Semi-commutative but not commutative ring.

Question

So, P.P Nielsen created an example of a semicommutative ring which is not McCoy (or say commutative). He took, $k=\mathbb{F}_2\left\langle a_0,a_1,a_2,a_3,b_0,b_1\right\rangle$ be the free algebra (with $1$) over $\mathbb{F}_2$ generated by six indeterminates. I am having difficulty writing the general element of this free associative algebra. Also, there is one more question; he also took an ideal $I$ generated by some relations like, $\left\langle a_0b_0, a_0b_1+a_1b_0,a_1b_1+a_2b_0,a_2b_1+a_3b_0,a_3b_1,...\right\rangle$. So actually what is $I$ here in simple terms and then he construct a ring $R=k/I$, so this is the ring we need, but actually without knowing what $k$ and $I$ are, it is difficult to understand what $k/I$ is.

Answer 1

without knowing what and are, it is difficult to understand what / is.

The free algebra $k$ is quite simple: it's all $F_2$ linear combinations of finite products of the variables. In this case things are made even easier by working over $F_2$, so that the coefficients are $0$ and $1$, and so you're basically looking at finite sums of finite products of the variables.

As for $I$, really it is not necessary to understand what exactly its contents are. The relations themselves are meant to give you simplifications in $k$ to customize the ring.

The trick is, of course, keeping your the quotient from collapsing too far to something trivial, and that is part of the art of constructing examples this way. However Pace Nielsen is very adept with these things and as far as I know this example has no problems.

In your case the generators of $I$ (which I'm pretty sure are the following)

$$ a_0b_0 \\ a_1b_0+a_0b_1 \\ a_1b_1+a_2b_0 \\ a_2b_1+a_3b_0 \\ a_3b_1 \\ a_0a_j \quad (0\leq j\leq 3) \\ a_3a_j \quad(0\leq j\leq 3) \\ a_1a_j+a_2a_j \quad(0\leq j\leq 3)\\ b_ib_j \quad(0\leq i,j,\leq 1)\\ b_ia_j \quad(0\leq i\leq 1, 0\leq j\leq 3) $$

are meant to produce specifically the behavior needed to prove the ring is not McCoy. I do not have the original argument in front of me but of course you can look it up in P. P. Nielsen. Semi-commutativity and the McCoy condition. (2006) @ Section 3 p 138

The relations suggest the relevant polynomials might be $f(X)g(X)=(a_0+a_1X+a_2X^2+a_3X^3)(b_0+b_1X)$

It multiplies out to be $a_3 b_1 X^4 + a_3 b_0 X^3 + a_2 b_1 X^3 + a_2 b_0 X^2 + a_1 b_1 X^2 + a_1 b_0 X + a_0 b_1 X + a_0 b_0$. The relations already make

$$ a_0 b_0 = 0 \\ a_1 b_0 X + a_0 b_1 X = 0 \\ a_2 b_0 X^2 + a_1 b_1 X^2 = 0 \\ a_3 b_0 X^3 + a_2 b_1 X^3 = 0 \\ a_3 b_1 X^4 = 0 $$

So in fact $f(X)g(X)=0$ in $R[X]$. The remainder is there to ensure there is no nonzero $r\in R$ such that $f(X)r=0$ and/or ensure the ring is semicommutative.