Exercise 2.6 (General form of the Chinese remainder theorem) from Eisenbud's Commutative Algebra

Question

Exercise 2.6 (General form of the Chinese remainder theorem): Let $R$ be a ring, and let $Q_1,\dots,Q_n$ be ideals of $R$ such that $Q_i+Q_j = R$ for all $i\neq j$. Show that $R/(\bigcap_i Q_i) \cong \prod_{i} R/Q_i$.

1. Consider the map of rings $\varphi:R\to\prod_i R/Q_i$ obtained from the $n$ projection maps $R\to R/Q_i$. Show that $\ker\varphi = \bigcap_iQ_i$.

2. Let $\mathfrak{m}$ be a maximal ideal of $R$. Show that the hypothesis that $Q_i +Q_j=R$ for all $i\neq j$ means that at most one of the $Q_i$ is contained in $\mathfrak{m}$. Now use Corollary 2.9 to show that $\varphi$ is surjective.

Corollary 2.9 is stated in a previous post that I made.

I'm still learning about localization. The hardest part about this proof for me was considering localization of some object as both the localization of a ring and the localization of a module. I understand the main ideas of the proof, but if someone can check for me that everything I'm saying is technically correct, I would appreciate it. In other words, I'm at the stage where I understand the ideas of localization, but I'm still sometimes getting lost in all of the notation. Thanks!

Part (1) is obvious. For part (2), it is clear that any maximal ideal $\mathfrak{m}$ contains at most one of the ideals $Q_i$. I will show that localizing $\varphi$ at each maximal ideal is surjective, whence it follows by Corollary 2.9 that $\varphi$ is surjective.

Let $\mathfrak{m}$ be a maximal ideal of $R$. First, we treat $R$ as an $R$-module, and $\prod_i R/Q_i$ as an $R$-module by writing it as a direct sum $\bigoplus_i R/Q_i$ with the action of $R$ induced by the map $\varphi$; in other words, the action is given by: $$ r\cdot ([r_1],\dots,[r_n]) = \varphi(r)([r_1],\dots,[r_n]) = ([r],\dots,[r])([r_1],\dots,[r_n]) = ([rr_1],\dots,[rr_n]) $$ However, this is the standard direct sum $R$-module structure on $\bigoplus_iR/Q_i$ induced by the usual $R$-module structure on each of the summand $R$-modules $R/Q_i$ with action $r\cdot[r']=[rr']$. So localizing $\varphi$ at $\mathfrak{m}$ we obtain: $$ \varphi_m:R_\mathfrak{m}\to \left(\bigoplus_i R/Q_i\right)_{\mathfrak{m}} \cong \bigoplus_i (R/Q_i)_\mathfrak{m} $$ Here, in localizing $(R/Q_i)_\mathfrak{m}$ we are treating $R/Q_i$ as an $R$-module. However, this is the same as the $(R/Q_i)$-module $(R/Q_i)_{\overline{\mathfrak{m}}}$ where $\overline{\mathfrak{m}}$ is the image of $\mathfrak{m}$ under the quotient $R\to R/Q_i$. It follows that $(R/Q_i)_{\overline{\mathfrak{m}}}$ has a ring structure.

Since localization commutes with quotients, we have that $(R/Q_i)_{\overline{\mathfrak{m}}} \cong R_\mathfrak{m}/{(Q_i)}_\mathfrak{m}$. Since all except for perhaps one of the $Q_i$ are not contained in $\mathfrak{m}$ we have that $(Q_i)_{\mathfrak{m}} = R_\mathfrak{m}$ for all $Q_i$ which are not contained in $\mathfrak{m}$ and hence the summand is zero. So at most one of the $Q_j\subset \mathfrak{m}$ in which case $\varphi_\mathfrak{m}: R_\mathfrak{m}\to \left(\bigoplus_i R/Q_i\right)_\mathfrak{m} \cong \bigoplus_i \left(R/Q_i\right)_\mathfrak{m} = (R/Q_j)_\mathfrak{m}$ which is clearly surjective. So we are done by Corollary 2.9.

Answer 1

There's actually a more constructive way to show part $\phi$ is surjective (not requiring the existence of a maximal ideal, aka not requiring the axiom of choice).

Note that every element $(x_1, x_2, ..., x_n) \in \prod_i R / Q_i$ can be written as an $R$-linear combinination of basic elements $e_i$, where $e_i$ has a 1 in the $i$th position and a 0 everywhere else. So it suffices to show that each $e_i$ is in the range of $\phi$.

Indeed, consider that for all $j \neq i$, we have $1 \in R = Q_i + Q_j$. That is, we can write $1 = w_j + k_j$ for some $w_j \in Q_i$ and $k_j \in Q_j$. Then consider $x := \prod\limits_{j \neq i} (1 - w_j)$. We see that this element is in each $Q_j$ for $j \neq i$ because $1 - w_j \in Q_j$. And we see that, reducing this value modulo $Q_i$, we get $0$, since $\pi_i(x) = \pi_i(\prod\limits_{j \neq i} (1 - w_j)) = \prod\limits_{j \neq i} \pi_i (1 - w_j) = \prod\limits_{j \neq i} (1 - \pi_i(w_j)) = \prod\limits_{j \neq i} 1 = 1$. So $\phi(x) = e_i$. Here, $\pi_i : R \to R / Q_i$ is the canonical quotient map. Therefore, we see that $\phi$ is surjective.

Your proof works too, of course.