# Thread: Show that Z_p is intgrally closed

1. ## Show that Z_p is intgrally closed

Hello,

Let $Z_p =${ $\frac{a}{b}:a,b\in Z, b\ne 0, gcd(a,b)=1, p$ does not divide $b$ }
$Z_p$ is a subring of $Q$
1) Show that for each $x\in Q$ either $x\in Z_p$ or $x^{-1} \in Z_p$

2) use 1) to show that $Z_p$ is integrally closed.

I answered the first part of the question but I need your help in solving part 2, how can we show that $Z_p$ is integrally closed.

All helps and guidance are highly appreciated

2. Originally Posted by student2011
Hello,

Let $Z_p =${ $\frac{a}{b}:a,b\in Z, b\ne 0, gcd(a,b)=1, p$ does not divide $b$ }
$Z_p$ is a subring of $Q$
1) Show that for each $x\in Q$ either $x\in Z_p$ or $x^{-1} \in Z_p$

2) use 1) to show that $Z_p$ is integrally closed.

I answered the first part of the question but I need your help in solving part 2, how can we show that $Z_p$ is integrally closed.

All helps and guidance are highly appreciated
suppose that $a \in \mathbb{Q} \setminus Z_p$ is integral over $Z_p$. then $a^{-1} \in Z_p$. let
$p(x) =x^n + c_1x^{n-1} + \ldots + c_{n-1}x + c_n \in Z_p[x]$
be a polynomial of minimal degree such that $p(a)=0$. if $n = 1$, then $a \in Z_p$, which is a contradiction. so $n > 1$. now let
$q(x)=x^{n-1} + c_1x^{n-2} + \ldots + c_{n-2}x + c_{n-1} + c_na^{-1}$.
then $q(x) \in Z_p[x]$ because $a^{-1} \in Z_p$. see that $q(a)=0$, which contradicts the minimality of $n$.