Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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.

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer
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.

share|cite|improve this answer

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 $\pm 1$.

share|cite|improve this answer
Glad to see you're back! – Tyler Feb 26 '12 at 0:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.