The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.14.4 (Chinese remainder). Let $R$ be a ring.

  1. If $I_1, \ldots , I_ r$ are ideals such that $I_ a + I_ b = R$ when $a \not= b$, then $I_1 \cap \ldots \cap I_ r = I_1I_2\ldots I_ r$ and $R/(I_1I_2\ldots I_ r) \cong R/I_1 \times \ldots \times R/I_ r$.

  2. If $\mathfrak m_1, \ldots , \mathfrak m_ r$ are pairwise distinct maximal ideals then $\mathfrak m_ a + \mathfrak m_ b = R$ for $a \not= b$ and the above applies.

Proof. Let us first prove $I_1 \cap \ldots \cap I_ r = I_1 \ldots I_ r$ as this will also imply the injectivity of the induced ring homomorphism $R/(I_1 \ldots I_ r) \rightarrow R/I_1 \times \ldots \times R/I_ r$. The inclusion $I_1 \cap \ldots \cap I_ r \supset I_1 \ldots I_ r$ is always fulfilled since ideals are closed under multiplication with arbitrary ring elements. To prove the other inclusion, we claim that the ideals

\[ I_1 \ldots \hat I_ i \ldots I_ r,\quad i = 1, \ldots , r \]

generate the ring $R$. We prove this by induction on $r$. It holds when $r = 2$. If $r > 2$, then we see that $R$ is the sum of the ideals $I_1 \ldots \hat I_ i \ldots I_{r - 1}$, $i = 1, \ldots , r - 1$. Hence $I_ r$ is the sum of the ideals $I_1 \ldots \hat I_ i \ldots I_ r$, $i = 1, \ldots , r - 1$. Applying the same argument with the reverse ordering on the ideals we see that $I_1$ is the sum of the ideals $I_1 \ldots \hat I_ i \ldots I_ r$, $i = 2, \ldots , r$. Since $R = I_1 + I_ r$ by assumption we see that $R$ is the sum of the ideals displayed above. Therefore we can find elements $a_ i \in I_1 \ldots \hat I_ i \ldots I_ r$ such that their sum is one. Multiplying this equation by an element of $I_1 \cap \ldots \cap I_ r$ gives the other inclusion. It remains to show that the canonical map $R/(I_1 \ldots I_ r) \rightarrow R/I_1 \times \ldots \times R/I_ r$ is surjective. For this, consider its action on the equation $1 = \sum _{i=1}^ r a_ i$ we derived above. On the one hand, a ring morphism sends 1 to 1 and on the other hand, the image of any $a_ i$ is zero in $R/I_ j$ for $j \neq i$. Therefore, the image of $a_ i$ in $R/I_ i$ is the identity. So given any element $(\bar{b_1}, \ldots , \bar{b_ r}) \in R/I_1 \times \ldots \times R/I_ r$, the element $\sum _{i=1}^ r a_ i \cdot b_ i$ is an inverse image in $R$.

To see (2), by the very definition of being distinct maximal ideals, we have $\mathfrak {m}_ a + \mathfrak {m}_ b = R$ for $a \neq b$ and so the above applies. $\square$


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