$A\neq \{0\}$ is a unit commutative ring. If $a\in A-\{0\}$ is such that for all $b\in A$, $a\ast b=0$ or $a\ast b=a$. Then $(1+a)$ is prime ideal

Question

I have this problem for homework

Let $(A, +, \ast)$ be a ring with unit such that $A \neq \{0\}$. Suppose that each element of A is idempotent. Also, suppose that there exists an element $a \in A-\{0\}$ such that for all $b \in A$ we have $a \ast b = 0$ or $a \ast b = a$. Prove that $(1+a)$, i.e. the ideal generated by $1 + a$, is a prime ideal of A.

I tried to use this proposition:

Let $(A, +, \ast)$ be a commutative ring with unit and $A \neq \{0\}$. Then an ideal P is prime if and only if $A/P$ is an itegral domain.

I proved that $(A, +, \ast)$ is commutative. Let $a,b \in A$ be arbitrary, then since every element of A is idempotent, we have $[a+1] = [a+1]^2 = a^2 + 1^2 + 2a = a + 1 + 2a$, whence $2a = 0$. Also $[a+b] = [a+b]^2 = a^2 + b \ast a + a \ast b + b^2 = a + b \ast a + a \ast b + b$, whence $-b \ast a = a \ast b$ and finally, adding $a \ast b$ on both sides, \begin{align*} a \ast b - b \ast a &= 2[a \ast b] \\ a \ast b - b \ast a &= 0 \\ a \ast b &= b \ast a. \end{align*} Thus, A is commutative and we can apply the stated proposition. So I have to prove that $A/(1+a)$ is and integral domain. To do this I have to prove that $A/(1+a) \neq \{0 + (1+a)\}$ and that it doesn't have cero divisors. The first one is easy, but I have tried the second one and I couldn't prove it. Please, any suggestion to finish this prove or another way to prove the problem is welcome. Thanks in advance.

Answer 1

Your proof of commutativity looks good to me, so let's show that $P:=(1+a)$ is prime. Since maximal ideals are always prime, it suffices to show that $P$ is maximal, so let $c\in A\setminus P$; we wish to show $1\in P+(c)$. By hypothesis, we know that either $ac=0$ or $ac=a$. In the first case, we have $c=(1+a)c$, meaning $c\in P$, a contradiction, and thus we must have $ac=a$. On the other hand, as you have shown, $A$ has characteristic $2$, so $a+a=c+c=0$. In particular, we have $$1=1+a+c+a+c=(1+a)(1+c)+c\in P+(c),$$ as desired. Thus $P$ is maximal and is hence in particular a prime ideal. (As a further exercise, try proving that every prime ideal in $A$ is in fact maximal.)

Answer 2

Let $\mathfrak p := (1+a)$, and take $x \in A$.

• If $a*x=0$, then $x=(1+a)*x \in \mathfrak p$.
• If $a*x=a$, then $x+1 = (1+a)*(x+1+a) \in \mathfrak p$.

Thus we have $x+\mathfrak p=\mathfrak p$ or $x+\mathfrak p=1+\mathfrak p$, and since $x$ was arbitrary, $A/\mathfrak p = \{\mathfrak p,1+\mathfrak p\} \cong \mathbb Z/2\mathbb Z$, meaning that $A/\mathfrak p$ is an integral domain (in fact, a field).