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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?

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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

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: $\pm1,\pm1+2^{\ell-1}$.

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.

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.

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    But they have no division, yes? – 2017-01-25
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    Correct @Anixx: The presence of zero divisors means that division by all non-zero elements becomes impossible. So if you want more than two square roots of $1$ **and** want to divide by everything non-zero, you need to give up on at least some of the ring axioms. – 2017-01-25
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    The point about zero divisors is great. I am investigating the consequences. Can you please give a non-zero element division by which is impossible as an expression of u? – 2017-01-25
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    @Anixx: Consider the ring $\Bbb{Z}_{15}$ again. With $u=4$ we get $u-1=3$ and $u+1=5$. Division by either of these is usually not possible in this ring. For example $2/3$ does not exist. If $2/3=a$ for some $a$, then we want $2=3a$. But in $\Bbb{Z}_{15}$ only $\{0,3,6,9,12\}$ are of the form $3a$ for some $a$, so we won't get $2$. Another problem is that all of $3\cdot0$, $3\cdot5$ and $3\cdot10$ are equal to zero in this ring. So which of the alternatives $0,5,10$ should be $0/3$? When we attempt to divide by a zero-divisor, the answer either does not exist, or is not unique. – 2017-01-25