 Solution

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

(r_{1}, \dots, r_{n})

has a two-sided inverse exactly when each

r_{j}

has a two-sided inverse in

R_{j}

, and then

(r_{1}, \dots, r_{n})^{- 1} = (r_{1}^{- 1}, \dots, r_{n}^{- 1}) .

U (R_{1} \times \dots \times R_{n}) = U (R_{1}) \times \dots \times U (R_{n}) .

The identification preserves multiplication because both sides use coordinatewise multiplication. This is the unit group of a direct product of rings formula.

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