Question 1.
Find the subring of the ring Z×Z that is not an ideal of Z×Z.
Solution. Consider the subset
D={(n,n)∣n∈Z}⊂Z×Z.
It is a subring of Z×Z. Indeed, it is a subgroup of additive group of Z×Z, because
(n,n)−(m,m)=(n−m,n−m)∈D
for arbitrary (n,n),(m,m)∈I. Furthermore, D is closed under multiplication:
(n,n)⋅(m,m)=(nm,nm)∈D
for all (n,n),(m,m)∈I. Finally, D contains the identity (1,1) of Z×Z. The last fact also shows that D cannot be an ideal of Z×Z because D=Z×Z. This can be demonstrated by the following observation:
(1,1)⋅(1,0)=(1,0)∈D,
while (1,1)∈D.
Answer: for example, D={(n,n)∣n∈Z}⊂Z×Z. □
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