If a commutative ring is connected, then its reduced ring is connected.

Question

Let $R$ be a commutative ring with unit and $N$ its nilradical. If $R$ is connected then $R/N$ is also connected; the quotient map induces a homeomorphism between the spectra. I'd like to see a more hands-on proof, but I'm unable to make it work. That is to say, I'd like to show that if $e^2-e\in N$ for some $e\in R$, then $e\in N$ or $e-1\in N$. After scribbling full quite a couple of pages going around in circles, any idea is welcome.

Answer 1

See http://www.math.hawaii.edu/~lee/algebra/idempotent.pdf, but I'll repeat the argument here.

Suppose $e-e^2\in N$ and set $f=1-e$. Then $e^kf^k=0$, for some $k\ge1$.

Note that $e^k+N=e+N$ and $f^k+N=f+N$, so we can replace $e$ and $f$ with $e^k$ and $f^k$, so now $ef=0$.

Set $x=1-e-f\in N$; then $x^l=0$, for some $l\ge1$. Hence $1-x$ as the inverse $u=1+x+x^2+\dots+x^{l-1}$. Also $u-1\in N$ and $$ ue+N=e+N,\qquad uf+N=f+N $$ so we can replace $e$ and $f$ by $ue$ and $uf$.

Now $e+f=1$. So $$ e=e(e+f)=e^2+ef=e^2 $$

Answer 2

It is well known that every idempotent modulo $N$ can be lifted to an idempotent of $R$. Thus, if $R$ has no non trivial idempotent, then $R/N$ also has no non trivial idempotent. (See Anderson's "Rings and Categories of modules")