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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 $\mathbb{Z}/2\mathbb{Z}$?

egreg
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Makoto Kato
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4 Answers4

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  • $a+a=0$ 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 $R$.

Marc van Leeuwen
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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.

Jared
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  • I'm sure it is simple, but how does the first condition imply commutativity? – Daniel Fischer Jul 25 '13 at 22:20
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    @DanielFischer Expand $a^2+b^2=(a+b)=(a+b)^2$. – Karl Kroningfeld Jul 25 '13 at 22:21
  • The first condition makes the ring into a Boolean ring, and these are always commutative. To see this, consider the implications of the two equalities $a+a=(a+a)^2$ and $a+b=(a+b)^2$. – Jared Jul 25 '13 at 22:22
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    Dear Jared, At least at face value your argument seems to assume that $R$ admits an identity (since it uses concepts like Artinianness, Artin--Wedderburn theory, etc, which prime facie apply to rings with $1$). Regards, – Matt E Jul 25 '13 at 22:23
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    @user1 That gives me $ab + ba = 0$. Setting $b = a$ in that gives $a + a = 0$ (for all $a$), and therefore $ba = -ab = ab$. Yup, I was right, it is simple. Thanks for the pointer. – Daniel Fischer Jul 25 '13 at 22:25
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    Dear Jared, What if you add an identity by considering $\mathbb F_2 \oplus R$. Then you still get a Boolean and Artinian ring, and your argument probably goes through. What do you think? Regards, – Matt E Jul 25 '13 at 22:31
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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).

Boris Novikov
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$R$ 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.

Community
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Makoto Kato
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