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I like to be thorough, but if you feel confident you can skip the first paragraph.

Review: A ring is a set $R$ endowed with two operations of + and $\cdot$ such that $(G,+)$ is an additive abelian group, multiplication is associative, $R$ contains the multiplicative identity (denoted with 1), and the distributive law holds. If multiplication is also commutative, we say $R$ is a commutative ring. A ring that has no zero divisors (non-zero elements whose product is zero) is called an integral domain, or just a domain.

We want to show that for a domain, the equation $x^2 = 1$ has at most 2 solutions in $R$ (one of which is the trivial solution 1).

Here's what I did:

For simplicity let $1,a,b$ and $c$ be distinct non-zero elements in $R$. Assume $a^2 = 1$. We want to show that letting $b^2 = 1$ as well will lead to a contradiction. So suppose $b^2 = 1$, then it follows that $a^2b^2 = (ab)^2 = 1$, so $ab$ is a solution as well, but is it a new solution? If $ab = 1$, then $abb = 1b \Rightarrow a = b$ which is a contradiction. If $ab = a$, then $aab = aa \Rightarrow b = 1$ which is also a contradiction. Similarly, $ab = b$ won't work either. So it must be that $ab = c$. So by "admitting" $b$ as a solution, we're forced to admit $c$ as well.

So far we have $a^2 = b^2 = c^2 = 1$ and $ab = c$. We can procede as before as say that $(abc)^2 = 1$, so $abc$ is a solution, but once again we should check if it is a new solution. From $ab = c$, we get $a = cb$ and $b = ac$, so $abc = (cb)(ac)(ab) = (abc)^2 = 1$. So $abc$ is not a new solution; it's just one.

At this point I'm stuck. I've shown that it is in fact possible to have a ring with 4 distinct elements, namely $1,a,b$ and $c$ such that each satisfies the equation $x^2 = 1$ and $abc = 1$. What am I missing?

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Have you tried factoring $x^2-1$? –  user641 Feb 24 '12 at 0:40
I hope I'm not over simplifying things, but since the ring is commutative, if $a$ is any solution, then $a^2-1=0$, which implies $(a-1)(a+1)=0$. Since you're working in an integral domain, then either $a-1=0$ or $a+1=0$, which implies $a$ can only be $1$ or $-1$. –  Buble Feb 24 '12 at 0:43

5 Answers 5

up vote 2 down vote accepted

More generally, every element $\ell\in R$ of an integral domain $R$ cannot have more than two square roots. To see this, let $a$ be such that $a^2=\ell$, and suppose $b^2=\ell$ also for some $b\in R$. Then we can subtract one from the other and factor as $(a-b)(a+b)=0$, and deduce $b=\pm a$ via integrality.

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Hint: $x^2 - 1 = (x-1)(x+1)$. If this is $0 \ldots$

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You’ve shown that if $R$ has two distinct elements other than $1$ whose squares are $1$, then their product is a third such element. But in fact $R$ can’t have two distinct elements other than $1$ whose squares are $1$ in the first place. To see this, show that $x^2=1$ can have at most two solutions by factoring $x^2-1$ and using the fact that $R$ has no zero-divisors.

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Thanks, that explains it. –  mahin Feb 24 '12 at 1:27
+1 This is the only answer that actually mentions the OP's argument, and where it fails. –  M Turgeon Mar 23 '12 at 17:44

In any integral domain, a polynomial of degree $d$ has at most $d$ roots, which implies your result. If $aT + b$ is a degree-1 polynomial with coefficients in an integral domain and $a \ne 0$, then if $x$ and $y$ are roots, we see that $a(x-y) = 0$ which implies $x = y$. Now proceed by induction.

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More generally, overy any ring, a nonzero polynomial has no more roots than its degree if the difference of any two distinct roots is not a zero-divisor (which is true in any integral domain).

THEOREM $\ $ Let $\rm R$ be a ring and let $\rm\:f\in R[x].\:$ If $\rm\:f\:$ has more roots than its degree, and if the difference of any two distinct roots is not a zero-divisor, then $\rm\: f = 0.$

Proof $\ $ Clear if $\rm\:deg\: f = 0\:$ since the only constant polynomial with a root is the zero polynomial. Else $\rm\:deg\: f \ge 1\:$ so by hypothesis $\rm\:f\:$ has a root $\rm\:s\in R.\:$ Factor Theorem $\rm\Rightarrow\: f(x) = (x-s)\:g ,$ $\rm\: g\in R[x].$ Every root $\rm\:r\ne s\:$ is a root of $\rm\: g\:$ by $\rm\: f(r) = (r-s)g(r) = 0\ \Rightarrow\ g(r) = 0, \ by\ \ r-s\ $ not a zero-divisor. So $\rm\:g\:$ satisfies hypotheses, so induction on degree $\rm\:\Rightarrow\:g=0\:\Rightarrow\:f=0.\ $ QED

Here's a nice constructive application: if a polynomial $\rm\:f(x)\:$ over $\rm\:\mathbb Z/n\:$ has more roots than its degree, then we can quickly compute a nontrivial factor of $\rm\:n\:$ by a simple $\rm\:gcd\:$ calculation.

The quadratic case of this is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:n\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/n,\:$ e.g. a square root of $1$ that's not $\pm 1$.

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Glad to see you're back! –  Tyler Feb 26 '12 at 0:28

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