In Algebraic Geometry, chapter 3, Robin Hartshorne proves that for an affine variety (irreducible variety) $Y$, it follows that the ring of regular functions $\mathcal{O}(Y)$ is isomorphic to the coordinate ring of that variety $A(Y)$.

This he proves by first proving that both $\mathcal{O}(Y)$ and $\mathcal{O}_p$ are subrings of the function field $K(Y)$ whose elements are the rational functions on $Y$, and then proving that $A(Y)_{\mathfrak{m}_p} \cong \mathcal{O}_p$. Then embedding both $A(Y)_{\mathfrak{m}_p}$ and $A(Y)$ in $K(Y)$ he gets the chain \begin{equation*} A(Y) \subseteq \mathcal{O}(Y) \subseteq \bigcap_{p \in Y} \mathcal{O}_p = \bigcap_{\mathfrak{m}_p} A(Y)_{\mathfrak{m}_p} = A(Y) , \end{equation*} and concludes that $A(Y) \cong \mathcal{O}(Y)$.

My question is, does $A(Y) \cong \mathcal{O}(Y)$ hold true if $Y$ is an arbitrary affine algebraic set that is not necessarily a variety, (does it hold true if $Y$ is a reducible variety)?

The problem, such as I'm seeing it, is that $K(Y)$ is not necessarily well-defined as a ring if $Y$ is not a variety (irreducible variety), and then you cannot embed $A(Y)$ or $A(Y)_{\mathfrak{m}_p}$ in it through a ring morphism.

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    $\begingroup$ $A(Y) \simeq \mathcal{O}(Y)$, for any affine algebraic set. And you are also correct about the fact $K(Y)$, if $Y$ is not irreducible. But to prove that $A(Y) \simeq \mathcal{O}(Y)$, you require something more: locally ringed spaces etc. This is a terminology that Hartshorne avoided in the first chapter. If you want, you can look at other books/note, e.g. Milne's note on Algebraic Geometry (can be downloaded from his homepage jmilne.org/math) $\endgroup$ – Krish Aug 31 '17 at 6:51
  • $\begingroup$ Ah! Beginning to understand just how powerful those "higher" definitions really are if you want to do algebraic geometry in earnest. Thanks for letting me know how this problem can be resolved! $\endgroup$ – StormyTeacup Aug 31 '17 at 14:12

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