totally ramified in $\mathbb K$. This is a lemma in my textbook. So the link is good to show irreducibility but not good for solving this problem. – user341609 Jun 23 '16 at 08:58
  • user341609 Irreducibility implies that $a^{k/p}, k=0,1,2,\ldots,p-1$, is then a basis consisting of algebraic integers of the bigger field. Computing the discriminant of that basis is easy enough. But (as is often the case with $p=2$) this may not be an integral basis, so may be we won't get the discriminant of the field. However, the difference is a square factor. So should the discriminant here have $p$ to an odd power as a factor, we can still conclude that $p$ is a factor of the true discriminant as well. I don't know if this is the case here, but for lack of other ideas ... – Jyrki Lahtonen Jun 23 '16 at 09:06
  • Ignoring the sign, the formula $N_{L/K}(f'(\alpha))=N_{L/K}(p\alpha^{p-1})$ along with the irreducibility is enough. Thank you! – user341609 Jun 23 '16 at 09:12
  • 1
    user341609 Good job! In that case I would like to encoourage you to also flesh that out as an answer. That way you get more feedback, any lingering unclear steps in your argument will be exposed, and last but not least, this question gets removed from the unanswered queue :-) Particularly future readers will want to see where the argument fails when $p=2$ (and the result is known to be false). – Jyrki Lahtonen Jun 23 '16 at 10:19

1 Answers1

3

Using the irreducibility of $X^p-a$ as discussed in the comments write $K=\mathbb{Q}(\sqrt[p]{a})$ and $R=\mathbb{Z}[\sqrt[p]{a}]$. Then $\Delta(R)=\Delta(X^p-a)=(-1)^{\frac{1}{2}p(p-1)}p^pa^{p-1}$, and if we write $a=p^kb$ for $0\leq k<p$ and $(p,b)=1$ then we see that $$ \Delta(\mathbb{Z}[\sqrt[p]{a}])=(-1)^{\frac{1}{2}p(p-1)}p^{p+k(p-1)}b^{p-1}. $$ As $p$ is odd we know that $p+k(p-1)$ is also odd, hence the relation $$ \Delta(\mathbb{Z}[\sqrt[p]{a}])=[\mathcal{O}_K:R]^2\cdot \Delta(\mathcal{O}_K) $$ shows that we must have $p|\Delta(\mathcal{O}_K)$, and hence $p$ ramifies in $K$.

The statement indeed fails for $p=2$. For a quadratic field $\mathbb{Q}(\sqrt{d})$ with $d\equiv 1\bmod 4$ we have $\Delta(\mathcal{O}_K)=d$, which is odd and hence $2$ is unramified in $K$.

Tim.ev
  • 1,618