 Solution

ID: past-exam-of-the-algebra-course-of-the-university-of-oxford/2011/1/a/solution

Z (A) = {z \in A : z a = a z for every a \in A} .

(z_{1}, z_{2})

, commuting with every

(a_{1}, a_{2})

z_{1} a_{1} = a_{1} z_{1}

and

z_{2} a_{2} = a_{2} z_{2}

. Varying the two coordinates independently proves both inclusions, hence

Z (A_{1} \times A_{2}) = Z (A_{1}) \times Z (A_{2}) .

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