8

I am attending a first course in algebraic number theory. We have learned the basics of field extensions, integral closures, norm and trace. I am trying to solve the following problem:

Let $K=\mathbb{Q}(a)$ be a primitive field extension of $\mathbb{Q}$ of degree $n$ with minimal polynomial $f \in \mathbb{Z}[x]$ of $a \in \mathcal{O}_\mathcal{K}$. Show that a prime number $p$ divides $d(1, a, a^2, \dots , a^{n-1})$ iff $\overline{f} \in \mathbb{Z}_p [x]$ has a zero of multiplicity 2.

What I have tried: I have shown that in this case $d(1, a, a^2, \dots , a^{n-1}) = +- N(f^\prime (a))$. Furthermore I know that because $a$ is a root of $f$, $f^\prime(a)=\prod_{\beta \neq a} (a-\beta)$. Where the $\beta$ are the roots of $f$ in an algebraic closure. Knowing that $N_{K/\mathbb{Q}}(\gamma) = \prod_{i=1}^n \sigma^i(\gamma)$ yields a product for the discriminant. However I do not see how this can help because the factors are in $\mathcal{O}_\mathcal{K}$ and not $\mathbb{Z}$

I have already seen that there is a similar result about ramified primes. However we just introduced Dedekind domains. Therefore I assume there must be a more elementary approach.

janbmull
  • 103
  • 2

1 Answers1

1

This is probably not the most elementary approach, but it follows from basic properties of discriminants and one simple computation.

In the following let $R$ be a ring. Recall that the discriminant can be defined for any $R$-algebra $A$, which is free and of finite rank as an $R$-module, ie. $A \cong R^n$ for some $n$ as $R$-modules. This property gives us the existence of the trace (and norm). The discriminant of $A$ is defined as

$$\Delta(A):=\det(\mathrm{Tr}(x_ix_j))_{i,j}, $$ where $x_1,\dots,x_n$ is an $R$-basis of $A$ and $\mathrm{Tr}=\mathrm{Tr}^A_R$ is the trace. Note that this gives a well-defined element $$\Delta(A) \in R/(R^\times)^2,$$ where $(R^\times)^2$ acts on $R$ by multiplication. In particular it makes sense to talk about when $\Delta(A)$ is zero. In your case we obtain your discriminant by taking the $\mathbb{Z}$-algebra $\mathcal{O}_K$, which yields an integer because $(\mathbb{Z}^\times)^2=1$.

We are going to record two properties of the discriminant, which are easy to prove and will be very useful.

Lemma 1. Let $\phi:R\to S$ be a ring homomorphism and $f \in R[X]$ be a (monic) polynomial. Set $$\Delta(f):=\Delta_R(f):=\Delta(R[X]/f).$$ Then we have $$\phi(\Delta_R(f))=\Delta_S(\phi(f)).$$

Lemma 2. Let $A_1,A_2$ be $R$-algebras, which are free and of finite rank as $R$-modules. Then $A_1\times A_2$ is also free and of finite rank and

$$\Delta(A_1\times A_2)=\Delta(A_1)\Delta(A_2).$$

Note that with your assumptions and my notation, we have $\Delta(\mathcal{O}_K)=\Delta_\mathbb{Z}(f)$.

Let $p$ be a prime. Applying Lemma 1 to the projection $\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}$ we obtain

$$p \mid \Delta_\mathbb{Z}(f) \iff \Delta_\mathbb{Z}(f)\bmod p = 0 \iff \Delta_{\mathbb{Z}/p\mathbb{Z}}(\bar{f})=0, $$ where $\bar{f}$ is the reduction of $f$ mod $p$. We factor $\bar{f}$ in $\mathbb{Z}/p\mathbb{Z}[X]$ as $$\bar{f}=\prod_{i=1}^{r}g_i^{n_i}, $$ where $g_i \in \mathbb{Z}/p\mathbb{Z}[X]$ are irreducible and $n_i>0$. Using Lemma 2 and the Chinese Remainder Theorem, we obtain that $$p \mid \Delta_\mathbb{Z}(f) \iff \exists j : \Delta_{\mathbb{Z}/p\mathbb{Z}}(g_j^{n_j})=0.$$

Note that $\bar{f}$ has a double zero if and only if there is some $n_j>0$. Therefore we have reduced the proof to the following statement.

Lemma 3. Let $K$ be a field, $f \in K[X]$ be irreducible and $n>0$. Then $$\Delta_{K}(f^n)=0 \iff n>1.$$

Proof: We apply Lemma 1 to the homomorphism $\phi:K\to \bar{K}$, where $\bar{K}$ is the algebraic closure of $K$. Using the fact that $\phi$ is injective, factoring $f$ in $\bar{K}[X]$ and using Lemma 2, we can assume that $K$ is algebraically closed. In that case we have $$f=X-\alpha,$$ for some $\alpha \in K$. In particular $$K[X]/(X-\alpha)^n \cong K[X]/X^n. $$ Therefore we have reduced to the case of $f=X$. In that case one can easily compute $\Delta_K(X^n)$ by hand and verify the statement. This completes the proof.

Remark: Lemma 1 can be formulated more generally, as follows. Let $\phi:R\to S$ be a ring homomorphism and $A$ a finite and free $R$-algebra. Then $$A_S:=A\otimes_RS$$ is a finite and free $S$-algebra and we have $$\phi(\Delta(A))=\Delta(A_S).$$

Ben
  • 1,310