07-06. The chinise remainder theorem

Thorughout this section we assume

• All rings are commutative with indentity $1 \neq 0$,

Definition. comaximal

• $R$,
  • ring
• $A, B \subseteq R$,
  • ideals

$A$ and $B$ are said to be comaximal if $A + B = R$.

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Proposition.

• $R$,
  • ring
• $A, B \subseteq R$,
  • ideals

The followings are equivalent.

• (i)

\[\exists a \in A, \ b \in B, \ \text{ s.t. } \ a + b = 1 ,\]

• (ii) $A$ and $B$ are comaximal.

proof.

(i) $\Rightarrow$ (ii)

Let $c \in R$. By assumpation,

\[ca + cb = c .\]

Since $A$ and $B$ are ideals, $ca \in A$ and $cb \in B$.

(i) $\Leftarrow$ (ii) We know $1 \in R$.

$\Box$

Theorem 17. Chinise Remainder Theorem

• $R$,
• $A_{1}, \ldots, A_{k}$,
  • ideals
• $\phi: R \rightarrow R/A_{1} \times \cdots \times R/A_{k}$,

\[\phi(r) := (r + A_{1}, \ldots, r + A_{k}) .\]

$\phi$ is a ring homomorphisim with kernel $A_{1} \cap A_{2} \cap \cdots \cap A_{k}$. If \(\forall i, j \in \{1, \ldots, k\}\) with $i \neq j$ the ideals $A_{i}$ and $A_{j}$ are comaximal, then

• (i) $\phi$ is surjective,
• (ii)

\[A_{1} \cap \cdots A_{k} = A_{1} \cdots A_{k} .\]

That is,

\[R/(A_{1}\cdots A_{k}) \cong R/A_{1} \times R/A_{2} \times \cdots \times R/A_{k} .\]

proof.

We first show that

\[\mathrm{Ker}\phi = A_{1} \cap \cdots \cap A_{k} .\]

It’s easy to see $\mathrm{Ker}\phi_{i} = A_{i}$ for all \(i \in \{1, \ldots, k\}\). Since $\phi = (\phi_{1}, \ldots, \phi_{k})$, $\mathrm{Ker}\phi = A_{1} \cap \cdots \cap A_{k}$.

Now we show the statement for comaximal part by induction. In case of $k = 2$. Let $A = A_{1}$ and $B := A_{2}$. Suppose that $A + B = R$.

(i) There exists $x \in A$, $y \in B$ such that

\[x + y = 1.\]

This equation show that

\[\begin{eqnarray} \phi(x) & = & (x + A, x + B) & = & (0, 1) \nonumber \\ \phi(y) & = & (y + A, y + B) & = & (1, 0) \nonumber \end{eqnarray}\]

Then

\[\begin{eqnarray} \forall [r_{1}]_{A} \in R/A, \forall [r_{2}]_{B} \in R/B, \ \phi(r_{1}x + r_{2}y) & = & \phi(r_{1})\phi(x) + \phi(r_{2})\phi(y) \nonumber \\ & = & (r_{1} + A, r_{1} + B) (0, 1) + (r_{2} + A, r_{2} + B) (1, 0) \nonumber \\ & = & (0, r_{1} + B) (0, 1) + (r_{2} + A, 0) (1, 0) \nonumber \\ & = & (r_{2} + A, r_{1} + B) \nonumber \\ & = & ([r_{2}]_{A}, [r_{1}]_{B}) . \end{eqnarray}\]

Thus, $\phi$ is surjective.

(ii)

$A$, $B$ are ideals so that \(\{0\} \in A \cap B \neq \emptyset\). It always holds $AB \subseteq A \cap B$ by a porposition in Section 7.3. Since $A$ and $B$ are comaximal, for any $c \in A\ cap B$,

\[c = c1 = cx + cy = xc + cy \in AB .\]

Thus, $A \cap B = AB$.

Suppose that the statement holds up to $k = n-1$. We show that in case of $k = n$. Let $A := A_{1}$, $B := A_{2}, \ldots A_{k}$. We only need to show that $A$ and $B$ are comaximal. By asusmpation that $A_{1}, \ldots, A_{k}$ are pair-wise comaximal, there are elements $x_{i} \in A_{1}$, $y_{i} \in A_{i}$

\[\forall i \in \{2, \ldots, k\}, \ x_{i} + y_{i} = 1 .\]

Since \([x_{i}]_{A_{1}}\), \(x_{i} + y_{i} \equiv y_{i} (\mathrm{mod}\ A_{1})\). Hence

\[1 = (x_{2} + y_{2}) (x_{3} + y_{3}) \cdots (x_{k} + y_{k})\]

By expanding RHS, we see that RHS is an element of \(A_{1} + (A_{2} \cdots A_{k})\).

$\Box$

Reference