Question on reducedness This is Exercise 5.3.E in the latest notes of Vakil. We have the ideal $I=(x_0^2,x_0x_1,...,x_0x_n)\subseteq k[x_0,...,x_n]$ and we want to see that $\operatorname{Proj}k[x_0,...,x_n]/I$ is reduced. This is a local condition, so we may look at $\operatorname{D}_+(x_i+I)=\operatorname{Spec}((k[x_0,...,x_n,x_i^{-1}]/I)_{(0)})$ and check that this is reduced. So it suffices to see that $k[x_0,...,x_n,x_i^{-1}]/I$ is reduced. However, if $x_i$ is a unit, then $x_0$ becomes an element of $I$ in the localization an we can throw away the $x_0^2$ thing, giving $I$ is radical. 
My question is: is this a reasonable argument or am I completely mistaken? 
Now that I searched for similar questions, I saw this one Working out $\operatorname{Proj} k[x_0,...x_n]/(x_0^2,x_0x_1)$ which gives me confidence that I am completely mistaken. So rather then asking if I‘m wrong, I ask where I‘m wrong. Thanks!
 A: Your argument is reasonable and correct. $\operatorname{Proj} R/I$ is naturally a closed subscheme of $\operatorname{Proj} R$, so an open affine covering of $\operatorname{Proj} R$ gives an open affine covering of $\operatorname{Proj} R/I$ by taking intersections, as you've done. Since localization commutes with taking quotients, we see that if $i\neq 0$, then $k[x_0,\cdots,x_n]/(x_0^2,x_0x_1,\cdots,x_0x_n)$ localized at $x_i$ gives $k[x_0,\cdots,x_i^\pm,\cdots,x_n]/(x_0^2,x_0x_1,\cdots,x_0,\cdots,x_0x_n)$ which is isomorphic to $k[x_0,\cdots,x_i^\pm,\cdots,x_n]/(x_0)\cong k[x_1,\cdots,x_i^\pm,\cdots,x_n]$, a reduced ring. (If $i=0$, then we inverted a nilpotent element, so we get the zero ring, which is again reduced.) So it's degree-zero portion is again reduced, and thus $\operatorname{Proj} R/I$ has an open cover consisting of reduced affine open subschemes. This means it is reduced.
The linked question is similar, but the key difference is that the nilpotents survive the localization process. Looking at the intersection of $X=\operatorname{Proj} k[x_0,x_1,x_2]/(x_0^2,x_0x_1)$ with the standard affine coordinate patches $U_i=D(x_i)$, we get the following:


*

*$U_0\cap X$: just like above, since we invert $x_0$ which is a zero-divisor, we get the zero ring. This means $U_0\cap X=\emptyset$.

*$U_1\cap X$: just like above, we get the spectrum of the degree-zero portion of $k[x_0,x_1^\pm,x_2]/(x_0)$, which is reduced.

*$U_2\cap X$: this time, things are different: we want the spectrum of the degree-zero portion of $k[x_0,x_1,x_2^\pm]/(x_0^2,x_0x_1)$. But writing $x=x_0/x_2$ and $y=x_1/x_2$, we get $k[x,y]/(x^2,xy)$ which is not reduced.
So our variety $V(x_0^2,x_0x_1)\subset\Bbb P^2$ has nilpotents hanging out at the point $[0:0:1]$. 
One way to think about it is that for any term of the form $x_0x_i$, this means that $x_0$ must honestly vanish on $D(x_i)$, because $x_i$ is invertible there and so $x_i$ is in our ideal. So the only way that $x_0$ can dishonestly vanish (ie, nilpotently) is if $x_0$ isn't in our ideal after localization at some $x_i$, which means that $x_0x_i$ couldn't have been in our ideal.
