I've been given the following question for some revision of algebraic number theory:

Let $K = \mathbb{Q}(\alpha)$, where $\alpha ^n = a$. If $p\mid a, p^2\nmid a$, then $p\nmid [\mathcal{O}_K : \mathbb{Z}[\alpha]]$

I have already proved that if $\alpha \in K$ is algebraic with $\mathbb{Q}(\alpha) = K$ and $\mathbb{Z}[\alpha]≠\mathcal{O}_K$, the ring of integers of $K$, then for all prime divisors $p$ of $[\mathcal{O}_K : \mathbb{Z}[\alpha]]$ there is an algebraic integer of the form $$\beta = \sum_{I=0}^{n-1} \frac{c_i}{p} \alpha^I$$ where the $c_i \in \mathbb{Z}$ are not all divisible by $p$.

What I have attempted so far is to go by contradiction: if it is the case that $p\mid[\mathcal{O}_K : \mathbb{Z}[\alpha]]$, then there is such a $\beta$. I then tried to compute an expression for the norm of $\beta$, and show that $\text{Norm}_{K/\mathbb{Q}}(\beta) \notin \mathbb{Z}$, but couldn't manage this.

Perhaps I am supposed to extract some other algebraic integer from the expression for $\beta$, then show that its norm is not an integer, but I can't see how I could do that.

Thanks for your help, any hints would be appreciated.


Here is one way of doing this problem that is a little different than the way you were going or the hint of @peter a g.

Let $D(\alpha)$ be the polynomial discriminant of the minimal polynomial for $\alpha$ and let $d(K)$ be the field discriminant for $K$.

The quantity $[\mathcal{O}_K : \mathbb{Z}[\alpha]]$ is sometimes called the index. It turns out that there is a formula relating the discriminant of the minimal polynomial, the discriminant of the field, and the index: $$ D(\alpha)=[\mathcal{O}_K : \mathbb{Z}[\alpha]]^2\cdot d(K). $$

Assuming that $a\in\mathbb{Q}$ we have that $x^n-a$ is the minimal polynomial for $\alpha$. In particular, since $p|a$ and $p^2\nmid a$ it follows that $x^n-a$ is $p$-Eisenstein. Now since the minimal polynomial of $\alpha$ is $p$-Eisenstein we have that $p$ is totally ramified in $\mathbb{Q}(\alpha)$, and more importantly for us that $p^{n-1}||d(K)$ if $p\nmid n$ and $p^n|d(K)$ if $p|n$ (this is theorem 3.6 here).

The polynomial discriminant of $x^n-a$ is known to be $D(\alpha)=\pm n^n a^{n-1}$.

So, when $p\nmid n$ we have $p^{n-1}||d(K)$ and that $p^{n-1}||D(\alpha)=\pm n^n a^{n-1}$ since $p\nmid n$ and $p||a$. But now by the `index-discriminant' formula $$D(\alpha)=[\mathcal{O}_K : \mathbb{Z}[\alpha]]^2\cdot d(K) $$ and since $p^{n-1}||D(\alpha)$ and $p^{n-1}||d(K)$ it follows that $p\nmid [\mathcal{O}_K : \mathbb{Z}[\alpha]]$.

When $p|n$ things are a little less strait forward. One way to proceed is to use the following theorem which is equivalent to part of Theorem 6.1.4 in Cohen's book A Course in Computational Algebraic Number Theory:

Theorem (Dedekind) Let $\alpha$ be an algebraic integer with minimal polynomial $m$ and set $K=\mathbb{Q}(\alpha)$. Let $p$ be a prime, and write $$ m(x)=\prod_{i=1}^r m_i(x)^{e_i} \pmod{p} $$ where $m_i\in\mathbb{Z}[x]$ are monic, irreducible lifts of the irreducible factors of $m$ modulo $p$. Set $$ g(x)=\prod_{1\leq i\leq r} m_i(x),\hspace{3mm} h(x)=\prod_{1\leq i\leq r} m_i(x)^{e_i-1}, \hspace{3mm} \text{and }\ f(x)=\frac{g(x)h(x)-m(x)}{p}. $$ Then $p|[\mathcal{O}_K : \mathbb{Z}[\alpha]]$ if and only if $\gcd(\bar{f},\bar{g},\bar{h})\neq 1$, where over-lines denote reduction modulo p.

In our situation we have $$m(x)=x^n-a\equiv x^n\pmod{p}.$$ Therefore $g(x)=x$ and $h(x)= x^{n-1}$ and also $$ f(x)=\frac{x^n-(x^n-a)}{p}.$$ So by Dedekind's Theorem we have $p|[\mathcal{O}_K : \mathbb{Z}[\alpha]]$ if and only if $\gcd(\bar{f},\bar{g},\bar{h})\neq 1$ which can only happen only if $x|\bar{f}$. However $x|\bar{f}$ only when $f(0)\equiv 0 \pmod{p}$ which in turn happens only if $p\cdot f(0)\equiv 0\pmod{p^2}$. But $p\cdot f(0)=a$ and therefore $a\not\equiv 0\pmod{p^2}$, since $p||a$. Thus $p\nmid [\mathcal{O}_K : \mathbb{Z}[\alpha]]$.

Of course one could use the Theorem of Dedekind which we used for the case when $p|n$ also in the case when $p\nmid n$.

A couple of references for this sort of stuff (besides those already mentioned) are a preprint of a paper by Alden Gassert which can be found here and chapters 7 and 10 of Alaca and Williams Introductory Algebraic Number Theory.


Hint: Suppose $\cal P$ is the prime ideal above $(p)$. The order of $\alpha$ at $\cal P$ is $1$: $(\alpha) = \cal P A$, where $\cal A$ is prime to $\cal P$. The order of $p$ is $n$. So look at your expression for $p\beta$ to conclude that the $c_i$ must be divisible by $p$.

Edit: For instance, $p\beta = c_0 + \alpha\cdot (\text{stuff})$, so $\cal P$ divides $c_0$, and hence $p$ divides $c_0$. Then (assuming $n>1$) $${\cal P }^2\ |\ p\beta -c_0 = \alpha \cdot ( c_1 + \alpha\cdot \text{stuff}),$$ so $\cal P$ must divide $c_1$...

  • $\begingroup$ @elDino - is this enough to be useful? $\endgroup$ – peter a g May 27 '17 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.