Yet another characterization of the field $\mathbb{Z}/2\mathbb{Z}$ Let $R \neq 0$ be a ring which may not be commutative and may not have an identity.
Suppose $R$ satisfies the following conditions.
1) $a^2 = a$ for every element $a$ of $R$.
2) $R$ has no two-sided ideals other than $0$ and $R$. 
Is $R$ isomorphic to the field $\mathbb{Z}/2\mathbb{Z}$?
 A: *

*$a+a=0$ for all $a\in R$, since $a+a=(a+a)^2=a+a+a+a$;

*$ab=ba$ for all $a,b\in R$, since $a+b=(a+b)^2=a+ab+ba+b$ so $ab=-ba=ba$;

*$a=b$ for $a,b\in R\setminus\{0\}$. Namely $b\in aR$ and $b\in aR$ because $\{0\}$ and $R$ are the only (two-sided) ideals, whence there exist $x,y$ with $ax=b$ and $a=yb$; then $a=yb=yb^2=ab=a^2x=ax=b$.


Since $R\neq\{0\}$ there is exactly one nonzero element of $R$.
A: I believe so.  By the first condition, the ring is commutative.  By the second, it is simple Artinian, and therefore semisimple.  Then, it must be a product of matrix rings over division rings.
Commutativity then shows that all matrix groups have dimension $1$, and that all division rings are fields.  Finally, with no two-sided ideals, we can't have the product of two or more fields, so we are left with just a field.
Finally, the polynomial $x^2-x$ can only have two solutions over this field, so the field must have only two elements.
Edit: As Matt E has pointed out, I've assumed that the ring is unital.  Finite boolean rings are always unital, so if a counterexample exists, it will be infinite.
Again, in the comments, Matt E has suggested we adjoin an identity element to the ring by considering $R'=\mathbb{F}_2\oplus R$ and defining multiplication by $(a,b)\cdot(c,d)=(ab,ad+bc+bd)$.  This makes $(1,0)$ the identity element.  Now, using a modified version of the argument above, we can show $R'$ is a direct sum of two fields, so that $R$ is a field.
A: Yes. In $R$ 
$$
(1+a)^2=1+a
$$
implies $a=-a$. Further, 
$$
(a+b)^2=a+b
$$
implies $ab=-ba=ba$, so $R$ is commutative. Since each ideal $aS$ coincides with $R$ then for every $b$ the equation $ax=b$ is soluble, so $R$ is a field.
Correction: If $R$ has not $1$, the commutativity was proved in the comment of Jared (thank to Matt E and  Jonas Meyer for the remarks).
A: $R$ is commutative as shown in the Marc van Leeuwen's answer.
Let $a, b$ be non-zero elements of $R$.
It suffices to prove $ab \neq 0$ thanks to this result.
Since $a = a^2 \in aR$, $aR \neq 0$.
Hence the ideal $I = \{x \in R |\ ax = 0\}$ cannot be $R$.
Hence $I = 0$ as desired.
