5

We have the definition of integral closure that all the integral elements of A in B. Could we just compute the integral closure of certain A in B. I am considering such a problem that given a prime p, what is the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[x]/(x^n–p)$.

After some trials, I find the answer maybe $\mathbb{Z}[x]/(x^n–p)$. Since $\mathbb{Z}\subset\mathbb{Z}[x]/(x^n–p)$, the integral closure of $\mathbb{Z}$ should be the subset of the closure of $\mathbb{Z}[x]/(x^n–p)$. But $\mathbb{Z}[x]/(x^n–p)$ is generated by $1, s, s^2, \dots, s^{n–1}$ which are integral over $\mathbb{Z}$. $s$ is the image of x modulo (x^n–p). Therefore $\mathbb{Z}[x]/(x^n–p)$ is integral over $\mathbb{Z}$.

Now I want to prove that $\mathbb{Z}[x]/(x^n–p)$ is normal. Assume that $u\in\mathbb{Q}[x]/(x^n–p)$ is integral over $\mathbb{Z}[x]/(x^n–p)$ with a monic poly $f$. Since $\exists d \in\mathbb{Z}$ and $e\in \mathbb{Z}[x]/(x^n–p)$ s.t du–e=0. Then f can be divided by dx-e. Assume that $f=(dx-e)g$, where $g$ is a poly with leading coefficient h in $\mathbb{Z}[x]/(x^n–p)$. Then we get $hd=1$. Since $d$ is a integer, $h$ has to be a integer too. This implies that $d=1$ or $–1$. Both sides implies $u\in \mathbb{Z}[x]/(x^n–p)$. So we got the integral closure of $\mathbb{Z}$ contains $\mathbb{Z}[x]/(x^n–p)$ which is exactly the closure of itself. Q.E.D

Is there any mistakes in the proof?

Watson
  • 23,793
Intoks Liobein
  • 857
  • In general the question of finding the integral closure is quite subtle, and involves tools such as trace, discriminant et cetera. I use the definitive article, because the closure is unique in this instance. – Jyrki Lahtonen Dec 15 '15 at 11:12
  • @JyrkiLahtonen could you tell me how to solve this problem. I think it may need something concerned cyclic extension. – Intoks Liobein Dec 16 '15 at 02:43
  • Cam McLeman is one of the practitioners of algebraic number theory. He is discussing the case $n=3$, $p=2$ here, and says that this problem is difficult in general (to be honest he may be thinking about a bit more general problem, when the polynomial you mod out is not a binomial). I have no reason not to believe him. He is an expert. Use that as ammo against people who think you should be able to solve this easily. User26857 was right in that the special case $n=2$ is easy. The answer does depend on the residue class of $p$ modulo $4$. – Jyrki Lahtonen Dec 16 '15 at 21:52
  • Then again, user26857 is very knowledgeable about commutative algebra and its application to algebraic geometry. So if you were asking about integral closure inside a more geometrically originating ring, he would know of ways to relate integral closure to geometric properties. – Jyrki Lahtonen Dec 16 '15 at 21:55
  • @user26857 could we get the integral closure z[x]/(x^n–p) is the subset of the integral closure of z from it contains z[x]/(x^n–p) – Intoks Liobein Dec 17 '15 at 10:26
  • The proposition after lemma 6 in this answer could be interesting for you. – Watson Jul 23 '16 at 12:04
  • You can read https://kconrad.math.uconn.edu/blurbs/gradnumthy/integersradical.pdf. It is not true that the ring of integers of $\Bbb Q(\sqrt[n]{2})$ is equal to $\Bbb Z[\sqrt[n]{2}]$ in general! This is true for $n=2, 3, 4$ and in fact for all $n \leq 1000$, but not for $n=1093$... ! – Watson Jun 04 '21 at 09:47

0 Answers0