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I am trying to find the number of solutions to the equation $X^2 = 1$ that could exist in any ring $R$. I am unsure how to approach this question but below is what I came up with:

If the ring is without unity, then there are no solutions. If the ring has $1 = 0$ then there is just one solution, namely $0$. If the ring has $1 \neq 0$ then we can write the equation as $X^2 - 1 = 0$ and factorise to $(X-1)(X+1)=0$. Now $R$ could be a domain, or it could have zero divisors. In the former case, we have two solutions ($1$ and $-1$) and in the latter there could possibly be infinitely many zero divisors of $R$ (e.g. if $R$ = $\mathbb{Z}\times\mathbb{Z}$) so there could be infinitely many solutions to this equation.

I'm not sure the argument or the answer is correct.

wrb98
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Indeed, if $R$ has no zero divisors, there are at most two solutions: $1$ and $-1$ (possibly the same).

If $R$ has infinitely many zero divisors, this does not necessarily imply that there are infinitely many solutions. E.g. for $R=\mathbb Z^2$, there are only four solutions $(\pm1,\pm1)$.

In general, if $P\in\mathbb Z[X]$ is a polynomial and $n_R$ denotes the number of zeroes of $P$ in $R$, then $n_{R\times S}=n_R\cdot n_S$ for any rings $R$ and $S$.

A dull example of a ring with infinitely many solutions would thus be $\mathbb Z^{\mathbb N}$.

Bart Michels
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  • @DonAntonio As far as I know the infinite number of solutions is of $x^2=-1$. In any ring without zero divisors, the solutions of $x^2=1$ are two, except if the characteristic is $2$, when there's just one. – egreg Dec 17 '16 at 15:19
  • Yes, perhaps I misunderstood the minus one there... – DonAntonio Dec 17 '16 at 15:26