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 $$ R(1-e) $$.