User:Kcbetancourt/AlgebraHW6

Let $ R\ $ be a ring with identity $ 1\ne 0 $.

7.6.1

An element $ e\in R\ $ is called an idempotent if $ e^2 = e $. Assume $ e $ is an idempotent in $ R $ and $ er=re $, $ \forall r\in R $. Prove that $ Re $ and $ R(1-e) $ are two-sided ideals of $ R $ and that $ R\cong Re\times R(1-e) $. Show that $ e $ and $ 1-e $ are identities for the subrings $ Re $ and $ R(1-e) $ respectively.

Note that $ (1-e)^2 = (1-e)(1-e) = 1^2-2e+e^2 = 1-2e+e = 1-e $. Therefore, $ 1-e $ is also idempotent. Also, for any $ r\in R $ , $ r(1-e) = r1 - re = 1r - er = (1-e)r $.

To prove that the subset $ Re $ is an ideal of $ R $ it suffices to show that $ Re $ is nonempty, closed under subtraction, and closed under multiplication by all elements of $ R $. Since $ e\in Re $, $ Re $ is nonempty.

Let $ a,b\in Re $ such that $ a = r_1e $ and $ b = r_2e $ where $ r_1, r_2 \in R $. Then $ a - b = r_1e - r_2e = (r_1 - r_2)e $. Since $ (r_1 - r_2) \in R $ , $ (r_1 - r_2)e \in Re $. Therefore, the subset $ Re $ is closed under subtraction.

Now consider $ a\in Re $ as defined above, and any element $ r_0\in R $. Then $ ar_0 = r_1er_0 = r_1r_0e $. Since $ r_1r_0\in R $ , $ r_1r_0e\in Re $. Therefore, $ ar_0\in Re $. Similarly, $ r_0a = r_0r_1e $ and since $ r_0r_1\in R $ , $ r_0r_1e = r_0a\in Re $. Therefore $ Re $ preserves product absorption and thus is an ideal of $ R\ $.

Now consider the subset $ R(1-e) $. This is nonempty because $ 1-e\in R(1-e) $. Let $ a,b\in R(1-e) $ such that $ a = r_1(1-e) $ and $ b = r_2(1-e) $ where $ r_1, r_2 \in R $. Then $ a - b = r_1(1-e) - r_2(1-e) = (r_1 - r_2)(1-e)\in R(1-e) $. Thus $ R(1-e) $ is closed under subtraction. Next, consider $ ar_0 = r_1(1-e)r_0 = r_1r_0(1-e)\in R(1-e) $ , and $ r_0a = r_0r_1(1-e)\in R(1-e) $. Therefore $ R(1-e) $ preserves product absorption and thus is an ideal of $ R\ $. More specifically, $ Re $ and $ R(1-e) $ are two-sided ideals of $ R $.

Let $ \phi : R\to Re\times R(1-e)$ such that $ \phi (r) = (re, r(1-e)) \forall r\in R $. Let $ r_1, r_2\in R $. Then $ \phi (r_1r_2) = (r_1r_2e, r_1r_2(1-e)) = (r_1r_2e^2, r_1r_2(1-e)^2) = (r_1e, r_1(1-e))(r_2e, r_2(1-e)) = \phi (r_1)\phi (r_2) $. Thus $ \phi $ is a homomorphism.

Let $ \phi(r_1) = \phi(r_2) \implies (r_1e, r_1(1-e)) = (r_2e, r_2(1-e)) \implies r_1e = r_2e $ and $ r_1(1-e) = r_2(1-e) \implies r_1 = r_2 $. Therefore, $ \phi $ is injective.

Let $ (a,b)\in Re\times R(1-e) $. Then there exists an element $ r\in R $ such that $ re = a $ and $ r(1-e) = b $ and therefore, $ \phi (r) = (a,b) $. Thus, $ \phi $ is bijective and $ R\cong Re\times R(1-e) $.

Let $ a\in Re $. We know $ a = r_0e $ for some $ r_0\in R $ , $ ae = ea $ and $ e^2 = e $. Then, $ ae = r_0ee = r_0e = a $. Therefore, $ e $ is the identity for $ Re $.

Let $ b\in R(1-e) $ such that $ b = r_0(1-e) $ for any $ r_0\in R $. Then $ b(1-e) = r_0(1-e)(1-e) = r_0(1-e) = b $. Therefore, $ 1-e $ is the identity for $ R(1-e) $.