# Thread: Show that Z_p is intgrally closed

1. ## Show that Z_p is intgrally closed

Hello,

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

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

All helps and guidance are highly appreciated

2. Originally Posted by student2011
Hello,

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

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

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