 Solution

ID: past-exam-of-the-rings-and-modules-course-of-the-university-of-oxford/2023/1/a/ii/solution

Write

T = \prod_{j = 1}^{n} R_{j}

, and let

ρ_{j} : T \to R_{j}

I_{j} = ρ_{j} (I)

. Each

I_{j}

is an ideal: it is an additive subgroup, and multiplying a lift in

I

on either side by a tuple with prescribed

j

th coordinate proves closure under left and right multiplication by

R_{j}

Certainly

I \subseteq \prod_{j} I_{j}

. Conversely, let

a = (a_{1}, \dots, a_{n}) \in \prod_{j} I_{j}

. For each

j

, choose

b^{(j)} \in I

with

j

th coordinate

a_{j}

, and let

e_{j}

have

1_{R_{j}}

in coordinate

j

and zero elsewhere. Since

I

is an ideal,

e_{j} b^{(j)} \in I

, and

a = \sum_{j = 1}^{n} e_{j} b^{(j)} \in I .

Therefore

I = I_{1} \times \dots \times I_{n} .

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