# Lemma in KCd's notes on Totally Ramified Primes and Eisenstein Polynomials

I am reading Keith Conrad's notes here on Totally Ramified Primes and Eisenstein Polynomials. I am trying to understand the proof of Lemma 3.1 on page 4 of his notes which I will reproduce here:

Lemma 3.1. Let $K=\Bbb{Q}$ be a number field with degree $n$. Assume $K = \Bbb{Q}(\alpha)$, where $\alpha \in \mathcal{O}_K$ and its minimal polynomial over $\Bbb{Q}$ is Eisenstein at $p$. For $a_0,a_1,\ldots, a_{n-1} \in \Bbb{Z}$, if $$a_0 + a_1\alpha + + a_{n-1}\alpha^{n-1} \equiv 0 \pmod{p\mathcal{O}_K}$$ then $a_i \equiv 0 \pmod{p\Bbb{Z}}$ for all i.

The proof is as follows. We multiply the expression above by $\alpha^{n-1}$. Since $\alpha$ is Eisenstein at $p$ this eliminates everything except $$a_0\alpha^{n-1}\equiv 0 \pmod{p\mathcal{O}_K}.$$ Now KCd then says that when we apply $N_{K/\Bbb{Q}}(-)$ we get that

$$a_0^n\alpha^{n-1}\equiv 0 \pmod{p^n\Bbb{Z}}.$$

My question (1) is: Why is the congruence mod $p^n$ and not mod $p$ ? Is it because $p^n | a_0^n$?

He then proceeds to say:

The norm of is, up to sign, the constant term of its characteristic polynomial for $K=\Bbb{Q}.$ Since $\alpha$ generates $K=\Bbb{Q}$, its characteristic polynomial is its minimal polynomial, which is Eisenstein. Therefore $N_{K/\Bbb{Q}}(\alpha)$ is divisible by $p$ exactly once, so the above congruence modulo $p^n$ implies $p|a_0^n$, so $p|a_0$.

My question (2) is: I get why the norm is divisible by $p$ exactly once, but what does he mean "the above congruence mod $p^n$ implies $p|a_0^n$? This is where I am confused because isn't the congruence $$a_0^n\alpha^{n-1}\equiv 0 \pmod{p^n\Bbb{Z}}.$$

taken mod $p^n$? Also, how did he deduce that $p|a_0^n$?

Thanks.

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I updated the file to clear up what was meant by taking norms on a congruence. – KCd May 28 '13 at 16:00
@KCd Thanks very much. – user38268 May 31 '13 at 5:06

For Question 1, $a_0\alpha^{n-1} \in p\mathcal{O}_K$, and we know that for any ideal $\alpha$, if $x \in \alpha$, then $N(\alpha)|N(x)$. Thus, $N(p\mathcal{O}_K)|N(a_o\alpha^{n-1})$. But, $N(p\mathcal{O}_K) = p^n$.
For Question 2, you get that $p^n | a^n_0N(\alpha)^{n-1}$, and you know that $p^{n-1}$ is the highest power of $p$ that divides $N(\alpha)^{n-1}$. Thus, you must have that $p|a^n_0$, and consequently, $p|a_0$.
Oh, that is because the map $R \rightarrow \alpha R$ which maps $r \mapsto \alpha r$ is an isomorphism of $\mathbb{Z}$-modules. Also, sorry, for some reason I got confused by your question. I have indeed seen if before. This being said, we should not discuss this second question here. – Rankeya Dec 23 '12 at 13:31