Prove that any conic in $\mathbb{P}^2$ is normal.

A variety $X$ is called normal if for any point $P$$\in$$X$ ,the local ring $\mathbb{O}_P$ is an integrally closed ring. Show that any conic in $\mathbb{P}^2$ is normal.

Here is my attempt: Use the the fact that any conic in $\mathbb{P}^2$ is isomorphic to $\mathbb{P}^1$. So now I need to show for any point $P$$\in$$\mathbb{P}^1$ ,the local ring $\mathbb{O}_P$ is an integrally closed ring. Over $\mathbb{P}^1$, the local ring $\mathbb{O}_P$ is isomorphic to $k[x_{0},x_{1}]_{(m_{P})}$ which is the zero degee piece of the localization at $m_{p}$=[$f$$\in$$k[x_{0},x_{1}]$|$f$ homogeneous and $f(P)$=$0$]. Thus, I have to show the local ring $k[x_{0},x_{1}]_{(m_{P})}$ is integrally closed. I know UFD is integrally closed. But how can I show that this local ring is a UFD? Or is there another way to prove the conclusion?

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You mean any smooth conic, right? (because no singular conic will be isomorphic to $\mathbb{P}^1$) – Fredrik Meyer Nov 25 '12 at 22:22
If $P\ne \infty$, then $O_P$ is a localization of the polynomial ring $k[x]$, hence is UFD. If $P=\infty$, then again $O_P$ is a localization $k[1/x]$ at the maximal ideal generated by $1/x$. – user18119 Nov 25 '12 at 22:49

You can prove that $k[x_0,x_1]$ is integrally closed. Then prove that if a ring $A$ is integrally closed, then any localization $S^{-1} A$ is integrally closed.

Edit: As pointed out by Fredrik Meyer, you should say you are considering a smooth conic in $\mathbb{P}^2$. The other type of conic in $\mathbb{P}^2$ is a union of two lines, so it's not even a variety (assuming you require a variety to be irreducible). (If you work with schemes, there is a third case - a double line, i.e. the zero scheme of the homogeneous ideal in $k[x_0,x_1,x_2]$ of the square of a linear form.)

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It is not true in general that a conic in $\mathbb P^2_k$ is normal, if by conic you mean the curve defined by a non-constant homogeneous polynomial $f(X,Y,Z)\in k[X,Y,Z]\setminus k$ of degreee $2$.
For example the polynomials $f_1(X, Y, Z)=X^2-Y^2$ or $f_2(X, Y, Z)=X^2$ or $f_3(X,Y,Z)=X^2+Y^2+Z^2=(X+Y+Z)^2$ in characteristic $2$ all define non-normal conics.

If however $k$ is an algebraically closed field and the polynomial $f$ is irreducible, then indeed the conic $C=V(f)\subset \mathbb P^2_k$ is normal.
The most geometric way to see this is simply to project the conic from one of its points $P\in C$ unto a line $l\subset \mathbb P^2_k$ not going through $P$ (i.e. $P\notin l$) and realize with delight that this yields an isomorphism $C\cong \mathbb P^1_k$. (Details in Mumford's red book, pages 11-12 and 15-16 ) .
Since $\mathbb P^1_k$ is normal, this implies that $C$ is normal too.

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