I'm looking for an example of a commutative ring with two elements $x,y$ such that $x^2=y^2=(x+y)^2=0$, and $xy \neq 0$. Obviously, $2$ must be a zero divisor in such a ring. I don't think products of $\mathbb{Z}/n\mathbb{Z}$ will do. Anyone has an idea?
$x^2=y^2=(x+y)^2 = 0$ but $xy \neq 0$?
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commutative-algebra
ring-theory
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0Since we're in a commutative ring $\,R\,$ , $\,(x+y)^2=x^2+2xy+y^2\,$ , so your conditions imply that it must be $\,char(R) = 2\,$ . Perhaps this can help to see why Georges's nice answer takes such a ring. – 2012-12-16
2 Answers
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An example of what you require is $\mathbb F_2[x,y]\stackrel {def}{=}\mathbb F_2[X,Y]/(X^2,Y^2)$
What might not be completely evident is that $xy\neq 0$.
This means that in $\mathbb F_2[X,Y]$ the monomial $XY$ is not in the ideal $(X^2, Y^2)$ or that an equation $ XY=X^2f(X,Y)+Y^2g(X, Y)\quad \text{with }f(X,Y),g(X,Y)\in \mathbb F_2[X,Y]$ is impossible.
But this is clear: taking the degree 2 components of both sides yields $XY=X^2f(0,0)+Y^2g(0,0)$, an obvious impossibility.
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0+1 Impressive: The key point is "not completely evident", but "clear" per an "obvious impossibility". ;) – 2012-12-16
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Let $E_2=\begin{pmatrix}1&1\\1&1\end{pmatrix}$. Consider the ring of matrices over $\mathbb{F}_2$ generated by $ 0,\ I_4,\ x=\begin{pmatrix}E_2&0\\0&E_2\end{pmatrix},\ y=\begin{pmatrix}I_2&I_2\\I_2&I_2\end{pmatrix} $ under usual matrix addition and multiplication.