TITLE: If $x \in R$ is non-invertible implies $x^2 \in \{\pm x\}$ and $|R| >9$ odd then $R$ is a field
QUESTION [8 upvotes]: Let $(R, +, \cdot)$ be a commutative ring with $2n+1$ elements, for some $n\neq 4$ a positive integer. Suppose also that $R$ also satisfies the following condition: If an element $x\in R$ is non-invertible, then $x^2 \in \{\pm x\}$. Prove that $(R,+,\cdot)$ is a field.

Here it's my attempt: Suppose $x\in R$ is a non-invertible element, with $x\neq 0$. Then also, $2x$ is non-invertible, so we have that $x^2 \in \{\pm x\}$ and $4x^2 \in \{\pm 2x\}$. It follows that $3x=0$, so $3$ is non-invertible which means that $9\in \{\pm 3\}$ $\Rightarrow$ $3=0$. How can i continue? I am trying to prove that the ring should have $9$ elements in this case.

REPLY [3 votes]: There is also a way to do this using elementary methods, without using Artin-Wedderburn.
We denote $R^*$ the invertible [by multiplication] elements in $R$. We say that an element $a$ is invertible if precisely $a \in R^*$. We call an element $c$ in $R$ a scalar if $c$ can be written $c=1+1+ \ldots +1$ [i.e., $c$ is obtained if one adds up $1 \in R$ precisely $m$ times for some positive integer $m$]. Let $x_1,x_2,\ldots,x_r \in R$ we define the set Span$\{x_1,x_2,\ldots,x_r\}$ to be the set of elements $\alpha \in R^*$ of the form $\alpha = \sum_{i=1}^r c_ix_i$ where the $c_i$s are scalars in $R$. As noted in the OP:

Claim 0: Let $R$ be as per the conditions of this problem, and let us assume that $R$ is not a field. In $R$ the scalars 1 and 2 is invertible, but 3 is 0. So in other words the scalars are isomorphic to $\mathbb{F}_3$.

We now note Claim 1:

Claim 1: Let $b \in R$ but not be a scalar. Then one of $\{b,b+1,b+2\}$ is in $R^*$.

Claim 1 follows because one can use Claim 0 to show that $(b+i)^2$ cannot be in $\{(b+i),-(b+i)\}$ for all $i=0,1,2$, unless $b$ is itself a scalar. In particular, if $b^2=-b$ then $(b+2)^2=b^2+4b+4$ $= -b+4b+4 = 1$, and if $b^2=b$ then $(b+1)^2=$ $b^2+2b+1 = 1$[working $\pmod 3$]. $\surd$

Claim 2: Let $a$ is in $R^*$. Then $a^2=1$.

Proof of Claim 2: Indeed, let $x$ be any nonzero nonscalar element in $R \setminus R^*$. Let us assume that
$x^2+x=0$. Then $x(x+1) =0$ so $x+1$ is not invertible either, and thus, $ax$ and $a(x+1)$ are nonzero [as $a$ is invertible] and not invertible either. On the one hand $[a(x+1)]^3 = a^3[x+1]^3 = a^3[x+1]$ and on the other hand,
$[a(x+1)]^3 = a[x+1]$. [Here I am using the result in rschwieb's answer that says $y^3=y$ for all $y \in R \setminus R^*$. To see this directly, if the equation $y^2=y$ holds, then so must the equation $y^3=y^2=y$; if $y^2=-y$ holds then so must $y^3=-y^2=y$.]
Likewise $[ax]^3 = a^3x= ax$. So subtracting the equation $a^3x=ax$ from $a^3[x+1]=a[x+1]$ gives $a^3=a$. As $a$ is in $R^*$ one can multiply both sides of $a^{-1}$ to get Claim 2.
If in fact $x$ is a nonzero nonscalar noninvertible element satisfying $x^2-x=0$ instead, then $x(x-1)=0$ so both $x$, $x-1$ are noninvertible. The above line of reasoning still holds. Thus Claim 2 follows here as well. $\surd$

Claim 3: Let $x$ be a nonzero nonscalar element in $R \setminus R^*$. Then if $x^2+x=0$
then both $x$ and $x+1$ are not invertible and every element $b \in R$ is in
Span$\{x,x+1\}$. If $x^2-x=0$ then both $x$ and $x-1$ are not invertible and every element $b \in R$ is in Span$\{x,x-1\}$.

Proof of Claim 3: Let us assume for now that $x^2+x=0$. Then $x(x+1)=0$ and so $x+1$ is not invertible either. Furthermore, $(x+1)^2=x+1$. Also, let $b$ be an element not in Span$\{x,x+1\}$. Then $b+1$ and $b+2$ are not in Span$\{x,x+1\}$ as well. [Indeed, if say $b+1$ were in, then as $1$ is in Span$\{x,x+1\}$, so would $(b+1)+2=b$.] Let us use Claim 1 then and let $a \in \{b,b+1,b+2\}$ be an invertible element; then
$a$ is not in Span$\{x,x+1\}$. We show next that in fact $a$ must be in Span$\{x,x+1\}$. And then from this Claim 3 will follow.
Now on the one hand $[ax]^2 = a^2x^2 = -x$ [by Claim 2 $a^2=1$ and $x^2=-x$ by assumption]. On the other hand, as $ax$ is nonzero noninvertible as well, $[ax]^2 \in \{ax,-ax\}$. So $-x \in \{ax,-ax\}$. Similarly, $[a(x+1)]^2 = a^2(x+1)^2 = x+1$, and also, $[a(x+1)]^2 \in
\{a(x+1),-a(x+1)\}$, so $(x+1) \in \{a(x+1),-a(x+1)\}$. This gives us 4 cases; $-x \in \{ax,-ax\}$, and $(x+1) \in \{a(x+1),-a(x+1)\}$. We work through 2 of these cases here:
Case 3.1: $-x=ax$, and $(x+1) = a(x+1)$. Then adding these equations together yields $1=2ax+a$, which plugging in $-x=ax$ gives $1=-2x+a$ which gives $a \in$ Span$\{x,x+1\}$ after all, which is a contradiction.
Case 3.2: $-x=ax$ and $(x+1)=-a(x+1)$. Then adding these together gives $1=-a$ which gives  $a \in$ Span$\{x,x+1\}$ after all, which is a contradiction.
One can handle the remaining two cases in a similar fashion as Cases 3.1 and 3.2 to show that $a$ must be in Span$\{x,x+1\}$ here as well if $x$ is a noninvertible nonscalar nonzero element  satisfying $x^2+x=0$. If $x$ is a noninvertible nonscalar nonzero element satisfying $x^2-x=0$ instead, then one can use the above line of reasoning again to show that $a$ so defined must be in Span$\{x,x-1\}$ with both $x,x-1$ noninvertible. So Claim 3 follows. $\surd$

Claim 4: Let $R$ be as in the problem and suppose that there is a $x$ that is nonzero and in in $R \setminus R^*$. Then $R$ has precisely 9 elements.

As shown above in Claim 3, if $x^2+x=0$ then $R \subseteq$ Span$\{x,x+1\}$, with $3 =0$. So it follows that $R$ has at most 9 elements. However, it is easy to see that any ring with 3,5, or 7 elements must be a field so $R$ must have exactly 9 elements.
Similarly if $x^2-x=0$.
And the result follows from Claim 4. $\surd$