Talk:Algebraic Geometry Study Group

Here is a solution to the difficult part of A&M Chapter 1, Exercise 22. I used a hint from Eisenbud's book to get it.

Suppose that $$ Spec(A) $$ is disconnected. Then, one can find two ideals $$ F,G $$ such that $$ A $$ surjects onto $$ A/F \times A / G $$, and $$ F + G = (1) $$, and $$ F \cap G $$ consists of nilpotent elements.

Let $$ f \in F, g \in G $$ be elements of $$ F,G $$ respectively, such that $$ f + g = 1 $$. Let $$ n $$ be a positive integer, satisfying $$ (fg)^n $$ = 0 (observing that the product $$ fg $$ is an element of $$ F \cap G $$).

Then, we find that (for constants $$ a,b $$ depending on $$ n $$):

$$ 1 = (f+g)^{2n} = f^{2n} + \cdots + a f^{n+1} g^{n-1} + b f^{n-1} g^{n+1} + \cdots + g^{2n}. $$

Define

$$ \tilde f = f^{2n} + \cdots + a f^{n+1} g^{n-1} $$, and

$$ \tilde g = b f^{n-1} g^{n+1} + \cdots + g^{2n} $$.

Then, it is not difficult to show that $$ \tilde f + \tilde g = 1 $$ and $$ \tilde f \tilde g = 0 $$. It follows that these are nontrivial idempotents.