Timeline for $Aut(\mathbb{Z}G)=?$ for $G=\mathbb{Z}^2\rtimes_n\mathbb{Z}$

6 events

when toggle format	what		by	license	comment
Nov 27, 2013 at 3:31	vote	accept	Jiang
Nov 13, 2013 at 14:43	comment	added	Jiang		But then, it should be clear the only units of this quotient ring consist of $ \pm 1, \pm x^{\pm 1}, \pm y^{\pm 1}$ and their products.
Nov 13, 2013 at 14:38	comment	added	Jiang		You are right， the quotient ring is the two variable Laurent polynomial ring, which I misunderstood to be the polynomial ring.
Nov 13, 2013 at 14:21	comment	added	Mariano Suárez-Álvarez		@Jiang, In the quotient algebra, $x$ is a unit but $x+y$ is not, so there is no such authomorphism.
Nov 13, 2013 at 14:01	comment	added	Jiang		there seems some tricky points concerning lifting automorphism from the quotient algebra to the original one, since $x\to x+y, y\to y$ gives us an automorphism of the quotient algebra $\mathbb{Z}G_2/I$, but we can check that $\alpha(x)=x+y, \alpha(y)=y, \alpha(z)=z$ is not an automorphism of $\mathbb{Z}G_2$ since it does not preserve the skew relation $yx=xyz^2$. So, in general, we have to determine two noncommutative polynomial $p, q$, s.t., $\alpha(x)=x+y+(z-1)p, \alpha(y)=y+(z-1)q, \alpha(z)=z$ and it preserves the above skew relation.
Nov 13, 2013 at 4:32	history	answered	Mariano Suárez-Álvarez	CC BY-SA 3.0