In a field $F$ and two elements $x$ and $y$ in $F$, if $x^2 = y^2$, then $x=y$ or $x=-y$. I am looking for an example of a ring $R$ such that there are two elements $x$ and $y$ such that $x^2 = y^2$, but neither $x=y$ nor $x=-y$.
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1How about quaternions? – user888379 May 22 '23 at 17:42
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As shown here a commutative ring $R$ is a domain iff nonzero polynomials $\in R[x],$ have no more roots than their degree. So if $R$ has a counterexample then it has a zero divisor, say $,ab = 0.,$ Then $,(a+b)^2 = (a-b)^2$ and $,a-b\neq \pm (a+b),$ if $,2a,2b\neq 0,$ (true in $\Bbb Z/n,$ for odd composite $,n).\ \ $ – Bill Dubuque May 22 '23 at 18:06
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Also true in $,\Bbb Z[a,b]/(ab),,$ a universal example. – Bill Dubuque May 22 '23 at 18:26
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A natural way to construct a counterexample would be to consider $$R := \mathbb{Z}[x,y] / \langle y^2 - x^2 \rangle.$$ In this ring, pretty much by construction, you have $[x]^2 = [y]^2$. So now, the point would be to see why you don't have in this ring that $[y] = [x]$ or $[y] = -[x]$.
One way to see this would be to observe that you can define a ring homomorphism $R \to \mathbb{Z}$ such that $[x] \mapsto 1, [y] \mapsto -1$; therefore, you cannot have $[y] = [x]$. Similarly, you can define a ring homomorphism $R \to \mathbb{Z}$ such that $[x] \mapsto 1, [y] \mapsto 1$; therefore, you cannot have $[y] = -[x]$. (More generally, if $S$ is another commutative ring, and you are given a pair of elements $a, b \in S$ such that $a^2 = b^2$, then there is a unique ring homorphism $R \to S$ such that $[x] \mapsto a, [y] \mapsto b$.)
Another way is: by division, you can see that every element of $R$ must be of the form $[p(x) + y q(x)]$ for some polynomials $p, q \in \mathbb{Z}[x]$. If you can show that this representative is unique, then this will imply that you cannot have either $[y] = [x]$ or $[y] = -[x]$. If this is true, then you should expect that $$[p_1(x) + y q_1(x)] [p_2(x) + y q_2(x)] = [(p_1(x) p_2(x) + x^2 q_1(x) q_2(x)) + y (p_1(x) q_2(x) + q_1(x) p_2(x))].$$ This suggests that you should have that $R$ is isomorphic to a ring with underlying set $\mathbb{Z}[x] \times \mathbb{Z}[x]$, with addition defined by $(p_1, q_1) + (p_2, q_2) = (p_1 + p_2, q_1 + q_2)$ and multiplication defined by $(p_1, q_1) \cdot (p_2, q_2) = (p_1 p_2 + x^2 q_1 q_2, p_1 q_2 + q_1 p_2)$. From here, it would be a tedious but straightforward exercise to show that this does define a commutative ring. Then, in this ring, you clearly have that $(x, 0)^2 = (0, 1)^2 = (x^2, 0)$, but $(x, 0) \ne \pm (0, 1)$.
Daniel Schepler
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This is the universal commutative ring with a nontrivial square root. $\ \ $ – Bill Dubuque May 22 '23 at 18:18
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You have $\begin{pmatrix} 0 &1 \\ 0 &0 \end{pmatrix}^2 = \begin{pmatrix} 0 &0 \\ 0 &0 \end{pmatrix} = \begin{pmatrix} 0 &0 \\ 0 &0 \end{pmatrix}^2$ in the ring of $2 \times 2$ real matrices.
balddraz
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