A boolean ring which is local must be isomorphic to $\Bbb{F}_2$.

Question

Recall: a Boolean ring is a (commutative) ring $R$ where $\forall x \in R: x^2=x$.

I don't really know how to proceed. I have tried some things like

• If $x,y \ne 0$ in $R$ such that $x^2=x$ and $y^2=y$, then by the fact that $R$ is local, we must have $x=y$, meaning $R$ would be a ring with 2 elements, or $\Bbb{F}_2$.
• If $x \ne 1$ with $x^2=x$ then by some steps we should conclude $x=1$, by which we conclude the only non-zero element in $R$ is 1, so $R = \Bbb{F}_2$.

I haven't managed to actually prove either of those. Could someone help me out?

Answer 1

Let $R$ be a local ring and $M=R\setminus R^{*}$ its maximal ideal.

• If $R$ is a boolean (local) ring, then it is a field
  It is sufficient to show $M=0$.
  Let $x\in M$. Note that this implies $x-1\notin M$, otherwise you get $1\in M$, which is clearly absurd. Thus you have that $x-1\in R^{*}$.
  But $x^2=x$ implies $x(x-1)=0$. Since $x-1$ is a unit, it can't be a zero divisor, so necessarily $x=0$. This shows that any boolean local ring is a field.
• Any boolean domain is isomorphic to $\mathbb{F}_2$
  Let $R$ be a boolean integral domain. We know that $R\neq 0$.
  Let $0\neq x\in R$. We want to show that $x=1$.
  As above we have $x(x-1)=0$ and, since $x\neq 0$ and $R$ is a domain, we get $x-1=0$, which proves the assertion.

Answer 2

In a local ring the only idempotents are $0$ and $1$. (This is easy: $eR+(1-e)R=R$ but $eR$ and $(1-e)R$ can't be proper at the same time.)

Everything in a boolean ring $R$ is idempotent.

So for every $x\in R$, $x=0$ or $x=1$.

That's $F_2$.