# Projective line is normal variety

A projective variety $X$ is normal iff its local rings $O_{x,X}$ are integrally closed in $\text{Quot}(O_{x,X})$. I'm trying to show the normality of $\mathbb P^1$ by just using the above definition, that means by showing the integral closure condition directly. The problem is, i always end with the condition that $O_{x,X}$ has to be an ideal in $\text{Quot}(O_{x,X})$, which is obviously not the case. Does anyone has a hint for me how to show it without using $O_{x,X}$ being an DVR?

• Since normalization commutes with localization, you only need to show that $k[x]$ is a normal ring. – Andrew Aug 31 '13 at 16:30

Since this is a local question, we only have to show that $\mathbb{A}^1$ is normal. After translating, we can assume that $x=0$. We need to check that $k[x]_{(x)}$ is integrally closed in the quotient ring of $k[x]_{(x)}$. Now, the quotient ring of $k[x]_{(x)}$ is easily seen to be $k(x)$. $k[x]_{(x)}$ consists of all quotients $f(x)/g(x)$ where $f$ and $g$ are polynomials such that $g(0)\neq0$.
Assume that $p(x)/q(x)\in k(x)$ such that $p$ and $q$ don't have common factors (that is, common roots), and such that it satisfies an equation $$\left(\frac{p(x)}{q(x)}\right)^n+\frac{f_{n-1}(x)}{g_{n-1}(x)}\left(\frac{p(x)}{q(x)}\right)^{n-1}+\cdots+\frac{f_1(x)}{g_1(x)}\frac{p(x)}{q(x)}+\frac{f_0(x)}{g_0(x)}=0,$$ where $f_i(x)/g_i(x)\in k[x]_{(x)}$. We wish to prove that $q(0)\neq0$. Multiplying everything by $q(x)^n$, we get that $$\left(p(x)\right)^n+\frac{q(x)f_{n-1}(x)}{g_{n-1}(x)}\left(p(x)\right)^{n-1}+\cdots+\frac{f_0(x)q(x)^n}{g_0(x)}=0.$$ If $q(0)=0$, then if we evaluate this whole expression in $0$, we get that $p(0)=0$, but this is a contradiction since $p$ and $q$ don't have common factors. Therefore $q(0)\neq0$ and so $p(x)/q(x)\in k[x]_{(x)}$.
• 1. The assumption of no common roots is wlog because of the equivalence relation of the localization, right? 2. Does it play any role here that $\frac{f_i}{g_i} \in k(x)_{(x)}$? – user83496 Sep 4 '13 at 8:29
• 1. For two polynomials in one variable, having common roots is the same thing as having a common factor. So we can assume that $p$ and $q$ don't have a common factor. This is because if they did, let's say $r$, then $p=r\tilde{p}$ and $q=r\tilde{q}$, and so $p/q=\tilde{p}/\tilde{q}$. Since the degree of $\tilde{p}$ and $\tilde{q}$ is lower than the degree of $p$ and $q$, respectively, this process must stop, and we can assume that $p$ and $q$ don't have common factors. This is analogous to taking a rational number $a/b$ and assuming that $\mbox{gcd}(a,b)=1$. – rfauffar Sep 4 '13 at 13:54
• 2. Yes, we are using the fact that $f_i/g_i\in k(x)_{(x)}$ if and only if $g_i(0)\neq0$. Therefore in the last equation up top we can evaluate at $x=0$ without having problems in the denominator. – rfauffar Sep 4 '13 at 13:55
As rfauffar’s argument ,the question is local,since a variety is normal if it is covered by open affine varieties which are normal,and $P^{1}$ can be covered by two $A^{1}$.Note that $A^{1}$ is affine, it is normal iff its affine coordinate ring $k[x]$ is integrally closed. Since any UFD is integrally closed and hence normal,so is $k[x]$.In general, $P^{n}$ is always normal.