The spectrum of a direct sum of rings

The following exercise is taken from Atiyah-Macdonald, chapter 1.

Let A be a commutative ring. Show that the following statements are equivalent:

(i) X=\mbox{Spec }A is disconnected;

(ii) A \cong A_1 \oplus A_2, where neither A_1 or A_2 is the zero ring;

(iii) A contains an idempotent \neq 0, 1.

Proof : first we do the easiest parts. It’s clear that (ii) \Rightarrow (iii); for example, the elements (1, 0) and (0,1) are such idempotents. To show that (iii) \Rightarrow (ii), let r \in A be an idempotent which is not 0 or 1. Let r'=1-r. Then r'^2 = 1-2r+r^2 = 1-2r+r = 1-r, so r' is also an idempotent which is not 0 or 1. Now since 1=r+r', we have a=ra+r'a for every a \in A; so A=(r)+(r'). Since rr'=r(1-r)=r-r^2=r-r=0, the sum is actually direct (as a sum of A-modules); indeed, if ra+r'b=0, then, multiplying by r we have r^2a=ra=0. Similarily r'b=0. So the sum is direct. Give (r) an internal ring structure by defining its unit to be r; this works, since r(ra)=ra for every ra \in (r). Then A \cong (r) \oplus (r') as a direct sum of rings.

Now clearly (ii) \Rightarrow (i); indeed, every prime ideal of A contains one of (0,1) or (1,0), but none contains both.

Now we show that (i) \Rightarrow (iii). Let V(R_1) and V(R_2) be closed sets in X such that X = V(R_1) \sqcup V(R_2) = V(R_1 \cap R_2). Suppose without loss of generality that R_1 and R_2 are ideals. Since the union is disjoint, we must have R_1+R_2=A, so there exists r_1 \in R_1 and r_2 \in R_2 with r_1+r_2=1. We must show that we can take r_1 and r_2 so that r_1r_2=0; this will cause the sum R_1+R_2=A to be direct. Since V(R_1 \cap R_2) = X, and since r_1r_2 \in R_1\cap R_2, we have V(r_1r_2) \supset V(R_1R_2)=X. Thus every prime ideal in A contains r_1r_2. Thus r_1r_2 is contained in the nilradical of A, so it is nilpotent.
Let n be such that r_1^nr_2^n=0. Then, by the binomial theorem, r_1^n+r_2^n = 1 +r_1r_2s for some s \in A. Since r_1r_2 is nilpotent, so is r_1r_2s; hence r_1^n+r_2^n is a unit, say \alpha(r_1^n+r_2^n)=1. Let r_1'=\alpha r_1^n and r_2'=\alpha r_2^n. Then r_1'+r_2'=1 and r_1'r_2'=0, as promised. This shows that A=r_1' A \oplus r_2' A as A-modules. We can, as before, define a ring structure on r_i'A be letting r_i' be the unit, since we have r_1'(r_1'+r_2')=r_1'=r_1'^2 since r_1r_2=0. Thus A \cong r_1' A \oplus r_2' A as rings.

A cute problem

I came up with this little problem last night. It’s not very difficult to prove but still fun (I think). Here it is : let A be a commutative ring, and let h(u,v) \in A[u,v]. Suppose that, for any polynomials f(u), g(u) \in A[u], we have h(f(u), g(u))=h(g(u), f(u)). Then h(u,v)=h(v,u).

I’ll post my solution in a couple of days to see if anyone can come up with an alternative solution in the meantime. :)

A nice problem in Galois theory

This problem was given to me by my research supervisor. This is the problem and my solution:

Let k be a field of characteristic zero, and k(x) the rational function field in one variable over k. Suppose F_1 and F_2 are subfields of k(x) such that [k(x):F_1] and [k(x):F_1] are finite. Is it possible for [k(x):F_1\cap F_2] to be infinite? Indeed, it is. Note that \mbox{Gal }(k(x)/k) = \mbox{PSL}_2(k), the projective special general linear group over k (we identify its elements with Möbius transformations). Let S, T denote the generators of the modular group (or rather, their image in \mbox{PSL}_2(k)). Note that they are of finite order. Hence, taking F_1=k(x)^{<S>} and F_2=k(x)^{<T>}, we have [k(x):F_i]<\infty for i=1,2. However, F_1 \cap F_2 is fixed by the whole modular group, which is infinite since k has characteristic zero. Hence k(x) cannot be of finite index over F_1 \cap F_2.

I don’t know whether such a construction is possible if k has prime characteristic. I’d be interested to know if you find out!