# 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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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, $p|a_0$.
 Thanks for your answer. I think I was confused before between taking the norm of an element and the norm of an ideal. Thanks. Also, I was getting tied up in knots for question 2. I will accept your answer. – BenjaLim Dec 23 '12 at 12:36 I am glad I was able to help. Others may have better advice to offer on how to think about norms of ideals vs. elements, but the norm map on ideals is essentially an extension of the norm map on elements. Indeed, the norm of an ideal is defined in such a way that the norm of a principal ideal equals the norm of the element generating the ideal. I find it helpful to keep this in mind. I think Milne's notes on ANT discusses this. – Rankeya Dec 23 '12 at 13:15 Yes that is true. The norm map defined for ideals that I have is: $N(I)$ where $I$ an ideal in a number ring $R$ is defined to be the cardinality of the finite group $R/I$. Then as you say it can be proved that $N((a)) = |N(a)|$. Just to ask in passing, do you know why very briefly every ideal in a number ring is free abelian of rank equal to that of the number ring? – BenjaLim Dec 23 '12 at 13:17 Interesting, I never heard of the question you ask. Where did you come across this? I know that the set of nonzero fractional ideals of a number ring form a free abelian group. – Rankeya Dec 23 '12 at 13:21 I think this is exercise 4 chapter 3 of Marcus. "Let $K$ be a number field of degree $n$ over $\Bbb{Q}$. Prove that every non-zero ideal $I$ in $R$ ( the integral closure of $\Bbb{Z}$ in $K$) is a free abelian group of rank $n$." The hint given is that for any $\alpha \in R$, we have $\alpha R \subset I \subset R$ but it is not immediately clear to me as to why the rank of $\alpha R$ is equal to $n$. – BenjaLim Dec 23 '12 at 13:28