Timeline for Zero divisors in the boolean polynomial ring $\mathbb{F}_2[x_1,x_2,...,x_n]/(x_1^2-x_1,x_2^2-x_2,...x_n^2-x_n)$

10 events

when toggle format	what		by	license	comment
Jan 25 at 6:49	vote	accept	joro
Jan 17 at 16:26	answer	added	Peter Taylor		timeline score: 2
Jan 17 at 11:57	comment	added	joro		@PeterTaylor Many thanks, I edited about 1.
Jan 17 at 11:56	history	edited	joro	CC BY-SA 4.0	added 25 characters in body
Jan 17 at 9:41	comment	added	Peter Taylor		1 isn't a zero divisor. That quibble apart, see mathoverflow.net/questions/42647/… for examples.
Jan 17 at 9:33	comment	added	joro		@YCor Thanks. Do you know another commutative ring or algebra where every element is zero divisor?
Jan 17 at 8:40	comment	added	Dave Benson		$x^2-x=x(x-1)$ so by Chinese remainders, $\mathbb{F}_2[x]/(x^2-x)$ is $\mathbb{F}_2[x]/(x) \times \mathbb{F}_2[x]/(x-1)$, which is $\mathbb{F}_2 \times \mathbb{F}_2$. So you're tensoring together $n$ copies of this, to get a ring isomorphic to a direct product of $2^n$ copies of $\mathbb{F}_2$. Yes, it has cardinality $2^{2^n}$.
Jan 17 at 8:11	comment	added	YCor		Yes, the free Boolean algebra on $n$ generators has dimension $2^n$ (thus cardinal $2^{2^n}$).
Jan 17 at 8:10	history	edited	YCor		edited tags
Jan 17 at 8:01	history	asked	joro	CC BY-SA 4.0