See II 4.20 in the book.

It looks like $\mathbb Z[SL_2]$ satisfies the cohomological criterion for good filtration as a $SL_2\times SL_2$ module. (Left and right action).

First use conjugation by $SL_2$ instead. The advantage is that then it becomes the restriction of the action of a big product of $2n$ copies of $SL_2$. I believe one has a filtration of the coordinate ring whose layers are tensor products $H^0(\mu)\boxtimes H^0(\nu)$. Compare 2.2a in Donkin, On Schur Algebras and Related Algebras, 1 , JOURNAL OF ALGEBRA 104, (1986). And the diagonal inclusion into the big product is a Donkin pair. Do not have books here.

@stupid_question_bot The trivial representation is $V(0)$ and $0$ is dominant.

Chapter B explains that it indeed is implied by the result over fields. You may also need that field extensions do not matter for good filtrations. So talking about algebraically closed fields is just meant to put the reader at ease.

It must be true because Donkin shows one has a good filtration over $\mathbb Z$. That implies base change for the group scheme invariants. Infinite fields are only needed when one wants to confuse a group $G(k)$ with the group scheme $G_k$.

It is not just $K_0$? it is the negative ones also? In arXiv:0903.3717 we find the statement: The reason for this is that negative K-theory is, in particular, an obstruction for the (triangulated) quotients to be Karoubian.

One must choose a form of the $V_n$. For instance, one may define them to be symmetric powers of the standard two dimensional representation. Then $V_n\otimes V_m$ has a "good filtration" whose layers are given by Clebsch-Gordan. See the book of Jantzen (Representations of Algebraic Groups: Second Edition)?

`Since $p_1$ is a prime factor of $s$' is the wrong reason.

$B$ is the push out. The hypothesis says there is an element of the Yoneda Ext $Ext^1(C,X_A)$ so that $B$ is the push out.

If everything is affine and $G$ is smooth, then it suffices to take for $B=R$.

Sadly, Paul Gunnells confirms that Jim has pased away.

It is my understanding that some people liked the $\varpi$ exactly because only the incrowd understood it stands for poid.