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I have seen that the notion of affine scheme is a generalization of the notion of affine varieties where the coordinate ring is replaced by any commutative unit ring, and the variety with the Zariski topology is replaced by any topological space. Does this mean that every affine variety is an affine scheme? I think that if we have an affine variety X and we construct its structure sheaf then the affine variety is a locally ringed space but how we conclude that it is isomorphic to a spectrum of a ring? Or there is another proof of this statement?

Mathematician
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    What historically was called an "affine variety", e.g. the space of zeroes of a set of polynomials $V(f_1,\dots ,f_r)\subseteq k\lbrack x_1,\dots ,x_n\rbrack$, is not an affine scheme because it is not the spectrum of a ring (it is typically not sober). Instead, it is (homeomorphic, or isomorphic as locally ringed spaces) to the subspace of closed points of the underlying space of the affine scheme $\mathrm{Spec}(k\lbrack x_1,\dots ,x_n\rbrack / (f_1,\dots ,f_r))$. – o.h. May 29 '23 at 12:02
  • To elaborate on the comment above, topologically these two things are not the same as the generic point (e.g. $(0)$ in $Spec k[x]$) is not something that exists in the classical way we define affine varieties as maximal ideals of a finitely generated reduced $k$-algebra. – daruma May 29 '23 at 12:32
  • But these two notions can be translated from one another. In particular, there is an equivalence of categories between the category of affine varieties and the category of spectrum of finitely generated reduced $k$-algebra ($k=\bar{k}$). – daruma May 29 '23 at 12:34

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