Timeline for Zero divisors of the form $1+x+y$ in the rational group algebra

7 events

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Jun 6, 2015 at 1:41	comment	added	Alireza Abdollahi		Again to record, for any two numbers $m,n$ (not both zero, of course) we have $(1+x+y)( n(x+x^h+\cdots+x^{h^6})+m(x^{-1}+y+y^h+\cdots+y^{h^5})-(n+m)(1+h+\cdots+h^6) )=0$
Jun 5, 2015 at 19:54	comment	added	Alireza Abdollahi		Using the above answer let me give an element $\beta$ of $\mathbb{Q}[G]$ such that $(1+x+y)\beta=0$ to record one. Let $G=\langle a,h \;\|\; a^3=h^7=1, h^a=h^4 \rangle$. Now take $x:=a$ and $y:=(a^h)^{-1}$. Then $(1+x+y)(x+x^h+\cdots+x^{h^6}-x^{-1}-y-y^{h}-\cdots-y^{h^5})=0$. The support of $\beta$ has size $14$. There is another $\beta$ with support of size $21$.
Jun 5, 2015 at 19:30	comment	added	Alireza Abdollahi		Very nice. Many thanks. Have you another example in which the orders of $x$ and $y$ are not $3$? One key in your example is that $(1+x+x^2)(1-x)=0$ if $x^3=1$. Anyway, thanks again.
Jun 5, 2015 at 19:27	vote	accept	Alireza Abdollahi
Jun 5, 2015 at 10:25	history	undeleted	Jeremy Rickard
Jun 5, 2015 at 10:24	history	deleted	Jeremy Rickard		via Vote
Jun 5, 2015 at 10:23	history	answered	Jeremy Rickard	CC BY-SA 3.0