Sufficient condition for a ring to be a product of two rings

Question

In his algebraic geometry notes, Vakil suggests the exercise (remark 3.6.3) of showing that a ring $A$ is a product $A = A_1 \times A_2$ iff $\operatorname{Spec} A$ is disconnected.

His hint is to show that both conditions are equivalent to the existence of two elements $a_1, a_2 \in A$ s.t. $a_1^2 = a_1$, $a_2^2 = a_2$ and $(a_1 + a_2) = 1$, and hence $a_1a_2 = 1$.

2 questions: 1) Does it really follow that $a_1a_2 = 0$ from the other assumptions on $a_1$ and $a_2$ I tried deducing it but the most that I succeeded in showing was:

$$1=(a_1+a_2)^2 = a_1^2 + a_2^2 + 2a_1a_2 = a_1 + a_2 +2a_1a_2$$ $$0=2(a_1a_2)$$

However, I fail to see why this implies that $a_1a_2 = 0$, since he didn't mention any assumption of $2$ being regular.

2) How does one get that $\operatorname{Spec} A$ disconnected implies the existence of such $a_1,a_2$?

I reasoned the following:

taking $U = V(a_1), W = V(a_2)$ to be clopen sets such that $U\cap W = \varnothing$ and $U\cup W = \operatorname{Spec} A$ we get: $$\cup_{i\in I} D(f_i) = V(a_2)$$ $$\cup_{j\in J} D(f_j) = V(a_1)$$

where the left hand side is the expression of an open set as a union of basis elements and the right hand side is the definition of a closed set. From the fact that $V(a_1) \cup V(a_2) = V(a_1a_2) = \operatorname{Spec} A$ we get that $a_1a_2$ is nilpotent, but I can't seem to get anything else.

Answer 1

For 2), assume that $X=\operatorname {Spec}(A)=U\sqcup W$ (disjoint union) with $U,W$ open , but not assumed affine and even less assumed of the form $V(a)$.

Then define $a_1\in \Gamma(X,\mathcal O)=A$ by the conditions $$a_1\mid U=1\in \Gamma(U,\mathcal O),\quad a_1\mid W=0\in \Gamma(W,\mathcal O)$$ (The fact that $\mathcal O$ is a sheaf allows you to define $a_1$ in such a local way, because there is no compatibility condition on $U\cap W=\emptyset$.)
If you define similarly $a_2\in \Gamma(X,\mathcal O)=A$ by the conditions$$a_2\mid U=0\in \Gamma(U,\mathcal O),\quad a_2\mid W=1\in \Gamma(W,\mathcal O)$$ you get your required idempotents $a_1,a_2$.

Note that it turns out now that $U=D(a_1), W=D(a_2)$ are principal affine open subsets of $X$ after all, but you were not allowed to assume that at the beginning.

Answer 2

$\left(a_1 \cdot a_2\right) \;\; = \;\; \left(a_1 \cdot a_2\right)+0 \;\; = \;\; \left(a_1 \cdot a_2\right)+a_1+\left(-\hspace{.03 in}a_1\right) \;\; = \;\; a_1 + \left(a_1 \cdot a_2\right)+\left(-\hspace{.03 in}a_1\right)$

$= \;\; a_1^2 + \left(a_1 \cdot a_2\right)+\left(-\hspace{.03 in}a_1\right) \;\; = \;\; \left(a_1 \cdot a_1\right)+\left(a_1 \cdot a_2\right)+\left(-\hspace{.03 in}a_1\right) \;\; =$

$\left(a_1 \cdot \left(a_1+a_2\right)\right)+\left(-\hspace{.03 in}a_1\right) \;\; = \;\; \left(a_1 \cdot 1\right)+\left(-\hspace{.03 in}a_1\right) \;\; = \;\; a_1+\left(-\hspace{.03 in}a_1\right) \;\; = \;\; 0$

Answer 3

There is already a proof of #2 submitted, but here is an uglier one. I believe it is the one I came up with for an actual homework assignment years ago. Seeing the sheaf-theoretic answer later on was what made me decide to study algebraic geometry.

Writing $\operatorname{Spec} A = U \cup W$. Since each of $U$ and $W$ are closed, they are compact. Since they are open, each is a finite union $U = \cup_i D(f_i)$, $W=\cup_j D(g_j)$.

I claim that $U = V(J)$ and $W = V(I)$, where $I = (\{f_i\})$ and $J=(\{g_j\})$. To see this, note that $\mathcal{P} \in U$ if and only if $\mathcal{P} \notin D(g_j)$ for all $j$, i.e. $g_j\in\mathcal{P}$ for all $j$. In other words, $J\subset \mathcal{P}$. For $I$, the proof is identical.

Note that $I+J = A$, as every prime contains either $I$ or $J$. Furthermore, every prime contains $IJ$, so every element of $IJ$ is nilpotent. In fact, $IJ$ itself is nilpotent, since $I$ and $J$ are finitely generated.

For any positive integer $n$, we have $I^n + J^n = A$, because if $i+j = 1$ with $i\in I$, $j\in J$, then every term of $(i+j)^{2n}$ lies in $I^n$ or $J^n$. By choosing $n$ so that $(IJ)^n = 0$, we may as well assume that $IJ=0$.

Finally, pick $i\in I$, $j\in J$ as above, with $i+j = 1$. Then $ij=0$ and $i$ and $j$ are our desired idempotents.