The spectrum of a direct sum of rings

Posted on May 31, 2011 by Bruno Joyal | Leave a comment

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

Posted on May 27, 2011 by Bruno Joyal | 2 Comments

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

Posted on May 23, 2011 by Bruno Joyal | Leave a comment

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 $F_1 \cap F_2$ .