Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.8
Solution:
(1) Note that $(1,1) \in D$. However, $(1,0)(1,1) = (1,0) \notin D$. Since $D$ does not absorb $R$, it is not an ideal.
(2) We claim that S is an ideal. To that end, we need to show that $S$ is a subring and absorbs $R$. (Absorption on one side is sufficient since $R$ is commutative.)
Let $(2a_1,2b_1)$, $(2a_2,2b_2) \in S$ and $(x,y) \in R$.
i) $(0,0) \in S$, so that $S$ is not empty.
ii) $(2a_1,2b_1) – (2a_2,2b_2) = (2(a_1-a_2), 2(b_1-b_2)) \in S$, so that $S$ is an additive subgroup.
iii) $(2a_1,2b_1)(2a_2,2b_2) = (4a_1a_2,4b_1b_2) \in S$, so that $S$ is a subring.
iv) $(2a_1,2b_1) (x,y) = (2a_1x,2b_1y) \in S$, so that $S$ absorbs $R$.
(3) We claim that $T$ is an ideal. To that end, we need to show that $T$ is a subring and absorbs $R$. (Absorption on one side is sufficient since $R$ is commutative.)
Let $(2a_1,0), (2a_2,0) \in T$ and $(x,y) \in R$.
i) $(0,0) \in T$, so that $T$ is not empty.
ii) $(2a_1,0) – (2a_2,0) = (2(a_1-a_2), 0) \in T$, so that $T$ is an additive subgroup.
iii) $(2a_1,0)(2a_2,0) = (4a_1a_2,0) \in T$, so that $T$ is a subring.
iv) $(2a_1,0)(x,y) = (2a_1x,0) \in T$, so that $T$ absorbs $R$.
(4) Note that $(1,-1) \in U$. However, $(1,-1)(1,-1) = (1,1) \notin U$. Since $U$ is not closed under multiplication, it is not even a subring.
