Am I correct in assuming that any projective scheme $X \subset \mathbb P^n$ can be written as $X \cong \mathrm{Proj} (\Bbbk [x_0, \dotsc, x_n] / I)$ for some homogeneous ideal $I = (f_1, \dotsc, f_r)$ so that $X$ can be described concretely by considering the restriction of affine charts of $\mathbb P^n$ to $X$ and affine charts of $X$ can be given by $\mathrm{Spec} (A_i)$ where $$ A_i = \Bbbk [\tfrac{x_0}{x_i}, \dotsc, \hat{\tfrac{x_i}{x_i}}, \dotsc, \tfrac{x_n}{x_i}] / (f_{1, i}, \dotsc, f_{r,i}) $$ where $[x_0 : \dotsb : x_n]$ are homogeneous coordinates of $\mathbb P^n$ and $i$ runs from $0$ to $n$ and $f_{k,i}$ is the dehomogenization of the polynomial $f_k$ that appears as a relation in $I$?

When comparing schemes and varieties from the point of view of affine coordinate charts is the only difference that for schemes the $A_i$'s may contain nilpotent elements?


Regarding the first question: yes, this is true. Given a closed subscheme $X$ of $\mathbb P^n_k$, on each of the standard affine opens $U_i = D_+(x_i)$ it is defined by some ideal $J_i$ of $(k[x_0,\dots ,x_n]_{x_i})_0$, by affineness. Each $J_i$ corresponds to a homogeneous ideal $I_i$ of the homogeneous coordinate ring $k[x_0,\dots ,x_n]$, and you can check that $I = \bigcap_i I_i$ does the job. In fact, this argument works for closed subschemes of any projective scheme $\operatorname{Proj} S_\bullet$ (even when $S_\bullet$ is not generated in degree 1, with slight modification).

As for the second question, a (not necessarily irreducible) variety can be defined as a reduced, separated scheme of finite type over a field. A projective scheme is always separated and finite type, so the two things which can be relaxed are "reduced" and "field" - in general, your $A_i$ could be quotients of $R[x_{0/i}, \dots ,\hat x_{i/i} ,\dots ,x_{n/i}]$ for $R$ some ring, with $X\subset\mathbb P^n_R$. If you insist on working over $k$, then nilpotents are the only new thing that you will see in $A_i$.

Note also that if $X$ is a variety, then it is not necessarily true that $I$ is a radical ideal - consider $\operatorname{Proj} k[x_0,x_1]/(x_0^2,x_0 x_1)$. So, it's possible that the coordinate rings of the standard open affines have no nilpotents, yet the homogeneous coordinate ring of $X$ is nonreduced.

  • $\begingroup$ Thank you very much. I don't quite understand the last paragraph, which I guess pertains to the last paragraph of the question. So you can define a variety locally by any ideal, but the Nullstellensatz gives a correspondence between algebraic sets and radical ideals, so this means you can naturally associate a radical ideal to a variety. However, when considering schemes, it matters if the ideal is radical or not... Or did I miss your point? $\endgroup$ – Earthliŋ Apr 28 '20 at 19:54
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    $\begingroup$ Sorry, that was more of a remark about nilpotents in $A_i$ vs nilpotents in the homogeneous coordinate ring. I edited to add an actual answer to your second question and clarified my other statement. The Nullstellensatz gives a correspondence between algebraic subsets of $\mathbb A^n_k$ and radical ideals; there isn't such a correspondence between projective algebraic sets and homogeneous radical ideals, as this example shows. $\endgroup$ – Alex K Apr 28 '20 at 22:59

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