This is not really an explanation about the paper, but it is meant to help with the notion of quasiregularity here.
If you denote $r\circ s:=r+s-rs$, then you can show that $\circ$ is a binary operation on $R$ which is associative and has unity element $0$. (I read about this in Heatherly and Tucci's paper The circle semigroup of a ring, but I think Jacobson is usually credited for the idea.) If the ring has identity, then $1$ actually acts as an absorbing element! That is $a\circ 1=1\circ a=1$ for all $a\in R$. This is quite the reverse casting of 0 and 1 compared to the original operation!
Of course we can consider right and left ("quasi"-)inverses in this operation $\circ$. If you check, the definition you wrote for quasiregular was: $r\circ s=0$, that is, $s$ was a right inverse for $r$ (remember that 0 is the identity). As with any associative binary operation, if $r$ has any left inverse, then it has to be equal to $s$, but it is also possible that it has no left inverses.
Lam puts some exercises about this in First course in Noncommutative rings (and so you would also be able to see it in his Exercises in classical ring theory also)
- when $R$ has $1$, the circle semigroup is monoid isomorphic to the semigroup of the ring
- if a right ideal consists entirely of right quasiregular elements, then the elements are also left-quasiregular.
- $\{a\in R\mid aR \text{ is right quasiregular}\}=rad(R)$
The Artin-Wedderburn theorem tells us that the maximal semisimple quotient is a product of matrix rings over finite division rings, one for each irreducible representation. Furthermore, every finite division ring is a field, and the unit group of any finite field is cyclic. The only nontrivial homomorphism from $S_3$ to a cyclic group is the sign homomorphism $S_3\to\mathbb{Z}/2$. It follows that any homomorphism from $\mathbb{Z}S_3$ to a finite field lands in the prime subfield (since elements of $S_3$ can only map to $\pm 1$).
So, writing $\mathbb{F}$ for either $\mathbb{F}_2$ or $\mathbb{F}_3$, the maximal semisimple quotient of $\mathbb{F}S_3$ is a product of matrix rings $M_n(K)$ for finite extensions $K$ of $\mathbb{F}$, one for each irreducible representation, and in all the cases where $n=1$ the $K$ is just $\mathbb{F}$. The only $1$-dimensional representations are the trivial representation and the sign representation, and the sign representation is the same as the trivial representation in the case $\mathbb{F}=\mathbb{F}_2$.
For $\mathbb{F}=\mathbb{F}_3$, dimension-counting now tells us there can be no more irreducible representations: the two $1$-dimensional representations take up $2$ dimensions of the semisimple quotient, and the Jacobson radical is nontrivial since it contains $\sum_{g\in S_3} g$, so there are at most $3$ dimensions left. Another irreducible representation would give a copy of $M_n(\mathbb{F}_{3^d})$ in the semisimple quotient for some $d$ and some $n>1$, which is impossible since there aren't enough dimensions left. We conclude that the two $1$-dimensional representations are the only irreducible representations for $\mathbb{F}=\mathbb{F}_3$, and so the maximal semisimple quotient is $\mathbb{F}_3\times\mathbb{F}_3$. The Jacobson radical is then the kernel of the map $\mathbb{F}_3S_3\to\mathbb{F}_3\times\mathbb{F}_3$; explicitly, it is the set of elements $\sum_{g\in S_3} a_g g$ such that $\sum a_g=0$ and $\sum a_g \sigma(g)=0$, where $\sigma(g)$ is the sign of $g$.
Over $\mathbb{F}_2$, on the other hand, there are up to $4$ dimensions left after accounting for the single $1$-dimensional representation and the fact that the Jacobson radical is nontrivial, so there might be a $2$-dimensional irreducible representation. To find one, note that there is a permutation representation of $S_3$ on $\mathbb{F}_2^3$, and this splits as a direct sum of a trivial subrepresentation (generated by $(1,1,1)$) and a $2$-dimensional subrepresentation (consisting of $(a,b,c)$ such that $a+b+c=0$). (Note that this splitting of the permutation representation doesn't happen over $\mathbb{F}_3$, since $(1,1,1)$ is contained in the latter $2$-dimensional subrepresentation.) This $2$-dimensional representation can easily be verified to be irreducible (for another way of seeing it, note that $\mathbb{F}_2^2\setminus\{0\}$ has three elements, and every permutation of them gives a linear map, so in fact $GL_2(\mathbb{F}_2)\cong S_3$).
So over $\mathbb{F}_2$, we conclude that there is the trivial representation and also this $2$-dimensional irreducible representation; counting dimensions, we now see that we have accounted for all $6$ dimensions of $\mathbb{F}_2S_3$. We conclude that the Jacobson radical is only $1$-dimensional (generated by $\sum_{g\in S_3} g$), and the quotient is $\mathbb{F}_2\times M_2(\mathbb{F}_2)$.
Best Answer
For B1:
I have no idea what that is supposed to mean.
But you can use exactly the same approach mentioned in this similar question and this similar question and this similar question.
If $k$ is supposed to be a field, the radical will wind up being $\left\{\begin{bmatrix}0&0&0\\b&0&d\\0&0&0\end{bmatrix}\,\,\middle|\,\,b,d\in k\right\}$
for B2
Yeah, you should do that. $k[x]$ is a principal ideal domain, so its maximal ideals are easily understood.
The only question is how it factors over $k$. Fortunately for you, it has three "integer" roots. Unfortunately, that's still not enough information to get a single answer.
You'll have to ask whoever gave you the problem if $k$ is supposed to have characteristic $0$ or what. It could have two distinct maximal ideals, or just one maximal ideal.
Still, after you factor it, you can easily see the maximal ideals that contain the polynomial, and then easily compute their intersection.
These are all the hints I can give without knowing what specifically you are stuck with. You might consider editing that information into your question to improve its quality.