User:Kcbetancourt/Algebra

7.4.37 A commutative ring $$ R\ $$ is called a local ring if it has a maximal ideal. Prove that if $$ R\ $$ is a local ring with maximal ideal $$ M\ $$, then every element of $$ R-M\ $$ is a unit.

Since $$M \ $$ is a maximal ideal, it is prime. Let $$x,y\in R-M $$. If we had $$xy\in M \ $$, then since $$M \ $$ is prime, we have $$x\in M \ $$ or $$y \in M \ $$. Hence, $$R-M \ $$ is closed under multiplication.

Let $$ x\in R-M $$. Consider the ideal $$ (x)\ $$ generated by $$ x\ $$. Then, since $$ M\ $$ is the maximal ideal, $$ (x)\subseteq M $$. But $$ x\notin M $$ so we must have $$ M\subseteq (x) $$. But again, since $$ M $$ is the maximal ideal, this means that $$ (x) = R $$. Since $$ 1 \in R, \exists y \in R : xy = 1 $$. This implies that $$ x\ $$ is a unit. Therefore, every element of $$ R-M\ $$ is a unit.

<= Let R be a commutative ring with 1 and the set of all non-units be M. We want to show that M is a Unique Maximal in R. Assume that M is not maximal. Then $$ \exists \ $$ an ideal, I, in R such that M $$ \subset \ $$ I $$ \subset \ $$ R. And M $$ \ne \ $$ I $$ \ne \ $$ R. We know that 1 does not exist in M, because 1 is a unit. So if M is not equal to I, then $$ \exists \ $$ u $$ \in \ $$ I, where u is a unit. Then 1 $$ \in \ $$ I. But that implies I=R. But if I=R, then I is not a maximal idea. If I is not maximal, then that implies that M is maximal. Now assume that M is not unique. Then $$ \exists \ $$ M' such that M' is a maximal ideal in R. Then M' $$ \subset \ $$ R and M $$ \ne \ $$ R. So 1 $$ \notin \ $$ M'. Thus M' is the set of all non-units. That implies that M'=M. Therefore M is a unique maximal ideal in R.

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.

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 $$ a\in Re $$ and any element $$ r_0\in R $$