# Why we need $R$ to be radical at here?

Jacob Lurie used one example for the theorem of Scholism:

Let $R=k[X]$ be the coordinate ring of a variety $X$ in $\mathbb{C}^{n}$. Assume $X$ is reduced. Then $MaxSpecR$ is a union of irreducible components $X_{i}$, which are the closures of the minimal primes of $R$. The fields you get by localizing at the minimal primes depend only on the irreducible components, and in fact are the rings of meromorphic functions on $X_{i}$. Indeed we have a map $$k[X]\rightarrow \prod k[X_{i}]\rightarrow \prod k(X_{i})$$

If we do not assume $R$ is radical, this is not true.

I am really confused with what he wrote. I have some really elementary questions to ask:

1) Why $R$ is radical matters at here?

2) Why $MaxSpec(R)$ is the union of irreducible components $X_{i}$? If I am not mistaken I think he means the maximum spectrum at here. I understand that points in $X$ corresponds to elements in $MaxSpec(R)$, however why each irreducible component corresponds to a point in $MaxSpec(R)$?

3) How should I understand "..in fact are rings of meromorphic functions on $X_{i}$?" I feel very confused.

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What is Scholism?? –  Georges Elencwajg Jan 24 '13 at 9:25
What is a "radical" ring? –  user26857 Jan 24 '13 at 14:47
@Georges: From Akhil Mathew's notes (21.11), p. 92, it looks like a scholium met a porism. I suspect they indulged in holism and, unexpectedly, there resulted a scholism. We should proactively engage in action against such abus de langage... –  Martin Jan 24 '13 at 21:27
@YACP: "radical ring" seems to be a typo for reduced ring. –  Georges Elencwajg Jan 24 '13 at 22:05
@ Martin ${}$ :-) –  Georges Elencwajg Jan 24 '13 at 22:06