Idempotents in coordinate ring Let $k$ be an infinite field. Let $V$ be an algebraic set of $k^n$.
Let's suppose that $f(x)+I(V)\in k[x_1,\cdots,x_n]/I(V)$ $(x=(x_1,\cdots,x_n))$ is such that $f(x)^2+I(V)=f(x)+I(V)$. Then does $f(x)+I(V)$ have to be $0+I(V)$ or $1+I(V)$? And what if $k$ is algebraically closed?
 A: No. Consider the simple case of $V\subset \mathbb{A}^1_k$ being two points, say given by the ideal $I(V)=(x^2-x)$. Then $k[x]/I(V)$ admits non-trivial idempotents, $x+I(V)$ and $1-x+I(V)$. Algebraically closed field does not resolve anything here...
In general a ring $R$ admits non-trivial idempotents if an only if $R\simeq R_1\times R_2$ is a product of two rings. In our example $k[x]/(x^2-x)\simeq k\times k$. Observe that in this case our algebraic set $V$ is disconnected. This is a general feature, whenever the coordinate ring of an algebraic set $V$ is disconnected, then the coordinate ring of $V$ admits non-trivial idempotents. Also conversely if an algebraic set is connected its coordinate ring does not admit non-trivial idempotents.
In case you are dealing with an irreducible algebraic set (affine variety), then this does not happen because any irreducible topological space is connected. or equivalently since $I(V)$ is prime, the coordinate ring $k[V]$ is an integral domain, hence cannot be isomorphic to a product ring $R_1\times R_2$ (product rings are never integral domains).
