# 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'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 $c_1$...

• @elDino - is this enough to be useful? May 27, 2017 at 0:31