 Solution

ID: past-exam-of-the-algebra-2-course-of-the-university-of-oxford/2015/1/b/solution

For two factors, the source's

R_{1} \oplus R_{2}

is the direct product of rings with componentwise operations. Its principal ideals satisfy

((a_{1}, a_{2})) = (a_{1}) \times (a_{2}),

One inclusion follows by projection to the factors. For the reverse inclusion, multiplication by

(1, 0)

and

(0, 1)

puts

(a_{1}, 0)

and

(0, a_{2})

in the generated ideal; taking ideal combinations separately produces every pair in

(a_{1}) \times (a_{2})

Suppose

(a_{1}, a_{2})

and

(b_{1}, b_{2})

generate the same principal ideal. Projecting to the factors gives

(a_{i}) = (b_{i})

for

i = 1, 2

. Since each factor is an associator ring, there are units

u_{i} \in R_{i}

with

a_{i} = u_{i} b_{i}

. The pair

(u_{1}, u_{2})

(u_{1}^{- 1}, u_{2}^{- 1})

, and

(a_{1}, a_{2}) = (u_{1}, u_{2}) (b_{1}, b_{2}) .

Hence

R_{1} \oplus R_{2} is an associator ring .

Repeating this argument proves the same assertion for every finite direct product of rings.

 New to topics? Read the docs here!