Example of a ring with zero divisors but with the biunique squares property.

Question

This is a follow-up to my previous question, here: Example of a ring where two elements have the same square but neither is plus or minus the other. A ring $R$ is without zero divisors if, for all $x$ and $y$ in $R$, $x*y = 0$ implies $x=0$ or $y=0$. I define a ring $R$ to have the biunique square property if, for all $x$ and $y$ in $R$, $x^2 = y^2$ implies $x=y$ or $x=-y$. Does there exist a ring with zero divisors but which also satisfies the biunique square property? I would prefer a finite ring, if one exists.

Answer 1

Consider the ring $R=\mathbb{Z}/6\mathbb{Z}$ of integers modulo $6$. This ring has zero divisors, since $2\times 3 = 0$, with $2\neq 0$ and $3\neq 0$.

Now let $a$ and $b$ be integers such that $a^2\equiv b^2\pmod{6}$. Then $(a+b)(a-b)\equiv 0\pmod{6}$, and therefore $(a+b)(a-b)\equiv 0\pmod{2}$; and $(a+b)(a-b)\equiv 0\pmod{3}$. The latter requires $a+b\equiv 0\pmod{3}$ or $a-b\equiv 0\pmod{3}$. As $a+b\equiv a-b\pmod{2}$, in either case we have either $a\equiv b\pmod{6}$ or $a\equiv -b\pmod{6}$, showing that $x^2=y^2$ in $R$ implies either $x=y$ or $x=-y$.

More generally the same holds in any ring of the form $\mathbb{Z}/2p\mathbb{Z}$, where $p$ is an odd prime. Of the rings of the form $\mathbb{Z}/n\mathbb{Z}$, these are the only ones.

Theorem. Let $n\gt 1$ be an integer. The ring $R=\mathbb{Z}/n\mathbb{Z}$ has zero divisors if and only if $n$ is composite. The ring $R$ satisfies $x^2=y^2\iff x=\pm y$ if and only if $n$ is prime, or is of the form $2p$ for some odd prime $p$.

Proof. That $R$ has zero divisors if and only if $n$ is not prime is well known. That $x=\pm y$ always implies $x^2=y^2$ is also clear. We show that $x^2=y^2$ implies $x=\pm y$ if and only if $n$ is prime or twice an odd prime.

If $n$ is prime then $R$ is a field, so $x^2=y^2$ holds if and only if $(x-y)(x+y)=0$, if and only if $x=y$ or $x=-y$. If $n=2p$ for some odd prime $p$, and $x$ and $y$ are integers such that $x^2=y^2$, then by the Chinese Remainder Theorem we have $x^2=y^2\iff (x-y)(x+y)\equiv 0\pmod{2}$ and $(x-y)(x+y)\equiv 0\pmod{p}$. This means that $x\equiv y\pmod{p}$ or $x\equiv -y\pmod{p}$, and that $x\equiv y\equiv -y\pmod{2}$. Again by the Chinese Remainder Theorem, this implies that $x\equiv y\pmod{2p}$ or $x\equiv -y\pmod{2p}$, showing that $R$ has the desired property.

Conversely, if $n$ is divisible by the square of a prime $p$, write $n=p^aq$ with $\gcd(p,q)=1$, and $a\gt 1$. Using the Chinese Remainder Theorem, find $x$ such that $x\equiv p^{a-1}\pmod{p^a}$ and $x\equiv 0\pmod{q}$. Then $x^2\equiv 0\pmod{n}$, so $x^2=0^2$, but $x\neq 0$ since $p^a\nmid x$ (or just take $x=p^{a-1}q$). And if $n$ is squarefree and divisible by two odd primes $p\neq q$, write $n=pqr$ with $\gcd(r,pq)=1$, then using the Chinese Remainder Theorem find $x$ with $x\equiv 1\pmod{p}$, $x\equiv -1\pmod{q}$, and $x\equiv 1\pmod{r}$. Then $x\not\equiv \pm 1\pmod{n}$. But $x^2\equiv 1 = 1^2\pmod{n}$. $\Box$

Answer 2

Let $A$ be any nontrivial ring with the biunique square property, $m$ a positive integer, and let $R := \mathbb{F}_{2^m} \times A$. Then $R$ has zero divisors since $(1, 0) \cdot (0, 1) = 0$. On the other hand, if $(x_1, y_1)^2 = (x_2, y_2)^2$, then $y_1^2 = y_2^2$, so by assumption, we have $y_1 = \pm y_2$. Also, since $x_1^2 = x_2^2$, we have that $x_1 = \pm x_2$ with the same choice of $\pm$ as for the $y$'s (since for $x \in \mathbb{F}_{2^m}$, we have $-x = x$). This shows that $R$ also has the biunique square property.

Similarly, any product $\mathbb{F}_{2^{m_1}} \times \cdots \times \mathbb{F}_{2^{m_k}} \times A$ will work.

(Note that the examples given in the answer by Arturo Magidin are isomorphic to a special case: in those examples, $\mathbb{Z} / 2p \mathbb{Z} \simeq \mathbb{F}_2 \times \mathbb{F}_p$ if $p$ is an odd prime.)