Was there ever proposed an algebraic system that has more than 2 square roots of unity?

Question

Was there ever proposed an algebraic system that has more than 2 square roots of unity? What are the consequences of having more than 2 square roots of unity?

Answer 1

Such rings exist without any need to introduce new algebra. If $m$ is an integer with $k$ odd prime factors, then the ring $\Bbb{Z}_m$ has at least $2^k$ solutions of $x^2=1$. More specifically, if $m=2^\ell\prod_{i=1}^kp_i^{a_i}$ with $2<p_1<p_2<\cdots<p_k$, then these come from Chinese Remainder Theorem combining the $2^k$ combinations of $x\equiv\pm1\pmod {p_i^{a_i}}$. The even prime is a bit special: If $4\mid m$ you get an extra factor of two and a total of $2^{k+1}$ such roots. If $8\mid m$ you get an extra factor of $4$ and a total of $2^{k+2}$ such roots.

Examples:

• In $\Bbb{Z}_{15}$ we have four: $\pm1$ and $\pm4$. These occur because $4\equiv1\pmod3$ and $4\equiv-1\pmod5$.
• In $\Bbb{Z}_4$ we have only two: $\pm1$, but in $\Bbb{Z}_{2^\ell}$, $\ell>2$ we have four: $\pm1,\pm1+2^{\ell-1}$.

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Any ring with an element $u$ such that $u^2=1$ but $u\neq\pm1$ necessarily has zero divisors. This can be seen as follows. Because $u\cdot1=1\cdot u$ we have the factorization $$ 0=u^2-1=(u-1)(u+1). $$ So if $u$ is neither $1$ nor $-1$ we see that $u\pm1$ are zero divisors.

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The above examples are special cases of the construction that a direct product of two rings $R=R_1\times R_2$ (of characteristic $\neq2$) automatically has at least four solutions to $x^2=1$. Namely the pairs $(\pm1,\pm1)$. More factors gives more possible sign combinations.